Could this linear algebra proof be done without computation?

Question

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.

Answer 1

Suppose one of $A_{12}$, $A_{21}$ is zero. Can you reason that $T$ has an eigenvector in this case?

Answer 2

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 $X^2+1=0$.

Answer 3

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.)