How Do Polynomial Functions Transform When Applied to Matrices?

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    Polynomials
In summary, this conversation discusses the definition of a matrix function and its properties, as well as the Cayley-Hamilton theorem. It also includes a question about verifying the Cayley-Hamilton theorem for specific cases.
  • #1
mathmari
Gold Member
MHB
5,049
7
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? :unsure: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? :unsure: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? :unsure: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? :unsure:
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
mathmari said:
Question 2: Let $n\leq 2$. Show that $P_a(a)=0_{M_n(\mathbb{K})}$.
Could you give me a hint for that? :unsure:
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.
 
  • #3
Opalg said:
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. :unsure:
 
  • #4
mathmari said:
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. (Worried)

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.

mathmari said:
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? (Sweating)
 
  • #5
Klaas van Aarsen said:
You have them the wrong way around. (Worried)

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? :unsure:

Klaas van Aarsen said:
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? (Sweating)

So some eigenvalues must have a multiplicity of $\geq 1$, or not? :unsure:
 
Last edited by a moderator:
  • #6
mathmari said:
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 $-1_{\mathbb K[t]} \cdot f(a) = -f(a)\in U$. 🧐

mathmari said:
So some eigenvalues must have a multiplicity of $\geq 1$, or not?
If $k<n$ then yes, there must be duplicate eigenvalues. 🤔
 
  • #7
Klaas van Aarsen said:
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? :unsure:


Klaas van Aarsen said:
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? :unsure:
 
  • #8
mathmari said:
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 $\mathbb{K}[t]$. 🧐

mathmari said:
So is at my solution something missing? Do we have to consider the case $k<n$ alone?

mathmari said:
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.
Instead we can say:

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. 🤔
 
  • #9
Klaas van Aarsen said:
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? :unsure:
 

FAQ: How Do Polynomial Functions Transform When Applied to Matrices?

What are polynomials at a equal to 0?

Polynomials at a equal to 0 are equations in the form of ax^n + bx^(n-1) + ... + cx + d, where a, b, c, and d are constants and x is a variable. When the value of x is set to 0, the polynomial becomes a constant and is equal to d.

What is the degree of a polynomial at a equal to 0?

The degree of a polynomial at a equal to 0 is the highest exponent of the variable x. In other words, it is the highest power of x in the polynomial.

What is the value of a polynomial at a equal to 0?

The value of a polynomial at a equal to 0 is the constant term, or the value of d in the polynomial equation. This is because when x is set to 0, all terms with x disappear and only the constant term remains.

What is the graph of a polynomial at a equal to 0?

The graph of a polynomial at a equal to 0 is a horizontal line passing through the y-axis at the value of d. This is because when x is set to 0, the polynomial becomes a constant and the graph is a straight line.

How do you solve a polynomial at a equal to 0?

To solve a polynomial at a equal to 0, simply set the value of x to 0 and solve for the constant term, d. This can be done by simplifying the polynomial equation and isolating the constant term on one side of the equal sign.

Similar threads

Replies
5
Views
1K
Replies
15
Views
1K
Replies
2
Views
2K
Replies
52
Views
3K
Replies
4
Views
1K
Replies
19
Views
2K
Replies
16
Views
1K
Replies
8
Views
3K
Back
Top