Welcome to our community

Be a part of something great, join today!

Relations between map and matrix

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Hey!! 😊

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

For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by \begin{equation*}\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac\end{equation*}

I have show that $\mu_a$ is a linear operator on $\mathbb{K}$-vector space $\mathbb{K}^{m\times n}$ :

Let $\lambda \in \mathbb{K}$, $c, c_1, c_2\in \mathbb{K}^{m\times n}$.

$\mu_a$ is homogeneous : \begin{equation*}\mu_a\left (\lambda c\right )=a\left (\lambda c\right )=\lambda \left (ac\right )=\lambda \mu_a(c)\end{equation*}
$\mu_a$ is additive : \begin{equation*}\mu_a\left (c_1+c_2\right )=a\left (c_1+c_2\right )=\lambda c_1+\lambda c_2= \mu_a(c_1)+ \mu_a(c_2)\end{equation*}

Next I want to show that $\text{Sp}(\mu_a)=n\text{Sp}(a)$, $\det (\mu_a)=\det (a)^n$ and $P_{\mu_a}=P_a^n$, and that if $a$ is diagonalizable then $\mu_a$ is diagonalizable. ($P$ is the characteristic polynomial.)


The spectrum is the set of eigenvalues. What is meant by $\text{Sp}(\mu_a)=n\text{Sp}(a)$ ? The eigenvalues of the map $\mu_a$ are the same as the eigenvalues of the matrix of the map, or not? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
$\mu_a$ is additive : ... $a\left (c_1+c_2\right )=\lambda c_1+\lambda c_2$
Hey mathmari !!

That's not true is it? (Worried)

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

For $a\in M_m(\mathbb{K})$ we consider the map $\mu_a$ that is defined by \begin{equation*}\mu_a: \mathbb{K}^{m\times n}\rightarrow \mathbb{K}^{m\times n}, \ c\mapsto ac\end{equation*}

The spectrum is the set of eigenvalues. What is meant by $\text{Sp}(\mu_a)=n\text{Sp}(a)$ ? The eigenvalues of the map $\mu_a$ are the same as the eigenvalues of the matrix of the map, or not?
Let's take a look at an example. :geek:

Suppose $m=n=2$ and $a=\begin{pmatrix}2&0\\0&2\end{pmatrix}$.
We find it's eigenvalues and eigenvectors by solving $av=\lambda v$ with $v\in \mathbb K^m$.
What are the eigenvalues and eigenvectors of $a$? 🤔

Can we solve $\mu_a(c)=ac=\lambda c$ with $c\in \mathbb{K}^{m\times n}$ as well and find its eigenvalues and eigenmatrices? :unsure:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
That's not true is it? (Worried)
I meant
\begin{equation*}\mu_a\left (c_1+c_2\right )=a\left (c_1+c_2\right )=ac_1+a c_2= \mu_a(c_1)+ \mu_a(c_2)\end{equation*} Is that wrong? :unsure:


Let's take a look at an example. :geek:

Suppose $m=n=2$ and $a=\begin{pmatrix}2&0\\0&2\end{pmatrix}$.
We find it's eigenvalues and eigenvectors by solving $av=\lambda v$ with $v\in \mathbb K^m$.
What are the eigenvalues and eigenvectors of $a$? 🤔

Can we solve $\mu_a(c)=ac=\lambda c$ with $c\in \mathbb{K}^{m\times n}$ as well and find its eigenvalues and eigenmatrices? :unsure:
I got stuck right now. Why do we get the eigenvalues and eigenmatrices by $\mu_a(c)=ac=\lambda c$ ? Could you explain that further to me? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
I meant
\begin{equation*}\mu_a\left (c_1+c_2\right )=a\left (c_1+c_2\right )=ac_1+a c_2= \mu_a(c_1)+ \mu_a(c_2)\end{equation*} Is that wrong?
Nope. It's all correct. Must have been a typo. (Bandit)

I got stuck right now. Why do we get the eigenvalues and eigenmatrices by $\mu_a(c)=ac=\lambda c$ ? Could you explain that further to me?
Let's go back to the formal definition:


