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- Feb 5, 2012

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Here's another question I encountered recently. I am writing the question and my full solution. Many thanks if you can go through it and find a mistake, or confirm whether it's correct, or can contribute with any other useful comments.

**Problem:**

Find the matrix of a linear transformation \(f:M_{2}(\mathbb{R})\rightarrow M_{2}(\mathbb{R})\) given by \(f(X)=X^t\), with respect to a basis of matrix units \(E_{ij}\). Is \(f\) diagonalizable?

**My Solution:**

So first we shall represent the basis of matrix units by column vectors as follows.

\[\begin{pmatrix}1&0\\0&0\end{pmatrix}\rightarrow \begin{pmatrix}1\\0\\0\\0\end{pmatrix}\]

\[\begin{pmatrix}0&1\\0&0\end{pmatrix}\rightarrow \begin{pmatrix}0\\1\\0\\0\end{pmatrix}\]

\[\begin{pmatrix}0&0\\1&0\end{pmatrix}\rightarrow \begin{pmatrix}0\\0\\1\\0\end{pmatrix}\]

\[\begin{pmatrix}0&0\\0&1\end{pmatrix}\rightarrow \begin{pmatrix}0\\0\\0\\1\end{pmatrix}\]

Now we have,

\[f\begin{pmatrix}1\\0\\0\\0\end{pmatrix}= \begin{pmatrix}1\\0\\0\\0\end{pmatrix}\]

\[f\begin{pmatrix}0\\1\\0\\0\end{pmatrix}= \begin{pmatrix}0\\0\\1\\0\end{pmatrix}\]

\[f\begin{pmatrix}0\\0\\1\\0\end{pmatrix}= \begin{pmatrix}0\\1\\0\\0\end{pmatrix}\]

\[f\begin{pmatrix}0\\0\\0\\1\end{pmatrix}= \begin{pmatrix}0\\0\\0\\1\end{pmatrix}\]

Hence the matrix of linear transformation is,

\[M=\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\]

I have basically used the method outlined in the Wikipedia article.

Now to find whether \(M\) is diagonalizable we shall check whether it has two four linearly independent eigenvectors. Unfortunately this has only one linearly independent eigenvector (dimension of eigenspace is one). Therefore \(M\) is not diagonalizable.