4_hw_cp220_A.ipynb

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      "source": [
        "**Topics**\n",
        "\n",
        "Determinants<br>\n",
        "Minors & Cofactors<br>\n",
        "Block Matrices <br>\n",
        "Eigenvalues & Eigenvectors <br>\n",
        "Matrix Powers and Exponential <br>\n",
        "Trace<br>\n"
      ],
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      "source": [
        "---\n",
        "**Determinants**\n",
        "\n",
        "We will assume $A$ is a square $nxn$ matrix in the following unless otherwise stated.\n",
        "\n",
        "We will use the notation $|A| := determinant(A)$\n",
        "\n",
        "Geometrically, $|A|$ represents the volume of the parallelopiped formed by columns of A (or rows of A).\n",
        "\n",
        "So for the identity matrix $I$, the column vectors are the  unit orthogonal vectors which forms a hypercube of volume 1.\n",
        "\n",
        "So we will ask for the following properties for the determinant:\n",
        "\n",
        "I) Determinant of identity matrix is 1. i.e $|I| = 1$\n",
        "\n",
        "II) When exchanging two rows of a matrix, the determinant changes sign.\n",
        "\n",
        "*III*) The determinant is linear in the first row of the matrix.\n",
        "\n",
        "i.e if $A = \\begin{bmatrix} \\lambda a_1^T + \\mu b_1^T \\\\ a_2^T \\\\ ... \\\\ a_n^T \\end{bmatrix}$ then\n",
        "\n",
        "$|A| = \\lambda\\begin{vmatrix} a_1^T \\\\ a_2^T \\\\ ...\\\\a_n^T \\end{vmatrix} + \\mu \\begin{vmatrix} b_1^T \\\\ a_2^T \\\\ ... \\\\ a_n^T \\end{vmatrix}$\n",
        "\n",
        "\n",
        "using these properties, we can derive all other properties as well as the formula for the determinant!\n"
      ],
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      "source": [
        "---\n",
        "4.  $|A|$ is linear in any row.\n",
        "\n",
        "Proof: from II & III\n",
        "\n",
        "5. If $A$ has a row of all 0, then $|A|=0$\n",
        "\n",
        "Proof; from II & III\n",
        "\n",
        "6. If $A = diag(d_1, d_2, ..., d_n)$, $|A| = \\prod_{i=1}^n d_i$\n",
        "\n",
        "Proof: By 4, we can factor out $d_i$ from each row in a step by step fashion resulting in a matrix with product of diagonals and identity. the latter has determinant 1 by definition of I.\n",
        "\n",
        "7. If $A$ is triangular with non-zero diagonals $d_1, d_2, ..., d_n$ then $|A| = \\prod_{i=1}^n d_i$\n",
        "\n",
        "Proof: With row elimination, put $A$ in diagonal form $D=diag(d_1, d_2, ... ,d_n)$. Note since row elimination is linear combination of rows, from 4 & 6, |A| = |D|\n",
        "\n",
        "8. If two rows are equal, then determinant is 0.\n",
        "\n",
        "Proof: By row exchange of two equal rows, sign changes, but matrix remains same means $-|A| = |A|$. Hence $|A|=0$\n",
        "\n",
        "9. A matrix with linearly dependant rows has determinant 0.\n",
        "\n",
        "Proof: By replacing the dependent row as a linear combination of other rows and using III and 9 above, $|A|=0$\n",
        "\n",
        "10. A singular matrix (linearly dependent rows) has determinant 0.\n",
        "\n",
        "11. $|cA| = c^n|A|$\n",
        "\n",
        "Proof: $cA$ has each element multiplied by $c$. By II & III, we get the results.\n",
        "\n",
        "12. $|LB| = |BU|=|B|$ where L, U are lower and upper triangular with all 1s in the diagonal.\n",
        "\n",
        "Proof: $LB = B{'}$ where each row of $B^{'}$ has linear combination of rows of B above it. From 4 & 8 this follows. Same argument for multiplying with U.\n",
        "\n",
        "13. $|DB| = (\\prod_{i=1}^nd_{ii}) |B|$ where D is diagonal\n",
        "\n",
        "Proof: Follows from III\n",
        "\n",
        "14. If $P$ is a permutation matrix $|P| = \\pm 1$\n",
        "\n",
        "Proof: Any permutation matrix is derived from row exchanges of Identity matrix. With even row exchanges, sign is positive and with odd, it is negative.\n",
        "\n",
        "15. $|AB|=|A||B|$\n",
        "\n",
        "Proof:\n",
        "If $A$ is singular, $|A| =0$ from 9. Also $AB$ is singular and hence $|AB|=0$\n",
        "\n",
        "Otherwise we can uniquely write $PA=LDU$ where $P$ is a permutation matrix, L, U are lower and upper triangular with 1s in the diagonal and D is diagonal with non-zeros. Hence from 12, 13, 14, result follows.\n",
        "\n",
        "16. $|A| = |A^T|$\n",
        "\n",
        "Proof: If A is singular, so is $A^T$ and both determinant are 0.\n",
        "\n",
        "If A is non-singular, then $PA=LDU$ and $|A| = |D|$ and $|A^TP^T| = |U^TDL^T|$ and $|A^T|=|D|$\n",
        "\n"
      ],
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      "source": [
        "---\n",
        "**Formula for determinant**\n",
        "\n",
        "Let $A=\\begin{bmatrix} a_{11} & a_{12} & ... & a_{1n} \\\\ a_{21} & a_{22} & ... & a_{2n} \\\\ ... \\\\ a_{n1} & a_{n2} & ... & a_{nn} \\end{bmatrix}$\n",
        "\n",
        "Let $e_i$ be a row vector with 1 in $i^{th}$ column and zero everywhere.\n",
        "\n",
        "Then any row $a_{i:} = \\sum_{j=1}^n a_{ij}e_j$\n",
        "\n",
        "\n",
        "\n",
        "So from III  $|A| = \\sum_{j=1}^n a_{ij} |A_{ij}|$ where $A_{ij}$ has $j^{th}$ row replaced by the elementary row vector $e_j$\n",
