# Reversing orders does not change eigen-spaces (for non-zero eigen-values)

2022-12-30

This is a proof I found when testing one of the examination questions for a course that I serve as a teaching assistant.

## Statement

Suppose $$A$$ is a $$4\times3$$ matrix and $$B$$ a $$3\times 4$$ matrix with real coefficients such that

$AB = \begin{pmatrix} 1 & -2 & -1 & -1 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ -2 & -2 & -1 & 2 \\ \end{pmatrix}.$

Then $$BA = 3I_{3\times 3}$$, where $$I_{3\times 3}$$ is the $$3\times 3$$ identity matrix.

## The proof the teacher had in mind

The teacher wanted the student to notice that, if the statement is true, then we would have the equality $ABAB = A (BA) B = 3AB,$ by the associativity of matrix products. So we would first multiply the given matrix $$AB$$ by itself to verify that it is indeed three times itself.

Then the next step is to verify that $$B$$ is surjective and $$A$$ is injective, so that we can cancel out the left $$A$$ and the right $$B$$ in the above equation to obtain that $$AB=3I_{3\times3}$$.

## An alternative proof

I found an alternative proof, which is the main subject of this post.

First we prove a lemma.

### Lemma

Let $$F$$ be a field, $$V_1, V_2$$ vector spaces over $$F$$. Let $$T_1: V_1\rightarrow V_2$$ and $$T_2: V_2\rightarrow V_1$$ be linear transformations.

For any $$\lambda\in F$$, and for $$i=1,2$$, define $$E_{i,\lambda}$$ as the subspace of $$V_i$$ consisting of elements $$x\in V_i$$ such that $$T_j \circ T_i (x)=\lambda x$$, where $$j = \begin{cases} 1 & i=2\\ 2 & i = 1\end{cases}$$.

If $$\lambda \neq 0$$ in $$F$$, then $$E_{1,\lambda}\cong E_{2,\lambda}$$.

Let $$\lambda\in F$$ be non-zero in the following.

Define a linear transformation $f:E_{1,\lambda} \rightarrow E_{2,\lambda}$ by sending $$v\in E_{1,\lambda}$$ to $$T_1(v)$$. We first verify that $$T_1(v)\in E_{2,\lambda}$$.

Since $$v\in E_{1,\lambda}$$, we know $$T_2 \circ T_1 (v) = \lambda v$$, and hence $$T_1 \circ T_2 (T_1 (v)) = \lambda T_1 (v)$$. This shows that $$T_1(v)\in E_{2,\lambda}$$.

Since $$T_1$$ is a linear transformation, so is $$f$$.

Moreover, if $$f(v) = 0$$, then $$T_1(v)=0$$. But $$v\in E_{1,\lambda}$$ implies that $v = \frac1\lambda \cdot \lambda v = \frac1\lambda T_2 (T_1 (v)) = \frac1\lambda 0 = 0.$ Therefore $$f$$ is an injective linear transformatoin.

Similarly, we have an injective linear transformation $$g: E_{2,\lambda} \rightarrow E_{1,\lambda}$$ sending $$v\in E_{2,\lambda}$$ to $$T_2 (v)$$.

Then we see that $f\circ g(v) = T_1 \circ T_2 (v) = \lambda v,\,\forall v\in E_{2,\lambda},$ and hence $$f$$ is also surjective. Therefore $$f$$ is an isomorphism from $$E_{1,\lambda}$$ to $$E_{2,\lambda}$$.

### Proof

We can use the above lemma to prove our statement easily.

The above lemma tells us that the kernel of $$AB-3I_{4\times 4}$$ is isomorphic to the kernel of $$BA-3I_{3\times 3}$$.

By a simple calculation, we see that the kernel of $$AB-3I_{4\times 4}$$ has dimension $$3$$. Hence the kernel of $$BA-3I_{3\times 3}$$ also has dimension $$3$$. Since $$BA-3I_{3\times 3}$$ is a $$3\times 3$$ matrix, if its kernel has dimension $$3$$, it is the zero matrix. Therefore $$BA = 3I_{3\times 3}$$ indeed.

## Remarks

The second proof, albeit longer, is more general and more beautiful, in my eyes. For example, it does not assume that the vector-spaces have finite dimension.