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- Apr 14, 2013

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Hey!!

Let $\mathbb{K}$ be a field and $1\leq n\in \mathbb{N}$.

For a polynom $\displaystyle{\sum_{i=0}^mc_it^i\in \mathbb{K}[t]}$ and a matrix $a\in M_n(\mathbb{K})$ the $f(a)\in M_n(\mathbb{K})$ is defined by \begin{equation*}f(a):=\sum_{i=0}^mc_ia^i=c_ma^m+c_{m-1}a^{m-1}+\ldots +c_2a^2+c_1a+c_0u_n\end{equation*}

We have that

\begin{equation*}f(a)=a^2+2a-5u_n =\begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{pmatrix}^2+2\cdot \begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{pmatrix}-5\cdot \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix}-2 & 8 & 2 \\ 0 & -1 & 3 \\ 0 & 3 & -4\end{pmatrix}\end{equation*}

Could you give me a hint for that?

I have done the following : \begin{equation*}f=\sum_{i=0}^mc_it^i \ \text{ and } \ g=\sum_{j=0}^kd_jt^j\end{equation*}

The sum is equal to \begin{equation*}f+g=\sum_{i=0}^mc_it^i+\sum_{i=0}^kd_it^i=\sum_{i=0}^{\max (m,k)}\left (c_i+d_i\right )t^i\end{equation*}

with $c_i:=0$ and $d_j:=0$ for all $i>m$ and all $j>k$.

We have that

\begin{align*}&(f+g)(a)=\sum_{i=0}^{\max (m,k)}\left (c_i+d_i\right )a^i \\ &f(a)+g(a)=\sum_{i=0}^mc_ia^i+\sum_{i=0}^kd_ia^i=\sum_{i=0}^{\max (m,k)}\left (c_i+d_i\right )a^i \end{align*}

Therefore $(f+g)(a)=f(a)+g(a)$.

The product of $f$ and $g$ is defined by \begin{equation*}f g=\left (\sum_{i=0}^mc_it^i\right )\cdot \left (\sum_{i=0}^kd_it^i\right )=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_jt^{\ell}\end{equation*}

with $c_i:=0$ and $d_j:=0$ for all $i>m$ and all $j>k$.

We have that

\begin{align*}&(fg)(a)=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_ja^{\ell} \\ &f(a)g(a)=\left (\sum_{i=0}^mc_ia^i\right )\cdot \left (\sum_{i=0}^kd_ia^i\right )=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_ja^{\ell}\end{align*}

Therefore $(fg)(a)=f(a)g(a)$.

For $U:=\left \{f(a)\mid f\in \mathbb{K}[t]\right \}\subseteq \mathbb{K}[t]$ the following conditions hold:

- $\forall f(a),g(a)\in U : (f+g)(a)\in U$, since $(f+g)(a)=f(a)+g(a)$

- $\forall f(a),g(a)\in U : (fg)(a)\in U$, since $(fg)(a)=f(a)g(a)$

- $1_{\mathbb{K}[t]}\in U$, with $1_{\mathbb{K}[t]}=u_n$, and $-1_{\mathbb{K}[t]}\in U$, with $1_{\mathbb{K}[t]}=-u_n$

so it follows that $\left \{f(a)\mid f\in \mathbb{K}[t]\right \}$ is a commutative subring of $\mathbb{K}[t]$.

Is everything correct?

I have done the following:

We have that \begin{align*}&av=\lambda v \\ &a^2v=a\left (\lambda v\right )=\lambda \left (av\right )=\lambda \left (\lambda v\right )=\lambda^2 v \\ &a^3v=a\left (a^2 v\right )=a\left (\lambda^2 v\right )=\lambda^2 \left (av\right )=\lambda^2 \left (\lambda v\right )=\lambda^3 v \\ &\ldots \\ &a^iv=\lambda^i v\end{align*}

We get:

\begin{equation*}f(a)v=\left (\sum_{i=0}^mc_ia^i\right )v=\sum_{i=0}^mc_i\left (a^iv\right )=\sum_{i=0}^mc_i\left (\lambda^i v\right )=\left (\sum_{i=0}^mc_i\lambda^i \right ) v=f(\lambda )v\end{equation*}

Therefore $v$ is an eigenvector of $f(a)$ for the eigenvalue $f(\lambda)$.

Is everything correct?

Since $a$ is diagonalizable, it can be written in the form $a=sds^{-1}$, where $d$ is a diagonal matrix with diagonal elements the eigenvalus of the matrix $a$.

