# Polynomials at a equal to 0

#### mathmari

##### Well-known member
MHB Site Helper
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*}

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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#### Opalg

##### MHB Oldtimer
Staff member
Question 2: Let $n\leq 2$. Show that $P_a(a)=0_{M_n(\mathbb{K})}$.
Could you give me a hint for that?
You have not said what $P_a$ is. Is it meant to be the characteristic polynomial of $a$? If so, this question is asking you to verify the Cayley-Hamilton theorem for the cases $n=1$ and $n=2$. That Wikipedia link shows exactly how to do that.

#### mathmari

##### Well-known member
MHB Site Helper
You have not said what $P_a$ is. Is it meant to be the characteristic polynomial of $a$? If so, this question is asking you to verify the Cayley-Hamilton theorem for the cases $n=1$ and $n=2$. That Wikipedia link shows exactly how to do that.
Oh ok!! Is everything else correct? Especially at questions 3 and 5 I am not really sure if I have done that correctly.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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 $\mathbb{K}[t]$.

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 $(fg)(a)=f(a)g(a)$
Hey mathmari !!

You have them the wrong way around.

We have to prove that $f(a)+g(a)\in U$.
You have already shown that $f(a)+g(a)=(f+g)(a)$ with $(f+g)\in\mathbb K[t]$.
Therefore $f(a)+g(a)$ is an element of $U$.

Same for the product.

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 $f(a)=0_{M_n(\mathbb{K})}$.

From $d-\lambda_i u_n$ the $i$-the row of the resulting matrix is contains only $0$.
You are assuming that $\lambda_i$ is in row $i$ of $d$, which is fine. We can indeed assume that.
Still, that leaves rows $k+1$ up to $n$. Which entries will those rows have on the diagonal?

#### mathmari

##### Well-known member
MHB Site Helper
You have them the wrong way around.

We have to prove that $f(a)+g(a)\in U$.
You have already shown that $f(a)+g(a)=(f+g)(a)$ with $(f+g)\in\mathbb K[t]$.
Therefore $f(a)+g(a)$ is an element of $U$.

Same for the product.
So we have the following:

$\forall f(a),g(a)\in U : f(a)+g(a)\in U$, da $f(a)+g(a)=(f+g)(a)$ mit $(f+g)\in\mathbb K[t]$
$\forall f(a),g(a)\in U : f(a)g(a)\in U$, da $f(a)g(a)=(fg)(a)$ mit $(fg)\in\mathbb K[t]$

So that $U$ is a subring we have to show also that $-f(a)\in U$, or not? But how?

You are assuming that $\lambda_i$ is in row $i$ of $d$, which is fine. We can indeed assume that.
Still, that leaves rows $k+1$ up to $n$. Which entries will those rows have on the diagonal?
So some eigenvalues must have a multiplicity of $\geq 1$, or not?

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
So we have the following:

$\forall f(a),g(a)\in U : f(a)+g(a)\in U$, da $f(a)+g(a)=(f+g)(a)$ mit $(f+g)\in\mathbb K[t]$
$\forall f(a),g(a)\in U : f(a)g(a)\in U$, da $f(a)g(a)=(fg)(a)$ mit $(fg)\in\mathbb K[t]$

So that $U$ is a subring we have to show also that $-f(a)\in U$, or not? But how?
It's already covered although I have to admit you've done so in a fashion I haven't seen before.
You proved that $f(a)g(a)\in U$ and that $-1_{\mathbb K[t]}\in U$. Therefore $-1_{\mathbb K[t]} \cdot f(a) = -f(a)\in U$.

So some eigenvalues must have a multiplicity of $\geq 1$, or not?
If $k<n$ then yes, there must be duplicate eigenvalues.