If $T$ is a linear transformation from a vector space $V$ over a field $F$ into itself and $\mathbf v$ is a nonzero vector in $V$, then $\mathbf v$ is an eigenvector of $T$ if $T\mathbf v$ is a scalar multiple of $\mathbf v$. This can be written as
$$T(\mathbf{v}) = \lambda \mathbf{v},$$
where $\lambda$ is a scalar in $F$, known as the eigenvalue, characteristic value, or characteristic root associated with $\mathbf v$.
🧐

We have $T=\mu_a$, $V=\mathbb K^{m\times n}$, and $\mathbf v = c$, don't we?
Doesn't that mean that we can find the eigenvalues $\lambda$ and eigenvectors $c$ by solving $\mu_a(c)=\lambda c$? 🤔
 
Last edited:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Let's go back to the formal definition:


If $T$ is a linear transformation from a vector space $V$ over a field $F$ into itself and $\mathbf v$ is a nonzero vector in $V$, then $\mathbf v$ is an eigenvector of $T$ if $T\mathbf v$ is a scalar multiple of $\mathbf v$. This can be written as
$$T(\mathbf{v}) = \lambda \mathbf{v},$$
where $\lambda$ is a scalar in $F$, known as the eigenvalue, characteristic value, or characteristic root associated with $\mathbf v$.
🧐

We have $T=\mu_a$, $V=\mathbb K^{m\times n}$, and $\mathbf v = c$, don't we?
Doesn't that mean that we can find the eigenvalues $\lambda$ and eigenvectors $c$ by solving $\mu_a(c)=\lambda c$? 🤔
So from $\mu_a(c)=\lambda c$ we get $ac=\lambda c$, or not? :unsure:
But what does this mean? That $\lambda$ is also an eigenvalue of $a$ with eigenmatrix $c$ ? :unsure:
 
Last edited:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
We have that $Sp(a)$ is the trace, so it is equal to the sum of the eigenvalues counted with multiplicities.
Does this mean that at $\mu_a$ each eigenvalue $\lambda$ has the multiplicity $n$ ? :unsure:

Do we maybe have the following?

From $\mu_a(c)=\lambda c$ we get $ac=\lambda c$.
So if $\lambda$ is an eigenvalue of $\mu_a$, tthere is a non-zero $c\in\mathbb{K}^{m\times n}$ with $\mu_a(c)=\lambda c$.
The columns of $c$ are all eigenvectors of $a$ with eigenvalue $\lambda$.
The matrix $c$ has $n$ columns.

So for each eigenvalue $\lambda$ of $a$ there are $n$ eigenvectors, so the multiplicity of $\lambda$ is $n$.
The trace of a matrix is the sum of teh eigenvalues considering the multiplicity.
Since each eigenvalue of $\mu_a$ has a multiplicity of $n$, it follows that $\text{Sp}(\mu_a)=\sum_i n\cdot \lambda_i=n\cdot \sum_i\lambda $.
Since $\lambda_i$ is the eigenvalue of $a$, it follows that $\text{Sp}(a)=\sum_i\lambda_i$.
Therefore we get $\text{Sp}(\mu_a)=n\cdot \text{Sp}(a)$.

Is everything correct? :unsure:


If the above is correct, then we get in a similar way the relation about the determinant:

The determinant is equal to tthe product of the eigenvalues.
Since each eigenvalue of $\mu_a$ has a multiplicity of $n$, it follows that $\det(\mu_a)=\left (\prod_i \lambda_i\right )^n $.
Since $\lambda_i$ are the eigenvalues of $a$, it follows that $\det(a)=\prod_i\lambda_i$.
Therefore we get $\det(\mu_a)=\det(a)^n$.