We have that \begin{align*}f(a)&=(a-\lambda_1u_n)(a-\lambda_2u_n)\cdots (a-\lambda_ku_n)\\ & =(sds^{-1}-\lambda_1u_nss^{-1})(sds^{-1}-\lambda_2u_nss^{-1})\cdots (sds^{-1}-\lambda_ku_nss^{-1})\\ & =(sds^{-1}-s\lambda_1u_ns^{-1})(sds^{-1}-s\lambda_2u_ns^{-1})\cdots (sds^{-1}-s\lambda_ku_ns^{-1})\\ & =[s(d-\lambda_1u_n)s^{-1}][s(d-\lambda_2u_n)s^{-1}]\cdots [s(d-\lambda_ku_n)s^{-1}] \\ & =s(d-\lambda_1u_n)(d-\lambda_2u_n)\cdots (d-\lambda_ku_n)s^{-1}\end{align*}

From $d-\lambda_i u_n$ the $i$-the row of the resulting matrix is contains only $0$.

A zero-row in $M$ makes at the product $MM_0$ also a zero-row, at the same row.

Therefore the product $(d-\lambda_1u_n)(d-\lambda_2u_n)\cdots (d-\lambda_ku_n)$, ans so also $s(d-\lambda_1u_n)(d-\lambda_2u_n)\cdots (d-\lambda_ku_n)s^{-1}$, is a zero matrix, so $f(a)=0_{M_n(\mathbb{K})}$.

Is everything correct?

Let $\mathbb{K}$ be a field and $1\leq n\in \mathbb{N}$.

For a polynom $\displaystyle{\sum_{i=0}^mc_it^i\in \mathbb{K}[t]}$ and a matrix $a\in M_n(\mathbb{K})$ the $f(a)\in M_n(\mathbb{K})$ is defined by \begin{equation*}f(a):=\sum_{i=0}^mc_ia^i=c_ma^m+c_{m-1}a^{m-1}+\ldots +c_2a^2+c_1a+c_0u_n\end{equation*}

**Question 1:**For $\displaystyle{a=\begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{pmatrix}\in M_3(\mathbb{K})}$ and $f=t^2+2t-5$ calculate $f(a)$.We have that

\begin{equation*}f(a)=a^2+2a-5u_n =\begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{pmatrix}^2+2\cdot \begin{pmatrix}1 & 2 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{pmatrix}-5\cdot \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix} =\begin{pmatrix}-2 & 8 & 2 \\ 0 & -1 & 3 \\ 0 & 3 & -4\end{pmatrix}\end{equation*}

**Question 2:**Let $n\leq 2$. Show that $P_a(a)=0_{M_n(\mathbb{K})}$.Could you give me a hint for that?

**Question 3:**For all $f,g\in \mathbb{K}[t]$ show that $(f+g)(a)=f(a)+g(a)$ and $(fg)(a)=f(a)g(a)$ and that $\left \{f(a)\mid f\in \mathbb{K}[t]\right \}$ is a commutative subring of $\mathbb{K}[t]$.I have done the following : \begin{equation*}f=\sum_{i=0}^mc_it^i \ \text{ and } \ g=\sum_{j=0}^kd_jt^j\end{equation*}

The sum is equal to \begin{equation*}f+g=\sum_{i=0}^mc_it^i+\sum_{i=0}^kd_it^i=\sum_{i=0}^{\max (m,k)}\left (c_i+d_i\right )t^i\end{equation*}

with $c_i:=0$ and $d_j:=0$ for all $i>m$ and all $j>k$.

We have that

\begin{align*}&(f+g)(a)=\sum_{i=0}^{\max (m,k)}\left (c_i+d_i\right )a^i \\ &f(a)+g(a)=\sum_{i=0}^mc_ia^i+\sum_{i=0}^kd_ia^i=\sum_{i=0}^{\max (m,k)}\left (c_i+d_i\right )a^i \end{align*}

Therefore $(f+g)(a)=f(a)+g(a)$.

The product of $f$ and $g$ is defined by \begin{equation*}f g=\left (\sum_{i=0}^mc_it^i\right )\cdot \left (\sum_{i=0}^kd_it^i\right )=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_jt^{\ell}\end{equation*}

with $c_i:=0$ and $d_j:=0$ for all $i>m$ and all $j>k$.