        "\n",
        "This can be done recursively to get |A| as a linear combination of $n^n$ scale factors times determinants of matrices with only elementary row vectors. Of these, only $n!$ of these matrices have non-zero determinants - precisely all the permutation matrices. All others will ahve repeated elementary rows and hence zero determinant.\n",
        "\n",
        "Hence $|A| = \\sum_\\sigma sign(\\sigma) \\prod_{i=1}^n a_{i\\sigma(i)}$\n",
        "\n",
        "where $\\sigma : \\{1,2,..,n\\} → \\{1,2,...,n\\}$ is a permutation and $sign(\\sigma)$ is the positive(negative) if a composition of even(odd) number of binary exchanges give the permutation.\n",
        "\n",
        "Example: $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = ac-bd$\n"
      ],
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    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q1**\n",
        "\n",
        "If $P$ is a $nxn$ permutation matrix with '1s' in the reverse diagonal, what is $|P|$"
      ],
      "metadata": {
        "id": "z7yAPA0l5vcV"
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    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "If $n=2k$ or $2k+1$, $k$ row exchanges (first and last, second and second-last etc) will get $I$. Hence $|P| = (-1)^{\\lfloor{n}\\rfloor}$"
      ],
      "metadata": {
        "id": "d-KhxwSz5_8N"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q2**\n",
        "\n",
        "Let $v = [v_1, v_2, ..., v_n]^T$. Find |vv^T|.\n"
      ],
      "metadata": {
        "id": "eyjS-zv17T4R"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "0 (as it is a rank deficient matrix)"
      ],
      "metadata": {
        "id": "g0dJIAlB7ok-"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q3**\n",
        "\n",
        "If $|A| \\neq 0$, what is $|A^{-1}|$"
      ],
      "metadata": {
        "id": "iP_oy1jP70Ez"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "$|AA^{-1}| = |A||A^{-1}| = |I| = 1$. Hence, $|A^{-1}|=\\frac{1}{|A|}$"
      ],
      "metadata": {
        "id": "0bmJozbp7_xN"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q4**\n",
        "\n",
        "If $Q$ is a orthogonal matrix, find $|Q|$"
      ],
      "metadata": {
        "id": "Dbfzc5j28auk"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "$|QQ^T| = |Q||Q^T| = |Q|^2 = |I|=1$. hence $|Q| =\\pm 1$"
      ],
      "metadata": {
        "id": "2MNV6Tsi8hqL"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q5**\n",
        "\n",
        "What is $\\begin{vmatrix} 1 & a & a^2 \\\\ 1 & b & b^2 \\\\ 1 & c & c^2\\end{vmatrix}$ the vandermonde matrix?"
      ],
      "metadata": {
        "id": "0tzRcPkK830w"
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    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "By row operations, we can put it into a lower triangular form with diagonals $1, (b-a), (c-a)(c-b)$. Hence its determinant is $(a-b)(b-c)(c-a)$"
      ],
      "metadata": {
        "id": "y9Firot29EOS"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q6**\n",
        "\n",
        "For which values of $\\lambda$ is $|A-\\lambda I| = \\begin{vmatrix} 4-\\lambda & 2 \\\\ 1 & 3 - \\lambda \\end{vmatrix}=0$"
      ],
      "metadata": {
        "id": "9uz8t018-TM9"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "$|A-\\lambda I| = 9 - 7\\lambda + \\lambda^2 = 0$ for $\\lambda=\\frac{7 \\pm \\sqrt{13}}{2}$"
      ],
      "metadata": {
        "id": "AlMZWwjS-k_7"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q7**\n",
        "\n",
        "Write python code to find a 4x4 matrix with only 0 or 1 and having the largest determinant"
      ],
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        "id": "mlQuIM7U__9H"
      }
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      "cell_type": "markdown",
      "source": [
        "---\n",
        "We will use the popular numerical linear algebra package numpy\n",
        "https://numpy.org/doc/stable/user/index.html\n",
        "The above website has documentation, tutorials etc.\n"
      ],
      "metadata": {
        "id": "s1T9jGeBbMeD"
      }
    },
    {
      "cell_type": "code",
      "source": [
        "#import the package for numerical linear algebra computations in Python\n",
        "import numpy as np"
      ],
      "metadata": {
        "id": "fJZykwE8aNrm"
      },
      "execution_count": null,
      "outputs": []
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    {
      "cell_type": "code",
      "source": [
        "#Code to find largest determinant for a 4x4 matrix with only 1's and 0's\n",
        "d = 4\n",
        "max = 2**d - 1\n",
        "maxdet=0\n",
        "for i in range(1,max):\n",
        "  for j in range(i+1, max):\n",
        "    for k in range(j+1,max):\n",
        "      for l in range(k+1, max):\n",
        "        x = np.array([\n",
        "            list(\"{0:04b}\".format(i)),\n",
        "            list(\"{0:04b}\".format(j)),\n",
        "            list(\"{0:04b}\".format(k)),\n",
        "            list(\"{0:04b}\".format(l))\n",
        "            ],np.int8)\n",
        "        if (maxdet < abs(np.linalg.det(x))):\n",
        "          maxdet = np.linalg.det(x)\n",
        "print(maxdet)"
      ],
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "uYADJfdNdpYp",
        "outputId": "5916aba6-804f-44d7-876e-2a553c33c158"
      },