:unsure:
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
We have that $Sp(a)$ is the trace, so it is equal to the sum of the eigenvalues counted with multiplicities.
Does this mean that at $\mu_a$ each eigenvalue $\lambda$ has the multiplicity $n$ ?
Not exactly.
It means that we basically have to prove that for each eigenvalue $\lambda$ of $a$ with algebraic multiplicity $i$, that $\lambda$ is also an eigenvalue of $\mu_a$ and that it has algebraic multiplicity $i\cdot n$. 🤔

From $\mu_a(c)=\lambda c$ we get $ac=\lambda c$.
So if $\lambda$ is an eigenvalue of $\mu_a$, tthere is a non-zero $c\in\mathbb{K}^{m\times n}$ with $\mu_a(c)=\lambda c$.
The columns of $c$ are all eigenvectors of $a$ with eigenvalue $\lambda$.
The matrix $c$ has $n$ columns.
Just to be clear: the $c$ we have here is an eigenvector of $\mu_a$. Let's call it an "eigenmatrix" to avoid confusion.
The columns of $c$ are not eigenvectors of $\mu_a$, but instead they are eigenvectors of $a$.

So for each eigenvalue $\lambda$ of $a$ there are $n$ eigenvectors, so the multiplicity of $\lambda$ is $n$.
This is not exactly true.
Again we need to be careful when talking about eigenvectors. Are they eigenvectors of $a$ or eigenvectors of $\mu_a$?
More specifically we have the following.
For each eigenvalue $\lambda$ of $a$, the matrix $a$ has up to $m$ eigenvectors.
For each eigenvalue $\lambda$ of $a$, the transformation $\mu_a$ has the same eigenvalue $\lambda$ with up to $m\cdot n$ eigenmatrices.
That is because an eigenmatric $c$ has $n$ columns. We can pick one of them to be an eigenvector of $a$ and set the other columns to zero.


Here's a different approach.
We can "unroll" $\mu_a$ into a regular matrix.
To do so we rewrite each matrix $c\in \mathbb K^{m\times n}$ as a vector in $\mathbb K^{mn}$ by writing each column below the previous column.
And we construct a new matrix $\tilde a$ that is a block matrix with $a$ repeated $n$ times along its diagonal.
Now we can find the eigenvalues and eigenvectors of $\tilde a$, and afterwards we can "roll" the eigenvectors back into eigenmatrices. 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Not exactly.
It means that we basically have to prove that for each eigenvalue $\lambda$ of $a$ with algebraic multiplicity $i$, that $\lambda$ is also an eigenvalue of $\mu_a$ and that it has algebraic multiplicity $i\cdot n$. 🤔
How could we do that? Is this because of the number of columns of $c$ ? :unsure:


Just to be clear: the $c$ we have here is an eigenvector of $\mu_a$. Let's call it an "eigenmatrix" to avoid confusion.
The columns of $c$ are not eigenvectors of $\mu_a$, but instead they are eigenvectors of $a$.

Again we need to be careful when talking about eigenvectors. Are they eigenvectors of $a$ or eigenvectors of $\mu_a$?
More specifically we have the following.
For each eigenvalue $\lambda$ of $a$, the matrix $a$ has up to $m$ eigenvectors.
For each eigenvalue $\lambda$ of $a$, the transformation $\mu_a$ has the same eigenvalue $\lambda$ with up to $m\cdot n$ eigenmatrices.
That is because an eigenmatric $c$ has $n$ columns. We can pick one of them to be an eigenvector of $a$ and set the other columns to zero.
Ahh ok!


Here's a different approach.
We can "unroll" $\mu_a$ into a regular matrix.
To do so we rewrite each matrix $c\in \mathbb K^{m\times n}$ as a vector in $\mathbb K^{mn}$ by writing each column below the previous column.
And we construct a new matrix $\tilde a$ that is a block matrix with $a$ repeated $n$ times along its diagonal.
Now we can find the eigenvalues and eigenvectors of $\tilde a$, and afterwards we can "roll" the eigenvectors back into eigenmatrices. 🤔
I haven't really understood this approach. Could you explain that further to me? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,736
I haven't really understood this approach. Could you explain that further to me?
Suppose we have $m=n=2$, $a=\begin{pmatrix}2&0\\0&3\end{pmatrix}$, and $c_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}$.
Then we have $ac_1=\begin{pmatrix}2&0\\0&3\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}2&0\\0&0\end{pmatrix}$ don't we? 🤔
So $c_1$ is an eigenmatrix of $\mu_a$.

