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12.18 Polynomial

  1. Let \(T: \mathbb{R}^3 \to \mathbb{R}^3\) linear map defined by \(T(x,y,z) = (x, z, -2y - z)\). Let \(p\) be the polynomial over \(\mathbb{R}\) defined by \(p(x) = -x^3 + 2\). Find \(p(T)\).

    Proof

    The idea is to find \([T]_C\) for \(C\) canonical basis and using the fact that \([T \circ \dots \circ T]_C = [T]_C \dots [T]_C\).

    We can evaluate \(p([T]_C)\).

    Now \([T]_C = \Big( [T(1,0,0)]_C \; [T(0,1,0)]_C \; [T(0,0,1)]_C \Big)\)​\(= \begin{pmatrix}1 & 0 & 0 \\0 & 0 & 1 \\0 & -2 & -1\end{pmatrix}\)​

    So \(p([T]_{C})=-[T]_{C}^3+2\mathrm{Id}=-\begin{pmatrix}1 & 0 & 0\\ 0 & 2 & -1\\ 0 & 2 & 3\end{pmatrix}+\begin{pmatrix}2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\end{pmatrix}= \begin{pmatrix}1 & 0 & 0 \\0 & 0 & -1 \\0 & -2 & -1\end{pmatrix}\)

    So \(p([T]_C) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -2 & -1 \end{pmatrix}\)​and now we take it to a linear map:

    \(p(T)(x, y, z) = (x, z, -2y - z) = T(x, y, z)\)​\(\implies p(T) = T\)​

  2. Let \(A \in M_n(\mathbb{F})\) be a diagonal matrix \((a_{ij} = 0 \, \text{if} \, i \neq j)\).

    Let \(p\) be the polynomial over \(\mathbb{F}\) defined by \(p(x) = (x - a_{11})(x - a_{22}) \dots (x - a_{nn}).\)

    Where \(A = \begin{pmatrix} a_{11} & 0 & \dots & 0 \\ 0 & a_{22} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & a_{nn} \end{pmatrix}\)​

    What is the matrix \(p(A)\)?

    We have \(p(A) = (A - a_{11} \mathrm{Id})(A - a_{22} \mathrm{Id}) \dots (A - a_{nn} \mathrm{Id})\)​

    For each \(i\):

    \(A - a_{ii} \mathrm{Id} = \begin{pmatrix}a_{11} & 0 & \dots & 0 \\0 & a_{ii} & \dots & 0 \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \dots & a_{nn}\end{pmatrix}- \begin{pmatrix}a_{ii} & 0 & \dots & 0 \\0 & a_{ii} & \dots & 0 \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \dots & a_{ii}\end{pmatrix}\)​\(= \begin{pmatrix}a_{11} - a_{ii} & 0 & \dots & 0 \\0 & 0 & \dots & 0 \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \dots & a_{nn} - a_{ii}\end{pmatrix}.\)​

    So each \((A - a_{ii} \mathrm{Id})\) is a diagonal matrix with a \(0\) in \((i,i)\).
    Since multiplying diagonal matrices gives you a diagonal matrix with the diagonal given by the product of the diagonal entries

    Then \(p(A)=(A-a_{11}\mathrm{Id})\dots(A-a_{nn}\mathrm{Id})=0\)

  3. Determine which of the following subsets of \(\mathbb{Q}[x]\) are ideals.

    1. \(S = \{ p \in \mathbb{Q}[x] \mid \deg(p) \text{ is even} \}\)​ .

      No, Counterexample: Take \(p(x) = x \in \mathbb{Q}[x]\), \(q(x) = 1 \in S\). Then \(p(x)q(x)=x\notin S\quad(\deg=1)\) 2. \(S = \{ p \in \mathbb{Q}[x] \mid p(0) = 0 \}\)

      Yes: \(S\) is a subspace (check).

      Now if \(p \in \mathbb{Q}[x]\) and \(q \in S\):

      \((pq)(0) = p(0)q(0) = p(0) \cdot 0 = 0.\)​\(\implies pq \in S.\) So \(S\) is an ideal. 3. \(R = \{ p \in \mathbb{Q}[x] \mid p \in \text{Range}(T) \}\)

      Where \(T: \mathbb{Q}[x] \to \mathbb{Q}[x]\) given by \(T\left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n a_i x^{i+1}.\)

      Let’s see that this is the exact same ideal as in exercise (b) \(R = \{ p \in \mathbb{Q}[x] \mid p(0) = 0 \}.\)

      \(\subseteq\)) Note that \(T\left(\sum_{i=0}^n a_i x^i\right) = \sum_{i=0}^n a_i x^{i+1}\) has no constant term.