      "execution_count": null,
      "outputs": [
        {
          "output_type": "stream",
          "name": "stdout",
          "text": [
            "3.0000000000000004\n"
          ]
        }
      ]
    },
    {
      "cell_type": "code",
      "source": [
        "#Code to find largest determinant for a 4x4 matrix with only 1's and 0's\n",
        "d = 5\n",
        "max = 2**d - 1\n",
        "maxdet=0\n",
        "for i in range(1,max):\n",
        "  for j in range(i+1, max):\n",
        "    for k in range(j+1,max):\n",
        "      for l in range(k+1, max):\n",
        "        for m in range(l+1, max):\n",
        "          x = np.array([\n",
        "              list(\"{0:05b}\".format(i)),\n",
        "              list(\"{0:05b}\".format(j)),\n",
        "              list(\"{0:05b}\".format(k)),\n",
        "              list(\"{0:05b}\".format(l)),\n",
        "              list(\"{0:05b}\".format(m)),\n",
        "              ],np.int8)\n",
        "          if (maxdet < abs(np.linalg.det(x))):\n",
        "            maxdet = abs(np.linalg.det(x))\n",
        "print(maxdet)"
      ],
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      "execution_count": null,
      "outputs": [
        {
          "output_type": "stream",
          "name": "stdout",
          "text": [
            "5.000000000000001\n"
          ]
        }
      ]
    },
    {
      "cell_type": "code",
      "source": [
        "d = 6\n",
        "max = 2**d - 1\n",
        "maxdet=0\n",
        "for i in range(1,max):\n",
        "  for j in range(i+1, max):\n",
        "    for k in range(j+1,max):\n",
        "      for l in range(k+1, max):\n",
        "        for m in range(l+1, max):\n",
        "          for n in range(m+1, max):\n",
        "            x = np.array([\n",
        "              list(\"{0:06b}\".format(i)),\n",
        "              list(\"{0:06b}\".format(j)),\n",
        "              list(\"{0:06b}\".format(k)),\n",
        "              list(\"{0:06b}\".format(l)),\n",
        "              list(\"{0:06b}\".format(m)),\n",
        "              list(\"{0:06b}\".format(n)),\n",
        "              ],np.int8)\n",
        "            if (maxdet < abs(np.linalg.det(x))):\n",
        "              maxdet = abs(np.linalg.det(x))\n",
        "print(maxdet)"
      ],
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      },
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            "\u001b[0;31mKeyboardInterrupt\u001b[0m                         Traceback (most recent call last)",
            "\u001b[0;32m<ipython-input-56-1117c7369f2e>\u001b[0m in \u001b[0;36m<cell line: 4>\u001b[0;34m()\u001b[0m\n\u001b[1;32m     11\u001b[0m               \u001b[0mlist\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0;34m\"{0:06b}\"\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mformat\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0mi\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m,\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m     12\u001b[0m               \u001b[0mlist\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0;34m\"{0:06b}\"\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mformat\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0mj\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m,\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0;32m---> 13\u001b[0;31m               \u001b[0mlist\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0;34m\"{0:06b}\"\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mformat\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0mk\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m,\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[0m\u001b[1;32m     14\u001b[0m               \u001b[0mlist\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0;34m\"{0:06b}\"\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mformat\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0ml\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m,\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n\u001b[1;32m     15\u001b[0m               \u001b[0mlist\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0;34m\"{0:06b}\"\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mformat\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0mm\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m,\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n",
            "\u001b[0;31mKeyboardInterrupt\u001b[0m: "
          ]
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    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Minors and Cofactors**\n",
        "\n",
        "$M_{ij}$ is a minor of $A$ and is a sub-matrix with $(n-1) x (n-1)$ with row $i$ and column $j$ of A removed.\n",
        "\n",
        "$C_{ij} = (-1)^{i+j}|M_{ij}|$ is the cofactor\n",
        "\n",
        "Determinant can be computed using any row (or column)\n",
        "\n",
        "For example with $i^{th}$ row, it is:\n",
        "$|A| = \\sum_{j=1}^n a_{ij}C_{ij}$\n",
        "\n",
        "\n"
      ],
      "metadata": {
        "id": "IFwUuC7Kn0o5"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "**Q8**\n",
        "\n",
        "Find the $|B|$ using cofactor approach\n",
        "$B = \\begin{bmatrix} 0 & 0 & 1 & 2 \\\\ 0 & 3 & 4 & 5 \\\\ 6 & 7 & 8 & 9 \\\\ 0 & 0 & 0 & 1\\end{bmatrix}$\n",
        "\n",
        "and check via numpy."
