# Matrix of a transformation in $M_2(\Bbb R)$ space

I'm working on a question and am looking for some advice:

$$T: M_{2}(\Bbb R) \to M_{2}(\Bbb R),\\ T\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} = A\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}$$

where $$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$.

Find the transformation matrix $$T$$ with respect to the basis:

$$e_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, e_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, e_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, e_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$

I have transformed each of the basis vectors according to the given transformation. Then expressed each as a linear combination of the basis vectors. Organizing the coefficients into a matrix I get

$$\begin{pmatrix} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \end{pmatrix}$$

The only way I can see this being correct is if a vector we wish to transform was written as a column vector instead of a $$2\times 2$$ matrix in order to satisfy matrix multiplication rules. This will then produce another column vector, rather than a $$2\times 2$$ matrix.

Does this seem correct? I don't seem to have any problems when we are working in $$\Bbb R^{n}$$, only with this $$M_{2}(\Bbb R)$$ space.

Thanks

The only way I can see this being correct is if a vector we wish to transform was written as a column vector instead of a $$2\times 2$$ matrix in order to satisfy matrix multiplication rules. This will then produce another column vector, rather than a $$2\times 2$$ matrix.
This is fine, and indeed, it's to be expected. If you have a linear transformation $$T : V \to W$$, and find a matrix representation $$M$$ of $$T$$ with respect to bases $$B$$ and $$C$$ for $$V$$ and $$W$$ respectively, then for any $$v \in V$$, $$[Tv]_C = M[v]_B.$$ This is the way that matrices for linear transformations are supposed to work: if you transform a vector into a coordinate column vector, and multiply it by the matrix for the transformation, then the resulting column vector is a coordinate vector with respect to the other basis.
In this case, our spaces $$V$$ and $$W$$ are both the space of $$2 \times 2$$ matrices. These are $$4$$-dimensional spaces, so any matrix for $$T$$ must be $$4 \times 4$$. In order to use this matrix, we have to multiply it to $$4 \times 1$$ coordinate vectors with respect to the basis. Similarly, the result will be a $$4 \times 1$$ coordinate vector.
• Thanks so much Theo. Just to confirm, you mention: In order to use this matrix, we have to multiply it to 4×1 coordinate vectors with respect to the basis. Similarly, the result will be a 4×1 coordinate vector. Does this imply that my matrix T for the transformation, being a $4\times 4$ made up of the $4\times 1$ column vectors is correct with respect to the given basis? And therefore multiplying it by a $4\times 1$ column vector results in a $4\times 1$ column vector that can then be re-written as a $2\times 2$ matrix? Thanks again. The help is much appreciated. – koot Sep 25 '19 at 9:25
• Thanks again for the clarification. I just wasn't sure whether the output of a $4\times 1$ column vector was acceptable, and can then be re-written as a $2\times 2$ matrix, even though it works. Cheers – koot Sep 26 '19 at 1:08