We have that

\begin{align*}&(fg)(a)=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_ja^{\ell} \\ &f(a)g(a)=\left (\sum_{i=0}^mc_ia^i\right )\cdot \left (\sum_{i=0}^kd_ia^i\right )=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_ja^{\ell}\end{align*}

Therefore $(fg)(a)=f(a)g(a)$.

For $U:=\left \{f(a)\mid f\in \mathbb{K}[t]\right \}\subseteq \mathbb{K}[t]$ the following conditions hold:

- $\forall f(a),g(a)\in U : (f+g)(a)\in U$, since $(f+g)(a)=f(a)+g(a)$

- $\forall f(a),g(a)\in U : (fg)(a)\in U$, since $(fg)(a)=f(a)g(a)$

- $1_{\mathbb{K}[t]}\in U$, with $1_{\mathbb{K}[t]}=u_n$, and $-1_{\mathbb{K}[t]}\in U$, with $1_{\mathbb{K}[t]}=-u_n$

so it follows that $\left \{f(a)\mid f\in \mathbb{K}[t]\right \}$ is a commutative subring of $\mathbb{K}[t]$.

Is everything correct?

**Question 4:**$v\in \mathbb{K}^n$ is an eigenvector of $a$ for the eigenvalue $\lambda$. Then I want to show that $v$ is also an eigenvector of $f(a)$ for the eigenvalue $f(\lambda)$, for all $f\in \mathbb{K}[t]$.I have done the following:

We have that \begin{align*}&av=\lambda v \\ &a^2v=a\left (\lambda v\right )=\lambda \left (av\right )=\lambda \left (\lambda v\right )=\lambda^2 v \\ &a^3v=a\left (a^2 v\right )=a\left (\lambda^2 v\right )=\lambda^2 \left (av\right )=\lambda^2 \left (\lambda v\right )=\lambda^3 v \\ &\ldots \\ &a^iv=\lambda^i v\end{align*}

We get:

\begin{equation*}f(a)v=\left (\sum_{i=0}^mc_ia^i\right )v=\sum_{i=0}^mc_i\left (a^iv\right )=\sum_{i=0}^mc_i\left (\lambda^i v\right )=\left (\sum_{i=0}^mc_i\lambda^i \right ) v=f(\lambda )v\end{equation*}

Therefore $v$ is an eigenvector of $f(a)$ for the eigenvalue $f(\lambda)$.

Is everything correct?

**Question 5:**Let $a$ be diagonalizable, $k=|\text{Spec}(a)|, \lambda_1, \lambda_2, \ldots , \lambda_k$ the eigenvalues of $a$ and $f:=(t-\lambda_1)(t-\lambda_2)\cdots (t-\lambda_k)\in \mathbb{K}[t]$. Show that $f(a)=0_{M_n(\mathbb{K})}$.Since $a$ is diagonalizable, it can be written in the form $a=sds^{-1}$, where $d$ is a diagonal matrix with diagonal elements the eigenvalus of the matrix $a$.

We have that \begin{align*}f(a)&=(a-\lambda_1u_n)(a-\lambda_2u_n)\cdots (a-\lambda_ku_n)\\ & =(sds^{-1}-\lambda_1u_nss^{-1})(sds^{-1}-\lambda_2u_nss^{-1})\cdots (sds^{-1}-\lambda_ku_nss^{-1})\\ & =(sds^{-1}-s\lambda_1u_ns^{-1})(sds^{-1}-s\lambda_2u_ns^{-1})\cdots (sds^{-1}-s\lambda_ku_ns^{-1})\\ & =[s(d-\lambda_1u_n)s^{-1}][s(d-\lambda_2u_n)s^{-1}]\cdots [s(d-\lambda_ku_n)s^{-1}] \\ & =s(d-\lambda_1u_n)(d-\lambda_2u_n)\cdots (d-\lambda_ku_n)s^{-1}\end{align*}

From $d-\lambda_i u_n$ the $i$-the row of the resulting matrix is contains only $0$.

A zero-row in $M$ makes at the product $MM_0$ also a zero-row, at the same row.

Therefore the product $(d-\lambda_1u_n)(d-\lambda_2u_n)\cdots (d-\lambda_ku_n)$, ans so also $s(d-\lambda_1u_n)(d-\lambda_2u_n)\cdots (d-\lambda_ku_n)s^{-1}$, is a zero matrix, so $f(a)=0_{M_n(\mathbb{K})}$.

Is everything correct?

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