      ],
      "metadata": {
        "id": "C-s6_WFLo5Y-"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "$b_{44} * (-1)^{4+4} * a_{13}*(-1)^{1+3}*\\begin{vmatrix} 0 & 3 \\\\ 6 & 7 \\end{vmatrix} = -18$"
      ],
      "metadata": {
        "id": "9lIW6b-Wp0wX"
      }
    },
    {
      "cell_type": "code",
      "source": [
        "#your code here\n",
        "x = np.array([\n",
        "        [0,0,1,2],\n",
        "        [0,3,4,5],\n",
        "        [6,7,8,9],\n",
        "        [0,0,0,1]\n",
        "        ])\n",
        "print(np.linalg.det(x))"
      ],
      "metadata": {
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          "base_uri": "https://localhost:8080/"
        },
        "id": "S_lN1AKLpn__",
        "outputId": "896ab40a-19e8-4171-b708-20f5e1940d54"
      },
      "execution_count": null,
      "outputs": [
        {
          "output_type": "stream",
          "name": "stdout",
          "text": [
            "-17.999999999999996\n"
          ]
        }
      ]
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q9**\n",
        "\n",
        "$A_1 = [1]$\n",
        "\n",
        "$A_2 = \\begin{bmatrix} 1 & 1 \\\\ 1 & 1 \\end{bmatrix}$\n",
        "\n",
        "$A_3 = \\begin{bmatrix} 1 & 1 & 0 \\\\ 1 & 1 & 1 \\\\ 0 & 1 & 1 \\end {bmatrix}$\n",
        "\n",
        "and so on, with the matrices being tri-diagonal (i.e only 1s in the main and upper and lower diagonal\n",
        "\n",
        "a) Find $D_n = |A_n|$ by expanding via co-factors\n",
        "\n",
        "b) Find $D_{1000} = |A_{1000}|$ by finding $D_1, D_2, ... D_8$ and guessing the pattern\n"
      ],
      "metadata": {
        "id": "3K2JBysvq97P"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "a) By expanding via cofactor of row 1\n",
        "$D_n = 1|A_{n-1}| -1*1* |A_{n-2}|$\n",
        "\n",
        "Hence $D_n = D_{n-1} - D_{n-2}$\n",
        "\n",
        "b)\n",
        "$D_1 = 1$\n",
        "\n",
        "$D_2 = 0$\n",
        "\n",
        "$D_3 = -1$\n",
        "\n",
        "$D_4 = -1$\n",
        "\n",
        "$D_5 = 0$\n",
        "\n",
        "$D_6 = 1$\n",
        "\n",
        "$D_7 = 1$\n",
        "\n",
        "$D_8 = 0$\n",
        "\n",
        "$D_9 = -1$\n",
        "\n",
        "So pattern repeats with period 6. Since 1000 = 6*166 +4, $D_{1000} = D_4 = -1$\n",
        "\n",
        "\n"
      ],
      "metadata": {
        "id": "XjxgQvPorr3l"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Block Matrices**\n",
        "\n",
        "We can consider a matrix having matrix sub-blocks and do operations in terms of the smaller sub-blocks.\n",
        "\n",
        "$M = \\begin{bmatrix} A & B \\\\ C & D \\end{bmatrix}$\n",
        "\n",
        "$N = \\begin{bmatrix} U & V \\\\ W & X \\end{bmatrix}$\n",
        "\n",
        "then $MN  = \\begin{bmatrix} AU+BW & AV + BX \\\\ CU+DW & CV+DX\\end{bmatrix}$\n",
        "\n",
        "Provided the dimensions of the sub-blocks are compatible for matrix multiplication.\n",
        "\n",
        "For example:\n",
        "Let A be a square invertible matrix. Then with $L = \\begin{bmatrix} I & 0  \\\\ -CA^{-1} & I \\end{bmatrix}$\n",
        "\n",
        "$LM = \\begin{bmatrix} A & B \\\\ 0 & D - CA^{-1}B \\end{bmatrix}$\n",
        "\n",
        "We have converted it into a lower block diagonal and $D-CA^{-1}B$ is called the schur complement of $A$\n",
        "\n",
        "Similarly, if $U = \\begin{bmatrix} I & -A^{-1}B \\\\ 0 & I \\end{bmatrix}$\n",
        "\n",
        "We have $LMU = \\begin{bmatrix} A & 0 \\\\ 0 & D-CA^{-1}B \\end{bmatrix}$\n",
        "\n",
        "\n",
        "For block triangular matrices $W=\\begin{bmatrix} A & B \\\\0 & D\\end{bmatrix}$ if $A, D$ are square, then $|W| = |A||D|$\n",
        "\n",
        "Therefore\n",
        "\n",
        "$|M| = |L||M||U| = |A||D-CA^{-1}B| = |D||A-BD^{-1}C|$\n",
        "\n",
        "The last is obtained by digonalizing $M$ as:\n",
        "\n",
        "$L^{'} = \\begin{bmatrix} I & -BD^{-1} \\\\ 0 & I \\end{bmatrix}$\n",
        "\n",
        "$L^{'}M = \\begin{bmatrix} A - BD^{-1}C & 0 \\\\ C & D \\end{bmatrix}$\n",
