Finding the determinant of linear transformations from the space of $2 \times 2$ upper triangular matrices

Question

Find the determinant of the linear transformation $$ T(M) = \begin{bmatrix} 3 && -3 \\ 0 && 9 \end{bmatrix}M $$ from the space $V$ of upper triangular $2 \times 2$ matrices to $V$.

Answer 1

A basis for the space of $2\times2$ upper triangular matrices is given by $$e_1:=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad e_2:=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad e_3:=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ and a few simple calculations show that $$T(e_1)=3e_1,\qquad T(e_2)=3e_2,\qquad T(e_3)=-3e_2+9e_3.$$ So with respect to this basis the linear map $T$ is given by $$\begin{pmatrix}3&0&0\\0&3&-3\\0&0&9\end{pmatrix},$$ so the determinant of $T$ is $81$.

Answer 2

Here is another way that relies on the equality $\det T = \prod_k \lambda_k$.

The eigenvalues are solutions to $T(M) = \lambda M $, with $M \neq 0$.

If we write $T(M) = AM$, we are looking for solutions to $(\lambda I -A ) M = 0$. Expanding shows that we must have $\lambda \in \{3,9\}$, and substituting these values shows that $\ker (3I -T)$ has dimension two and $\ker (9I -T)$ has dimension one, hence $\det T = 3^2 9 = 81$.

Note that $\ker (3I -T) = \operatorname{sp} \{ \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix} \}$, and $\ker (9I -T) = \operatorname{sp} \{ \begin{bmatrix} 0 & 1 \\ 0 & -2\end{bmatrix} \}$.

This follows from Servaes' observation that if $(\lambda I -A) M = 0$ then the columns of $M$ must be eigenvectors of $A$.

Keep in mind that the dimensions are computed with $M$ being upper triangular.