I'm given a vector space $V = M_{2\times2}(\mathbb{R})$ and a function $T(A) = 2A^t - A$ and told to find the determinant and characteristic polynomial for T.

I found that $T\begin{pmatrix}a&b \\ c&d\end{pmatrix} = \begin{pmatrix}a&2c-b \\ 2b-c & d \end{pmatrix}$, but I'm having a hard time finding a matrix to represent T. Is it possible for $T$ to be a $2\times2$ matrix or would I have to represent $A$ as a $4\times1$ vector $(a,b,c,d)^t$ and make T a $4\times4$ matrix?

  • 1
    $\begingroup$ "make $T$ a $4\times 4$ matrix". That's almost it. You have to represent it as such. For this, compute $T\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $T\begin{pmatrix}0&1\\0&0\end{pmatrix}$,$T\begin{pmatrix}0&0\\1&0\end{pmatrix}$,$T\begin{pmatrix}0&0\\0&1\end{pmatrix}$ and see which coefficients they have in the basis $\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix},\begin{pmatrix}0&0\\1&0\end{pmatrix},\begin{pmatrix}0&0\\0&1\end{pmatrix}\right\}$, respectively. For each one of them put the coefficients into a column. These 4 columns will be the representing matrix. $\endgroup$
    – amsmath
    Oct 23 '19 at 14:56
  • $\begingroup$ @amsmath so using what you said, I found the transformation of all the basis matrices and as a result got $[T] = \begin{pmatrix} 1&0&0&0 \\ 0&-1&2&0 \\ 0&2&-1&0 \\ 0&0&0&1\end{pmatrix}$ with $det([T]) = -3$. Does this seem like the right idea? $\endgroup$
    – Jon D.
    Oct 23 '19 at 15:11
  • $\begingroup$ Yes, that's it. Very good, dude! Now, compute the char. polynomial. $\endgroup$
    – amsmath
    Oct 23 '19 at 15:33
  • $\begingroup$ You should find that the characteristic polynomial is $p(x) = (x-1)^3(x+3)$. $\endgroup$ Oct 23 '19 at 15:59

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