        "\n",
        "$U^{'} = \\begin{bmatrix} I & 0 \\\\ -D^{-1}C & I \\end{bmatrix}$\n",
        "\n",
        "Hence $L^{'}MU{'} = \\begin{bmatrix}A-BD^{-1}C & 0 \\\\ 0 & D \\end{bmatrix}$\n"
      ],
      "metadata": {
        "id": "x7D0EDnSuIas"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q10**\n",
        "\n",
        "Show that $\\begin{vmatrix} A & B \\\\ C & D \\end{vmatrix} = |AD-CB|$ if $A^{-1}$ exists and $AC=CA$"
      ],
      "metadata": {
        "id": "y2oDqlvByqn4"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "If $A^{-1}$ exists, we can take the schur complement.\n",
        "\n",
        "Also $AC=CA => CA^{-1} = A^{-1}C$\n",
        "\n",
        "and hence\n",
        "\n",
        "$\\begin{vmatrix} A & B \\\\ C & D \\end{vmatrix} = |A||D-CA^{-1}B| = |A||A^{-1}AD-A^{-1}CB|=|A||A^{-1}||AD-CB| = |AD-CB|$\n"
      ],
      "metadata": {
        "id": "mFX_d2cS2OUq"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Matrix inverse from cofactors**\n",
        "\n",
        "\n",
        "Recall that Cofactor of a matrix $A$ is $C_{ij} = (-1)^{i+j}|M_{ij}|$ where $M_{ij}$ is the minor with row i and column j removed from matrix $A$\n",
        "\n",
        "Also $|A| = \\sum_{j=1}^n a_{ij}C_{ij}$ i.e we can expand using any row $i$.\n",
        "\n",
        "Let $C = [C_{ij}]$ the co-factor matrix.\n",
        "\n",
        "Then $AC^T = |A|I$\n",
        "\n",
        "Therefore $A^{-1} = \\frac{1}{|A|}C^T$ if $|A| \\neq 0$\n",
        "\n"
      ],
      "metadata": {
        "id": "2BY568SY3pG-"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q11**\n",
        "\n",
        "Let C be the cofactor matrix of A. The  show that $|C|=|A|^{n-1}$"
      ],
      "metadata": {
        "id": "K0upUz2MZAMS"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "$AC^T = |A|I => |AC^T| = |A||C^T| = |A|^n => |C|=|C^T| = |A|^{n-1}$"
      ],
      "metadata": {
        "id": "aL5Xv34TZLRD"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q12**\n",
        "\n",
        "Let C be the cofactor matrix of A. find A."
      ],
      "metadata": {
        "id": "k0e6wLgMZvhv"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "We know that\n",
        "\n",
        "$\n",
        "AC^T = |A|I\n",
        "$\n",
        "\n",
        "$\n",
        "|AC^T| = ||A|I| = |A|^n\n",
        "$\n",
        "\n",
        "$\n",
        "|AC^T|=|A||C^T|=|A||C|=|A|^n\n",
        "$\n",
        "\n",
        "$\n",
        "|C|=|A|^{n-1}\n",
        "$\n",
        "\n",
        "$\n",
        "|A| = e^{\\frac{2 i \\pi k}{n-1}}|C|^{\\frac{1}{n-1}}\n",
        "$  \n",
        "\n",
        "Hence\n",
        "\n",
        "$A = |A|C^{-T} = e^{\\frac{2 i \\pi k}{n-1}} |C|^{\\frac{1}{n-1}}C^{-T}$\n",
        "\n",
        "Where $k=0,1,...,n-1$"
      ],
      "metadata": {
        "id": "LUkVPJMYZ1j1"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q13**\n",
        "\n",
        "Let $a=(0,0), b=(1,0), c=(2,1), c=(1,3), e=(0.5,4)$ be a polygon. Find its areas using determinants."
      ],
      "metadata": {
        "id": "kUPQslQna9Ca"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "If a,b,c is a triangle then its area is $\\frac{1}{2}|\\begin{bmatrix}b-a & c-a \\end{bmatrix}|$\n",
        "\n",
        "Therefore the area of the polygon is the sum of areas of the triangles, abc, acd, ade, (note the consistent counter clockwise orientation. Therefore\n",
        "\n",
        "Area = 0.5*($\\begin{vmatrix} 1 & 2 \\\\ 0 & 1\\end{vmatrix} + \\begin{vmatrix} 2 & 1\\\\ 1 & 3\\end{vmatrix} + \\begin{vmatrix} 1 & 0.5 \\\\ 3 & 4\\end{vmatrix}$) = 4.25\n"
      ],
      "metadata": {
        "id": "OthF78gJbUjl"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Eigenvectors and Eigenvalues**\n",
        "\n",
        "For a square matrix $A$ which is $n \\times n$, certain vectors in the domain are special.\n",
        "\n",
        "When $A$ acts on them, it preserves their direction, but scales them.\n",
        "\n",
        "i.e. $Ax = \\lambda x$\n",
        "\n",