We can also write it as: $\tilde a \tilde c_1=\begin{pmatrix}2&0\\0&3\\&&2&0\\&&0&3\end{pmatrix}\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=\begin{pmatrix}2\\0\\0\\0\end{pmatrix}$ can't we? Just by putting the columns of $c_1$ below each other, and by constructing a block matrix $\tilde a$.

Now we can see that $\mu_a$ has eigenvalue $\lambda=2$, which has indeed algebraic multiplicity $n=2$, and we can also find the eigenmatrices that belong to it, can't we? 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Suppose we have $m=n=2$, $a=\begin{pmatrix}2&0\\0&3\end{pmatrix}$, and $c_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}$.
Then we have $ac_1=\begin{pmatrix}2&0\\0&3\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}2&0\\0&0\end{pmatrix}$ don't we? 🤔
So $c_1$ is an eigenmatrix of $\mu_a$.

We can also write it as: $\tilde a \tilde c_1=\begin{pmatrix}2&0\\0&3\\&&2&0\\&&0&3\end{pmatrix}\begin{pmatrix}1\\0\\0\\0\end{pmatrix}=\begin{pmatrix}2\\0\\0\\0\end{pmatrix}$ can't we? Just by putting the columns of $c_1$ below each other, and by constructing a block matrix $\tilde a$.

Now we can see that $\mu_a$ has eigenvalue $\lambda=2$, which has indeed algebraic multiplicity $n=2$, and we can also find the eigenmatrices that belong to it, can't we? 🤔
Is that the better approach for that exercise? Or could we also use the one I started? :unsure:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,036
Can we do that as follows?

Let $\lambda$ be the eigenvalues of $\mu_a$ then $\mu_a(c)=\lambda c$. Then we get $ac=\lambda c$.
So if $\lambda$ is an eigenvalue of $\mu_a$ there is a non-zero $c\in\mathbb{K}^{m\times n}$ with $\mu_a(c)=\lambda c$.
The columns of $c$ are eigenvectors of $c$ with eigenvalue $\lambda$.
The matrix $c$ has $n$ columns.
$c$ is an eigenmatrix of $\mu_a$ and teh columns of $c$ are eigenvectors of $a$.
For each eigenvalue $\lambda$ of $a$, the matrix $a$ has up to $m$ eigenvectors.
For each eigenvalue $\lambda$ of $a$, the operator $\mu_a$ has the same eigenvelue $\lambda$ with up to $m\cdot $ eigenmatrices.
So for each eigenvalue $\lambda$ with algebraic multiplicity $i$,tjis $\lambda$ is also an eigenvalue of $\mu_a$ with algebraic mutiplicity $i\cdot n$.

The trace is $\text{Trace}(\mu_a)=\sum_j i\cdot n\cdot \lambda_j=n\cdot \sum_ji\cdot \lambda_j $ and $\text{Trace}(a)=\sum_j i\cdot \lambda_j$.
So we get $\text{Trace}(\mu_a)=n\cdot \text{Trace}(a)$.



The determinant is equal to the product of eigenvalues.
So we have $\det(\mu_a)=\left (\prod_j \lambda_j\right )^{in}=\left (\prod_j \lambda_j^i\right )^{n} $ and $\det(a)=\prod_j\lambda_j^i$. Therefore we get $\det(\mu_a)=\det(a)^n$.


The characteristic polynomial of $a$ is $P_a=\prod_j \left (\lambda -\lambda_j\right )^i$, where $i$ is the algebraic multiplicity if the eigenvalues.
The characteristic polynomial of $\mu_a$ is $P_{\mu_a}=\prod_j \left (\lambda -\lambda_j\right )^{in}=\left (\prod_j \left (\lambda -\lambda_j\right )^{i}\right )^n$.
So we get $P_{\mu_a}=P_a^n$.



For the last question:
$a$ is diagonalizable. So the geometric multiplicity equals the algebraic one. In this case $P_a$ splits. The same holds also for $\mu_a$ due to the above results. And so it follows that $\mu_a$ is also diagonalizable.



:unsure: