# Dimension of a Linear Transformation Matrix

• I
patric44
TL;DR 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

Mentor
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.

patric44
Mentor
2021 Award
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
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

Mentor
2021 Award
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*}

Last edited:
patric44