        "Thus if we were to think of the matrix as a system and $x$ as its input, the output is same as input, but scaled (either amplified or attenuated). (compare with singular vectors and singular values)\n",
        "\n",
        "To find all such special vectors/directions - called eigenvectors and the corresponding scale values - called eigenvalues, we need to analyze the above equation. Rewriting it as\n",
        "\n",
        "$(A-\\lambda I)x = 0$\n",
        "\n",
        "We need to find non-trivial solutions $x$ and corresponding $\\lambda$s.\n",
        "\n",
        "This means, $x$ is in the nullspace of $A-\\lambda I$ which means we need to find $\\lambda$s such that $A-\\lambda I$ is singular or equivalently, its determinant is 0.\n",
        "\n",
        "So find $\\lambda$ such that $|A - \\lambda I| = 0 = \\begin{vmatrix} a_{11}-\\lambda & a_{12} & ... & a_{1n} \\\\ a_{21} & a_{22}-\\lambda & ... & a_{2n} \\\\ ... \\\\ a_{n1} & a_{n2} & ... & a_{nn}-\\lambda \\end{vmatrix}$\n",
        "\n",
        "This leads to a characteristic $n^{th}$ degree polynomial equation in $\\lambda$, which has $n$ roots that are the eigenvalues of $A$.\n",
        "\n",
        "Note that the roots might turn to be complex (in which case, for a real matrix, they will appear as conjugate pairs). The corresponding eigen vectors will also be complex.\n"
      ],
      "metadata": {
        "id": "U3DyEwohohHo"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Example calculation of Eigen values and vectors **\n",
        "\n",
        "$A = \\begin{vmatrix} 4 & -5 \\\\ 2 & -3\\end{vmatrix}$\n",
        "\n",
        "$|A-\\lambda I| = \\begin{vmatrix} 4-\\lambda & -5 \\\\ 2 & -3-\\lambda\\end{vmatrix} = \\lambda^2-\\lambda -2 = (\\lambda-2)(\\lambda+1)=0$. Therefore the eigenvalues are $\\lambda = 2$ and $\\lambda = -1$\n",
        "\n",
        "To find the eigen vectors, we need to find the nullspace for each eigen value.\n",
        "\n",
        "For $\\lambda=2$, $(A-2I)x = \\begin{bmatrix} 2 & -5\\\\2 & -5 \\end{bmatrix} \\begin{bmatrix} x_1 \\\\ x_2 \\end{bmatrix} = 0 $ clearly $x_1 = 5/2 x_2$ or $\\{a\\begin{bmatrix} 5 \\\\ 2 \\end{bmatrix}\\}$ defines the corresponding eigenspace, where $a$ is any real number.\n",
        "\n",
        "Similarly for $\\lambda=-1$, $(A+I)x = \\begin{bmatrix} 5 & -5 \\\\ 2 & -2 \\end{bmatrix}\\begin{bmatrix} x_1 \\\\ x_2 \\end{bmatrix} = 0 $. Clearly $x_1 = x_2$ or $\\{a\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}\\}$ where $a$ is any real number defines the other eigen space.\n",
        "\n",
        "Note that if $x$ is a eigenvector, then so is $ax$ for any scalar $a$. Also $Ax = \\lambda x$. In general the set of eigen vectors corrsponding to a eigenvalue forms a subspace.\n",
        "\n",
        "*Example*\n",
        "\n",
        "The identity matrix $I$ has $n$ eigen values, all equal to 1. The dimension of the eigenspace corresponding to this eigen value is $n$ - the entire vector space."
      ],
      "metadata": {
        "id": "QjqB_uSwq11r"
      }
    },
    {
      "cell_type": "code",
      "source": [
        "#check with numpy\n",
        "A = np.array([[4,-5],[2,-3]])\n",
        "print(np.linalg.eig(A))\n",
        "#Returns eigen values and corresponding eigen vectors as  columns and normalized to unity"
      ],
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "mCNrwRNYvRLq",
        "outputId": "01ab6322-98ed-49e5-ba23-8fbe10b01128"
      },
      "execution_count": null,
      "outputs": [
        {
          "output_type": "stream",
          "name": "stdout",
          "text": [
            "(array([ 2., -1.]), array([[0.92847669, 0.70710678],\n",
            "       [0.37139068, 0.70710678]]))\n"
          ]
        }
      ]
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Matrix Powers**\n",
        "\n",
        "If $Ax = \\lambda x$ , then $A^2x = A(Ax) = A(\\lambda x) = \\lambda Ax = \\lambda^2 x$ and in general, $A^nx = \\lambda^n x$ for integer\n",
        " n.\n",
        "\n",
        "Hence eigen vectors of $A^n$ is same as that of $A$, while Eigen values get exponentiated."