      \(\implies T(p)(0) = 0 \quad \forall p \in \mathbb{Q}[x]\)

      \(\implies q(0) = 0 \quad \forall q \in \text{Range}(T).\)

      \(\implies\) If \(p \in \mathbb{Q}[x]\) such that \(p(0) = 0\):

      \(\supseteq\)) \(p(x) = a_n x^n + \dots + a_1 x \quad (\text{no constant term}).\)

      We want to show that \(p \in \text{Range}(T)\).

      \(q(x)=a_1+2a_2x+3a_3x^2+\dots+na_{n}x^{n-1}\) where coefficient \(b_i\) of this is \(a_{i+1}(i+1)\)​

      \(\implies T(q)(x)=\sum_{i=0}^{n}\frac{a_{i+1}}{i+1}x^{i+1}\implies p(x)\implies T\left(q\right)=p\implies p\in\text{Range}T\)​

image​

Some properties of determinants:

  1. \(\det(AB) = \det(A)\det(B)\)​

  2. \(\det(A^{-1})=\det(A)^{-1}\quad\text{when A is invertible}\)​

  3. \(\det(A^T) = \det(A)\)​

  1. Let \(A, B, C \in M_n(\mathbb{R})\) such that \(\det(A) = -1\), \(\det(B) = 2\), and \(\det(C) = 3\). Calculate:

    \(\det(A^2 B C^T B^{-1})\)​

    \(\det(A^2BC^{T}B^{-1})=\det(A)^2\det(B)\det(C^{T})\det(B^{-1})=\det(A)^2\det(B)\det(C)\det(B^{-1})\)​

    \(=\det(A)^2\det(B)\det(C)\det(B)^{-1}=(-1)^2\cdot2\cdot3\cdot\frac12=3.\)

Note: We can use determinant formula to deal with \(2\times 2\) and \(3\times 3\), but if the matrices become bigger, we need to use row reduced operation and record the \(\det\) step by step.

Or we can use these three tricks

For example

Remember: If \(A\) is an upper triangular matrix: \(\det(A) = a_{11}a_{22} \dots a_{nn}\) where \(A = \begin{pmatrix}a_{11} & a_{12} & \dots & a_{1n} \\0 & a_{22} & \dots & a_{2n} \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \dots & a_{nn}\end{pmatrix}\)

So we can apply elementary row operations to a matrix to get it upper triangular and calculate the determinant.

We use the following: let \(A \in M_n(\mathbb{R})\) and suppose \(B \in M_n(\mathbb{R})\) is obtained from \(A\) by

  1. Interchanging 2 rows from \(A\) \(\implies \det(B) = -\det(A)\)

  2. Multiply \(\lambda \in \mathbb{R}\) to a row of \(A\) \(\implies \det(B) = \lambda \det(A)\)

  3. Add a multiple of one row of \(A\) to another \(\implies \det(B) = \det(A)\)

  4. Calculate the following determinants using row reductions:

    1. \(\det\begin{pmatrix}2 & 0 & 0 & 0\\ 0 & 0 & 3 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 4\end{pmatrix}\) \(= - \det \begin{pmatrix} 2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4 \end{pmatrix}\)​\(= - 2 \cdot (-1) \cdot 3 \cdot 4 = 24\)

    2. \(\det \begin{pmatrix} 1 & -1 & 2 & 0 & 0 \\ 3 & 1 & 4 & 0 & 0 \\ 2 & -1 & 5 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & -1 & 4 \end{pmatrix}\)

      Performing row operations: \(R_2 \to R_2 - 3R_1,R_3 \to R_3 - 2R_1,R_4 \to R_4 + 2R_5\)​

      \(=\det\begin{pmatrix}1 & -1 & 2 & 0 & 0\\ 0 & 4 & -2 & 0 & 0\\ 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 9\\ 0 & 0 & 0 & -1 & 4\end{pmatrix}\)​\(=\left\lbrack\text{R}_2\leftrightarrow\text{R}_3\quad\text{R}_4\leftrightarrow\text{R}_5\right\rbrack(-1)(-1)\det\begin{pmatrix}1 & -1 & 2 & 0 & 0\\ 0 & 1 & 1 & 0 & 0\\ 0 & 4 & -2 & 0 & 0\\ 0 & 0 & 0 & -1 & 4\\ 0 & 0 & 0 & 0 & 9\end{pmatrix}\)

      \(=\left\lbrack\text{R}_3-4\text{R}_2\right\rbrack\det\begin{pmatrix}1 & -1 & 2 & 0 & 0\\ 0 & 1 & 1 & 0 & 0\\ 0 & 0 & -6 & 0 & 0\\ 0 & 0 & 0 & -1 & 4\\ 0 & 0 & 0 & 0 & 9\end{pmatrix}=1\cdot1\cdot(-6)\cdot(-1)\cdot9=54\)

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