Suppose we have a vector space of 2x2 matrices with a basis { E11, E12, E21, E22] with Eij = a 2x2 matrix with a 1 in the i-th row and the j-th column.

In this lineair space, the transformation T : R2x2 -> R2x2 is defined by T(A) = AT

Define the matrix of T in respect to the given basis.

Now I'm struggling with that. T transposes a matrix so do I need to express it in terms of Eij or...? And if not what's the point of having that basis so excplicitely written down?

I also need the eigenvalues and eigenvectors of this transformation but without calculating. So even if I got the matrix, I wouldn't know how to extract eigenvalues unless it would be a lower triangular matrix.

A solution on a similar question was


Yet this is a 4x4 matrix so how could it be applied to 2x2 matrices? And more important. How do I extract the eigenvalues of this by reasoning.

  • $\begingroup$ Yes, you need to express $T$ in the basis $E_{i,j}$. $\endgroup$ – xbh Sep 5 '18 at 17:21
  • $\begingroup$ For eigenvalues, try definition, since you clearly knew what happen when you apply $T$ to matrices. $\endgroup$ – xbh Sep 5 '18 at 17:23
  • 1
    $\begingroup$ P.S. We recommend users to use MathJax to write math expressions. Here is the tutorial. $\endgroup$ – xbh Sep 5 '18 at 17:28

To write down the matrix of a linear transformation with respect to a given basis you have to see how the transformation acts on the base of the domain and write down the result as a linear combination of the basis of the image. Words! Here's a similar question where a gave better insights on how to solve this kind of problems.

Now let's see how does it work in your case: let us evaluate how the map acts on our base, remember that we have $$T: U\equiv \mathbb R^{2\times 2}\longrightarrow \mathbb R^{2\times2}\equiv V$$ and for both $U$ and $V$ we have the same base, the standard base $E_{ij}$. Now $$T\left(E_{11}\right) = E_{11}^T = E_{11} = \overbrace{\color{red}{1}E_{11}+ \color{blue}{0}E_{12}+\color{orange}{0}E_{21}+\color{green}{0}E_{22}}^{\text{Linear combination of base of }V}\\[5pt] T\left(E_{12}\right) = E_{12}^T = E_{21} =\color{red}{0}E_{11}+ \color{blue}{0}E_{12}+\color{orange}{1}E_{21}+\color{green}{0}E_{22}\\[5pt] T\left(E_{21}\right) = E_{21}^T = E_{12}=\color{red}{0}E_{11}+ \color{blue}{1}E_{12}+\color{orange}{0}E_{21}+\color{green}{0}E_{22}\\[5pt] T\left(E_{22}\right) = E_{22}^T = E_{22}=\color{red}{0}E_{11}+ \color{blue}{0}E_{12}+\color{orange}{0}E_{21}+\color{green}{1}E_{22} $$ remember that all that transposition does is to invert the indices $(a_{ij})^T = a_{ji}$. As you can see I wrote down the linear combination in the base of the image for every element mapped with $T$. The matrix is now to be written down as $$\begin{align}[T]=&\left(\begin{matrix}\color{red}{1}&\color{red}{0}&\color{red}{0}&\color{red}{0}\\ \color{blue}{0}& \color{blue}{0}& \color{blue}{1}& \color{blue}{0}\\ \color{orange}{0}&\color{orange}{1}&\color{orange}{0}&\color{orange}{0} \\ \color{green}{0}&\color{green}{0}&\color{green}{0}&\color{green}{1}\end{matrix}\right)\end{align}$$

So this is the matrix of the transformation! You could be disoriented by the fact that, by definition $$T(A) = [T]v$$ how is it possible that, if $v$ is a matrix, we have to do a $4\times4$ matrix times a $2\times 2$ matrix? That is because there's a little problem: $v$ is not a matrix! $v$ is a vector that contains the coefficients of the linear combination in a given base for a given element of $\mathbb R^{2\times 2}$. For example, if a wanted to evaluate $T$ on the identity matrix I should first evaluate the linear combination in the basis for the identity matrix which is $$v_{I} = (1,0,0,1) = 1E_{11}+0E_{12}+0E_{21}+1E_{22} \\[10pt] \implies T(I) = [T]v_{I} = \left(\begin{matrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{matrix}\right)\left(\begin{matrix}1\\0\\0\\1\end{matrix}\right) = \left(\begin{matrix}1&0\\0&1\end{matrix}\right)$$ as expected.

The eigenvectors to this transformations are clearly the matrices such that $$T(A) = A^T = \lambda A$$ which, as a trivial case, with $\lambda=1$ are symmetric matrices. Note that $T^2=I$ because $(A^T)^T = A$ so if $$T(A) = A^T = \lambda A$$ and $$T(A^T) = T(\lambda A) = \lambda T(A) = \lambda^2 A = I $$


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