I Dimension of a Linear Transformation Matrix

[patric44]

TL;DR

does the matrix of the linear transformation should have the same dimension of the vector space?

hi guys
I was trying to find the matrix of the following linear transformation with respect to the standard basis, which is defined as
$\phi\;M_{2}(R) \;to\;M_{2}(R)\;; \phi(A)=\mu_{2*2}*A_{2*2}$ ,
where $\mu = (1 -1;-2 2)$
and i found the matrix that corresponds to this linear transformation this 4*4 matrix , given by
$$\phi =(1\;0\;-2\;0;0\;1\;0-2;-1\;0\;2\;0;0 -1\;0\;2) $$
is there is some wrong here or the matrix should be 4*4

 

[Mark44]

Summary:: does the matrix of the linear transformation should have the same dimension of the vector space?

hi guys
I was trying to find the matrix of the following linear transformation with respect to the standard basis, which is defined as
$\phi\;M_{2}(R) \;to\;M_{2}(R)\;; \phi(A)=\mu_{2*2}*A_{2*2}$ ,
where $\mu = (1 -1;-2 2)$
and i found the matrix that corresponds to this linear transformation this 4*4 matrix , given by
$$\phi =(1\;0\;-2\;0;0\;1\;0-2;-1\;0\;2\;0;0 -1\;0\;2) $$
is there is some wrong here or the matrix should be 4*4

The matrix should be 4 x 4, since your transformation is a map from $M_2$ to itself. $M_2$, the space of 2 x 2 matrices, is of dimension 4, and any basis for this space will need to have 4 elements.

 

[fresh_42]

A linear transformation $\Phi: V\longrightarrow W$ of finite-dimensional vector spaces, say $\dim V = n$ and $\dim W = m$ has a matrix representation as an $m \times n$ matrix, $n$ columns and $m$ rows.

We have here $V=W=\mathbb{R}^2,$ the vector space of linear functions $\varphi : \mathbb{R}^2 \longrightarrow \mathbb{R}^2$. Since $\dim \mathbb{R}^2 = 2$, the dimension of $\mathbb{M}_2(\mathbb{R})$ is $2\cdot 2= 4.$

In case $V=W=\mathbb{M}_2(\mathbb{R})$ we get $\dim V = \dim W=\dim \mathbb{M}_2(\mathbb{R})=4,$ we have $\dim \mathbb{M}_4(\mathbb{R})=4\cdot 4=16$ and $\Phi$ is represented by a $4 \times 4$ matrix.

 

[patric44]

thanks guys i think i got it now, but i still have a question : how iam going to act with a 4*4 matrix on matrices that belong to $M_{2}(R)$ space

 

[fresh_42]

If you use them as a linear transformation of $\mathbb{R}^2$ then you simply consider the $2\times 2$ matrix.

If you want to deal with all those linear transformations and deal with $\mathbb{M}_2(\mathbb{R})$ as vector space instead, then you have to choose some ordered basis to make those matrices a linear vector. E.g. if $e_{ij}$ is the matrix with $1$ in the $i$-th row and $j$-th column, and zeros otherwise, then $(e_{11},e_{12},e_{21},e_{22})$ would be such a bases.

A $2 \times 2$ matrix would be
$\begin{bmatrix} 2&3\\4&5 \end{bmatrix}=2e_{11}+3e_{12}+4e_{21}+5e_{22}=(2,3,4,5)$

A linear transformation between those, e.g. $M_{\Phi}=\begin{bmatrix} 2&3&0&1\\0&0&0&5\\0&0&1&4\\0&0&0&0 \end{bmatrix}$
would map
\begin{align*}
e_{11}&\longmapsto 2e_{11}+3e_{12}+1e_{22}\\
e_{12}&\longmapsto 5e_{22}\\
e_{21}&\longmapsto 1e_{21}+4e_{22}\\
e_{22}&\longmapsto 0
\end{align*}Edit: This interpretation corresponds to $\vec{v} \longmapsto \vec{v}\cdot M_{\Phi}$. If we choose matrix multiplication from the left, i.e. $\vec{v} \longmapsto M_{\Phi}\cdot \vec{v}$ then
\begin{align*}
e_{11}&\longmapsto 2e_{11}\\
e_{12}&\longmapsto 3e_{11}\\
e_{21}&\longmapsto 1e_{21}\\
e_{22}&\longmapsto 1e_{11}+5e_{12}+4e_{21}
\end{align*}