      ],
      "metadata": {
        "id": "nYoHI6xEUZAP"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Relation to Determinant**\n",
        "\n",
        "The characteristic polynomial is $|A-\\lambda I| = (-\\lambda+\\lambda_1)(-\\lambda+\\lambda_2)...(-\\lambda+\\lambda_n)$ is a polynomial with roots $\\lambda_1, \\lambda_2, ... \\lambda_n$.\n",
        "\n",
        "When we put $\\lambda=0$ above, we get $|A| = \\prod_{i=1}^{n}\\lambda_i$\n",
        "\n",
        "Hence if any eigen value is 0, the determinant is 0 and hence $A$ is singular or rank(A) < n. Similarly any singular matrix will have determinant 0 and hence at least one of the eigenvalues is 0.\n",
        "\n",
        "From the characteristic equation, right side, the coefficient of $(-\\lambda)^{n-1}$ is $\\sum_{i=1}^n\\lambda_i$ and from left side it is $\\sum_{i=1}^na_{ii}$ i.e. the sum of diagonals - which is also called the trace. Hence, trace of a matrix is the sum of eigen values.\n",
        "\n",
        "Note that\n",
        "\n",
        "a) $Trace(A) = Trace(A^T)$ as diagonal remains same\n",
        "\n",
        "b) $Trace(AB) = Trace(BA)$\n",
        "\n",
        "as $Trace(AB) = \\sum_{i=1}^n \\sum_{j=1}^{m}a_{ij}b_{ji} = \\sum_{j=1}^{m} \\sum_{i=1}^n  a_{ij}b_{ji} = \\sum_{j=1}^{m} \\sum_{i=1}^n  b_{ji}a_{ij} = Trace(BA)$ (by changing the order of summation of index i and j\n",
        "\n",
        "\n",
        "\n",
        "Hence if $v$ is a vector, then $Trace(vv^T) = Trace(v^Tv) = v^Tv$\n",
        "\n",
        "\n",
        "The $nxn$ matrix $vv^T$ has rank 1 and one non-zero eigen value (=$v^Tv$ and $n-1$ of 0 eigen values. The eigen space for '0' eigenvalue is the hyperplane normal to $v$.\n"
      ],
      "metadata": {
        "id": "glIhyIm-3dp9"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Geometric and Arithmetic Multiplicity of Eigenvalues**\n",
        "\n",
        "Algebraic multiplicity of an eigen value is the number of repetitions as the root. For example, the eigen value '1' for Identity matrix $I$ has algebraic multiplicity of $n$\n",
        "\n",
        "Geometric multiplicity is the dimension of the corresponding eigensubspace and is less than equal to algebraic multiplicity. In case of identity matrix, the eigen value '1' also has geometric multiplicity of $n$\n",
        "\n",
        "Note that '0' can also be an eigen value/\n",
        "\n",
        "For now, lets assume that geometric multiplicity = arithmetic multiplicity. Then for each each eigenvalue with multiplicity $k$, the corresponding eigen subspace will have dimension $k$, and we can find $k$ linearly independant basis vectors for that subspace.\n"
      ],
      "metadata": {
        "id": "oXrS8Wv5uCjq"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Diagonalization of a square matrix (or Eigen Value Decomposition)**\n",
        "\n",
        "For any square matrix, if the eigenvalues have algebraic multiplicity same as geometric multipicity, then we can form a matrix of eigen vectors $S = [v_1, v_2, v_3, ...., v_n]$ where each column is a eigen vector for the corresponding eigen value in the diagonal. Then\n",
        "\n",
        "$AS = S\\Lambda$ where $\\Lambda$ is the diagonal matrix of eigen values.\n",
        "\n",
        "Since all column vectors are independent (either due to being with different eigen spaces or by construciton in case they fall in the same eigen space), then $S$ is invertible and\n",
        "\n",
        "$A = S\\Lambda S^{-1}$ or $\\Lambda = S^{-1}AS$\n",
        "\n",
        "Hence $A$ is diagonalizable. Clearly $S$ is not unique (for example any basis set for each eigensubspace can be chosen).\n",
        "\n",
        "Two matrices $A$ and $B$ are said to be similar, if they are digonalizable to the same Diagonal matrix, i.e. $\\Lambda = S_A^{-1}AS_A = S_B^{-1}BS_B$\n",
        "\n",
        "\n",
        "*Matrix Powers*\n",
        "\n",
        "If $A$ is diagonalizable, $A=S\\Lambda S^{-1}$, then $A^2 = S\\Lambda S^{-1}S\\Lambda S^{-1} = S\\Lambda^2S^{-1}$ and in general $A^n = S\\Lambda^nS^{-1}$"
      ],
      "metadata": {
        "id": "l85QL27c1ULw"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Cayley Hamilton Theorem**\n",
        "\n",
        "The characteristic equation of a matrix is\n",
        "\n",
        " $p(\\lambda) = |\\lambda I - A| = \\lambda^n + c_{n-1} \\lambda^{n-1} + ...+ c_1 \\lambda + c_0 = 0$.\n",
        "\n",
        "Then substituting $A$ for $\\lambda$ in this polynomial equation, we get a matrix equation in the powers of A which is also equal to the 0 matrix!\n",
        "\n",
        "$p(A) = A^n + c_{n-1} A^{n-1}+ ...+c_1 A + c_0 = 0$\n",
        "\n",
        "For $A$ diagonalizatble, this is easy to prove (though it is true for any square matrix A).\n",
        "\n",
        "Let $A=S\\Lambda S^{-1}$ therefore $p(S\\Lambda S^{-1}) = S\\Lambda^nS^{-1} + C_{n-1}S\\Lambda^{n-1}S^{-1} + ... + c_1 S\\Lambda S^{-1}+ cSS^{-1} = S(\\Lambda^n + c_{n-1}\\Lambda^{n-1} + ... + c_1\\Lambda + c_0I)S^{-1} = S0S^{-1} = 0$ since each eigen value is the root of the  characteristic equation - whichs is formed by summing  the diagonal entry across the matrix sum\n",
        "\n",
        "Hence and power of a matrix equal to or above $n$ can be replaces by a polynomial of matrix powers less than $n$\n",
        "\n"
      ],
      "metadata": {
        "id": "tB_RzLE-U16R"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Matrix Exponential**\n",
        "\n",
        "Formally, we can define $e^{A} = I + A + \\frac{A^2}{2!} + \\frac{A^3}{3!} ....$.\n",
        "\n",
        "From Cayley-Hamilton theorem, the above can be converted into a matrix sum with powers of $A$ upto $n-1$\n",
        "\n",
        "If $A$ is diagonalizable, then $e^{A} = SS^{-1} + S\\Lambda S^{-1} + ...  = S(I + \\Lambda + \\frac{\\Lambda^2}{2!} + ... )S^{-1} = Se^\\Lambda S^{-1}$"
      ],
      "metadata": {
        "id": "iSYo_b9NXmao"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Real-Symmetric Matrices or Hermitian Matrices have real eigen values**\n",
        "\n",
        "Let $A$ be a Hermitian (if complex,i.e. $A^H=A^{*T}=A$) or symmetric (if real,i.e.  $A^T = A$).<br>\n",
        "Let $Ax = \\lambda x$.<br>\n",
        "Then $x^{*T}A^{*T} = \\lambda^* x^{*T} = x^{*T}A$.\n",
        "\n",
        "Therefore $x^{*T}A^{*T}x = (x^{*T}A^{*T})x = \\lambda^*x^{*T}x = x^{*T}(A^{*T}x) = \\lambda x^{*T}x$.<br>\n",
        "Therefore $\\lambda = \\lambda^*$ is the only way this is possible for non-zero eigenvectors. Hence eigen values of Hermitian matrices are real."