#### mathmari

##### Well-known member
MHB Site Helper
It's already covered although I have to admit you've done so in a fashion I haven't seen before.
You proved that $f(a)g(a)\in U$ and that $-1_{\mathbb K[t]}\in U$. Therefore $-1_{\mathbb K[t]} \cdot f(a) = -f(a)\in U$.
Therefore, for $U:=\left \{f(a)\mid f\in \mathbb{K}[t]\right \}\subseteq \mathbb{K}[t]$ teh following hold:
- $\forall f(a),g(a)\in U : f(a)+g(a)\in U$, since $f(a)+g(a)=(f+g)(a)$ with $(f+g)\in\mathbb K[t]$.
- $\forall f(a),g(a)\in U : f(a)g(a)\in U$, since $f(a)g(a)=(fg)(a)$ with $(fg)\in\mathbb K[t]$
- It is $1_{\mathbb{K}[t]}:=u_n=\text{id}(u_n)\in U$ and so $-1_{\mathbb{K}[t]}:=-u_n=\text{id}(-u_n)\in U$. So $\forall f(a)\in U : -1_{\mathbb K[t]} \cdot f(a) \in U \Rightarrow -f(a)\in U$.

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

Since it holds also that \begin{equation*}f(a)g(a)=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_ja^{\ell}=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}d_jc_ia^{\ell}=g(a)f(a)\end{equation*} it follows that $\left \{f(a)\mid f\in \mathbb{K}[t]\right \}$ is a commutative subring of $\mathbb{K}[t]$.

Is everything correct?

If $k<n$ then yes, there must be duplicate eigenvalues.
So is at my solution something missing? Do we have to consider the case $k<n$ alone?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Therefore, for $U:=\left \{f(a)\mid f\in \mathbb{K}[t]\right \}\subseteq \mathbb{K}[t]$ teh following hold:
- $\forall f(a),g(a)\in U : f(a)+g(a)\in U$, since $f(a)+g(a)=(f+g)(a)$ with $(f+g)\in\mathbb K[t]$.
- $\forall f(a),g(a)\in U : f(a)g(a)\in U$, since $f(a)g(a)=(fg)(a)$ with $(fg)\in\mathbb K[t]$
- It is $1_{\mathbb{K}[t]}:=u_n=\text{id}(u_n)\in U$ and so $-1_{\mathbb{K}[t]}:=-u_n=\text{id}(-u_n)\in U$. So $\forall f(a)\in U : -1_{\mathbb K[t]} \cdot f(a) \in U \Rightarrow -f(a)\in U$.

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

Since it holds also that \begin{equation*}f(a)g(a)=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}c_id_ja^{\ell}=\sum_{\ell=0}^{m+n}\sum_{\substack{i,j \\ i+j=\ell}}d_jc_ia^{\ell}=g(a)f(a)\end{equation*} 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 just realized that $f(a)$ is not an element of $\mathbb K[t]$. Instead it's an element of $\mathbb K[a]$, which luckily happens to be isomorphic.
We may want to mention that.

That is, $\left \{f(a)\mid f\in \mathbb{K}[t]\right \}$ is a commutative subring of $\mathbb{K}[a]$.
Since $\mathbb{K}[a]$ is isomorphic with $\mathbb{K}[t]$, it follows that it is also a commutative subring of $\mathbb{K}[t]$.

So is at my solution something missing? Do we have to consider the case $k<n$ alone?
From $d-\lambda_i u_n$ the $i$-the row of the resulting matrix is contains only $0$.
I'd rephrase this a bit.
Let's not assume that row $i$ has $\lambda_i$ on its diagonal.

The matrix $d-\lambda_i u_n$ has only zeroes on every row that corresponds to a $\lambda_i$ on the diagonal of $d$.

Then the rest of your argument should hold.

#### mathmari

##### Well-known member
MHB Site Helper
I have just realized that $f(a)$ is not an element of $\mathbb K[t]$. Instead it's an element of $\mathbb K[a]$, which luckily happens to be isomorphic.
We may want to mention that.

That is, $\left \{f(a)\mid f\in \mathbb{K}[t]\right \}$ is a commutative subring of $\mathbb{K}[a]$.
Since $\mathbb{K}[a]$ is isomorphic with $\mathbb{K}[t]$, it follows that it is also a commutative subring of $\mathbb{K}[t]$.
So we consider $\phi: \mathbb{K}[t] \rightarrow \mathbb{K}[a]$ with $f(t)\mapsto f(a)$ and we have to show that $\phi$ is bijective and $\phi (fg)=\phi(f)\phi(g)$, or not?