# Relations between map and matrix

• MHB
• mathmari
In summary, the conversation discusses the definition of a linear operator $\mu_a$ on a vector space $\mathbb{K}^{m\times n}$ defined by the map $\mu_a(c) = ac$. It is shown that $\mu_a$ is a linear operator by proving its homogeneity and additivity. The conversation then moves on to discussing the spectrum of $\mu_a$, which is the set of eigenvalues. It is mentioned that the eigenvalues of $\mu_a$ are the same as the eigenvalues of the matrix $a$ that defines the map. The conversation also discusses the relationship between the eigenvalues and eigenvectors of $\mu_a$ and $a$, and how they can be found by solving $\ mathmari Gold Member MHB 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: mathmari said:$\mu_a$is additive : ...$a\left (c_1+c_2\right )=\lambda c_1+\lambda c_2$Hey mathmari! That's not true is it? (Worried) mathmari said: 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: Klaas van Aarsen said: 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: Klaas van Aarsen said: 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: mathmari said: 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) mathmari said: 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: Klaas van Aarsen said: 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 by a moderator: 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$. :unsure: Last edited by a moderator: mathmari said: We have that$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$. mathmari said: From$\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$. mathmari said: So for each eigenvalue$\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$, and afterwards we can "roll" the eigenvectors back into eigenmatrices. Klaas van Aarsen said: Not exactly. It means that we basically have to prove that for each eigenvalue$\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$? :unsure: Klaas van Aarsen said: Just to be clear: the$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$and set the other columns to zero. Ahh ok! Klaas van Aarsen said: Here's a different approach. We can "unroll"$\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$, and afterwards we can "roll" the eigenvectors back into eigenmatrices. I haven't really understood this approach. Could you explain that further to me? :unsure: mathmari said: I haven't really understood this approach. Could you explain that further to me? Suppose we have$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$, and we can also find the eigenmatrices that belong to it, can't we? Klaas van Aarsen said: Suppose we have$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$, and we can also find the eigenmatrices that belong to it, can't we? Is that the better approach for that exercise? Or could we also use the one I started? :unsure: Can we do that as follows? Let$\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:

## 1. What is the difference between a map and a matrix?

A map is a visual representation of data or information, typically in the form of a diagram or chart. It shows the relationships between different elements or variables. On the other hand, a matrix is a mathematical structure that contains rows and columns of numbers or symbols. It is often used to represent data or perform calculations.

## 2. How are maps and matrices related?

Maps and matrices are related in that they both represent data or information. A map can be seen as a visual representation of a matrix, with each element or variable in the matrix corresponding to a specific point or area on the map. Additionally, matrices can be used to perform calculations and analyze data that is presented in a map format.

## 3. Can a map be converted into a matrix?

Yes, a map can be converted into a matrix by assigning numerical values to the different elements or variables on the map. This allows for the data to be represented in a matrix format, which can then be used for calculations and analysis.

## 4. How do maps and matrices help in data analysis?

Maps and matrices are useful tools in data analysis as they allow for the visualization and organization of data. Maps can help identify patterns and relationships between different elements, while matrices can be used for calculations and statistical analysis.

## 5. What are some real-world applications of maps and matrices?

Maps and matrices have a wide range of applications in various fields, such as geography, economics, and computer science. They are commonly used in data analysis, risk assessment, network analysis, and image processing. For example, maps and matrices can be used to analyze population distribution, track stock market trends, and detect patterns in social media networks.

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