      ],
      "metadata": {
        "id": "WGJV5PlaMAv_"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Eigen spaces of Hermitian (including real symmetric) matrices are orthogonal to each other**\n",
        "\n",
        "Let $A$ be Hermitian - i.e. equal to its conjugate transpose $(A^H=A^{*T} = A$ or $A^* = A^T)$\n",
        "\n",
        "Let $Ax = \\lambda_x x$ and $Ay = \\lambda_y y$ be two different eigen values and subspaces.  \n",
        "\n",
        "Then $x^{*T}A^{*T}y = (x^{*T}A^{*T})y = \\lambda_x^{*}x^{*T}y = x^{*T}(A^{*T}y) = \\lambda_y  x^{*T}y$ and if $\\lambda_x \\neq \\lambda_y$, this can only happen if $x^{*T}y = 0$, which means that the eigen subspaces are orthogonal.\n",
        "\n",
        "Hence the eigen value decomposition of a Hermitian matrix is:\n",
        "\n",
        "$A = S\\Lambda S^T$ where $S$ is the orthogonal matrix of eigen vectors and $\\Lambda$ is the diagonal matrix of real eigen values.\n",
        "\n",
        "Note that in this case, the eigen values are also the singular values.\n",
        "\n",
        "For a any matrix, its SVD is  $A = U\\Sigma V^H $, and hence $AA^H = U \\Sigma V^H V \\Sigma^H U^H = U \\Sigma^2 U^H$\n",
        "\n",
        "Hence $\\Lambda(AA^H) = \\Sigma^2$\n",
        "\n",
        "If $A$ is Hermitian, then $A=Q \\Lambda Q^T$ hence the eigen values are also singular values.\n"
      ],
      "metadata": {
        "id": "fDyu9cF8OkN2"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q14**\n",
        "\n",
        "Find the three eigen values and vectors for $A=\\begin{bmatrix} 1 \\\\ 2 \\\\ 1\\end{bmatrix} \\begin{bmatrix} 2 & 1 & 2 \\end{bmatrix}$\n"
      ],
      "metadata": {
        "id": "epNeoyITb19B"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "$(uv^T)u = (v^Tu)u$, so $v^Tu$ is an eigenvalue and the eigenspace for this is $\\{ku\\}$. In this example, this is 6.\n",
        "\n",
        "Since $uv^T$ has rank 1, its other two eigen values are $0, 0$. The eignespace corresponding to eigenvalue $0$ is the null space $\\{x:uv^Tx=0\\}$, which is the $2D$ space perpendicular to $v$\n",
        "\n",
        "In this example, $[2\\ 1\\ 2]\\begin{bmatrix}x_1 \\\\ x_2 \\\\ x_3\\end{bmatrix} = 2x_1 + x_2 + 2x_3 = 0$ and $x_2 = -1/2(x_1 + x_3)$ therefore the eigen space is $span(\\begin{bmatrix}0\\\\ -0.5\\\\ 1\\end{bmatrix}, \\begin{bmatrix}1\\\\ -0.5\\\\ 0\\end{bmatrix})$\n",
        "\n",
        "\n"
      ],
      "metadata": {
        "id": "vS531OgufAE3"
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      "cell_type": "markdown",
      "source": [
        "---\n",
        "**Q15**\n",
        "when a+b = c+d, show that (1,1) is an eigen vector of $A=\\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}$. Find the other eigen value.\n"
      ],
      "metadata": {
        "id": "_16uG4COQX8c"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Ans:\n",
        "\n",
        "$A\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} = \\begin{bmatrix} a+b \\\\ a+b \\end{bmatrix} = (a+b)\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}$ since $a+b = c+d$\n",
        "\n",
        "Since sum of eigen values = trace of matrix, the other eigen value is $Tr(A)-(a+b) = d-b = a-c$\n",
        "\n",
        "The eigen vector this is null space of $(A-\\lambda I)$ for the specific value of $\\lambda = d-b = a-c$.\n",
        "\n",
        "i.e. we need to find null space of $\\begin{bmatrix} c & b \\\\ c & b \\end{bmatrix}$ which is $\\begin{bmatrix} 1 \\\\ -c/b \\end{bmatrix}$ if $b \\neq 0 $ or $\\begin{bmatrix} -b/c \\\\ 1 \\end{bmatrix}$ if $c \\neq 0 $.  If both c and b are zero, this second eigen value is same as first and the entire 2D space is the eigen space.\n",
        "\n",
        "\n",
        "\n"
      ],
      "metadata": {
        "id": "bp_dE3kTh-Qp"
      }
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}