Could this linear algebra proof be done without computation? From page 95 of Hoffman & Kunze's Linear algebra:

Let $T$ be the linear operator on $\mathbb{R}^2$ defined by

$T(x_1,x_2)=(-x_2,x_1)$

Prove that if $B$ is any ordered basis for $\mathbb{R}^2$ and $[T]_B=A$, then $A_{12}A_{21}\neq0.$

My approach was as follows: First find the matrix of $T$ relative to the standard ordered basis for $\mathbb{R}^2$:
$$A=\begin{bmatrix}
0 & -1\\
1 & 0\\ 
\end{bmatrix}$$
Then suppose that $B\prime$ is any other basis for $\mathbb{R}^2$ of the form $B\prime=\{\alpha_1,\alpha_2\}$, where $\alpha_1=(a,b), \alpha_2=(c,d)$. There exists a matrix $P$ such that for a given vector $\alpha$, $[\alpha]_B=P[\alpha]_{B\prime}$ Specifically,
$$P=\begin{bmatrix}
a & c\\
b & d\\ 
\end{bmatrix}$$
The matrix of $T$ in $B\prime$ is computed as follows:
$$
A=[T]_{B\prime}=P^{-1}[T]_BP=\frac{1}{ad-bc}\begin{bmatrix}
ab-bd & d^2+bc\\
a^2-bc & ac+cd\\ 
\end{bmatrix}
$$
In order for $A_{12}A_{21}=0$, it must be the case that
$$
(d^2+bc)(a^2-bc)=(ad)^2-(bc)^2+bc(a^2-d^2)=0
$$
By case analysis, it can be shown that any basis satisfying this condition does not span $\mathbb{R}^2$, contradicting the assumption that $B\prime$ spans $\mathbb{R}^2$.
My problem is the following: This proof seems very tedious. I feel that I am missing some insight that would lead to a much more elegant proof. Is this the case, and if so, what am I missing? I would also appreciate any feedback on my exposition - I am new to this and in high school so I don't really have anyone to get feedback from. Thanks.
 A: Hint: if either $A_{1,2} =0 $ or $A_{2,1}=0$ then $A$ is triangular.  The eigenvalues of $A$ are its diagonal elements and are real numbers.
But the eigenvalues of $A$ are roots of $X^2+1=0$.
A: Suppose one of $A_{12}$, $A_{21}$ is zero. Can you reason that $T$ has an eigenvector in this case?
A: You could note that $v\bullet Tv=0$ for any $v$; if $A_{12}=0$ then applying this fact with $v$ being the first basis vector will yield $A_{11}\|v\|^2=0$.  Since $v$ is part of a basis, it's not zero; so $A_{11}$ is zero, so $A$ has a zero row, which is inconsistent with the definition of $T$.  Similarly if $A_{21}=0$.
(Qualitatively, $T$ is a rotation, but triangular matrices represent shears.  The problem is just to find a concrete property that one has that the other doesn't.)
A: This answer uses the machinery developed by the authors before posing the question.
Let $\mathscr{F}$ be the standard ordered basis $\{\epsilon_1 , \epsilon_2\}$ and $\mathscr{B} '=\{(x_1,x_2),(y_1,y_2)\}=\{\alpha_1' ,\alpha_2'\}.$
Since $\mathscr{B} '$ is an ordered basis $x_1y_2-x_2y_1 \ne 0.$
If $\alpha \in \mathbb{R^2}$
$$[\alpha]_{\mathscr{F}}=\begin{bmatrix}x_1 & y_1\\x_2 & y_2\end{bmatrix}[\alpha]_{\mathscr{B'}}$$
$$\implies [\alpha]_{\mathscr{B'}}=\frac{1}{x_1y_2-x_2y_1}\begin{bmatrix}y_2 & -y_1\\-x_2 & x_1\end{bmatrix}[\alpha]_{\mathscr{F}}.$$
Hence 
$$[T\alpha_{1}^{'}]_{\mathscr{B'}}=\frac{1}{x_1y_2-x_2y_1}\begin{bmatrix}y_2 & -y_1\\-x_2 & x_1\end{bmatrix}\begin{bmatrix}-x_2 \\ x_1\end{bmatrix}=\frac{1}{x_1y_2-x_2y_1}\begin{bmatrix}-x_2y_2-x_1y_1 \\ x_2^2 + x_1^2\end{bmatrix}.$$
Also 
$$[T\alpha_{2}^{'}]_{\mathscr{B'}}=\frac{1}{x_1y_2-x_2y_1}\begin{bmatrix}y_2 & -y_1\\-x_2 & x_1\end{bmatrix}\begin{bmatrix}-y_2 \\ y_1\end{bmatrix}=\frac{1}{x_1y_2-x_2y_1}\begin{bmatrix}-y_2^2 - y_1^2 \\ x_2y_2 +x_1y_1\end{bmatrix}.$$
Thus we have $$A = [T]_{\mathscr{B'}}=\frac{1}{x_1y_2-x_2y_1}\begin{bmatrix}-x_2y_2-x_1y_1 & -y_2^2 - y_1^2 \\ x_2^2 + x_1^2 &  x_2y_2 +x_1y_1 \end{bmatrix}$$
It can be easily seen that $A_{12}A_{21}\ne 0.$
