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Let $V$ be the vector space of $2 \times 2$ matrices over a field $F$. Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in V$ and let $T : V \rightarrow V$ be the linear transformation defined by $T(X) = AX$. Compute $\det(T)$.

Can any one tell me if the answer would simply be the determinant of $A$ and, if not, how would you go about solving this problem?

Bernard
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Samantha Wyler
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2 Answers2

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First let me explain you what is the meaning of the determinant of a linear operator:

If $V$ is a finite-dimensional vector space, $n := \dim V$, and $T : V \to V$ is linear, in order to compute the determinant of $T$ choose any basis $v_1,v_2,\dots,v_n$ of $V$, and write each of the vectors $T(v_1),T(v_2),\dots,T(v_n)$ as a linear combination of the vectors $v_1,v_2,\dots,v_n$ like that: $$T(v_j) = a_{1j}v_1 + a_{2j}v_2 + \cdots + a_{nj}v_n \quad \textrm{for $j$ in $\{1,\dots,n\}$.}$$ Now, $$\det T = \det \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}.$$

In your specific example, do it by the easy way: choose $$v_1 = \begin{pmatrix} 1&0\\0&0\end{pmatrix}, \ v_2 = \begin{pmatrix} 0&1\\0&0\end{pmatrix}, \ v_3 = \begin{pmatrix} 0&0\\1&0\end{pmatrix} \textrm{ and } v_4 = \begin{pmatrix} 0&0\\0&1\end{pmatrix}.$$

azif00
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So using the $v_1, v_2, v_3, v_4$ that Azif$00$ suggested we get

$T(v_1) = \begin{pmatrix} a & 0 \\ c & 0 \end{pmatrix} = av_1 + 0v_2 + c v_3 + 0v_4 $,

$ T(v_2) = \begin{pmatrix} 0 & a \\ 0 & c \end{pmatrix} = 0v_1 + av_2 + 0 v_3 + cv_4$,

$T(v_3) = \begin{pmatrix} b & 0 \\ d & 0 \end{pmatrix} = bv_1 + 0v_2 + dv_3 + 0v_4$,

$T(v_4) = \begin{pmatrix} 0 & b \\ 0 & d \end{pmatrix} = 0v_1 + bv_2 + 0v_3 + dv_4$

and then det$(T) =$ det $\begin{pmatrix} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\0 & c & 0 & d \end{pmatrix} $

Samantha Wyler
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