prove that $BA=\begin{bmatrix}9 & 0\\ 0 & 9\end{bmatrix}$

Question

I want to prove the following question but I have no idea to start!

let $A$ be a $3\times2$ matrix and $B$ be a $2\times3$ one. if $AB=\begin{bmatrix}8 & 2 &-2 \\2 & 5 & 4 \\-2 & 4 & 5 \end{bmatrix}$

then prove that $BA=\begin{bmatrix}9 & 0\\ 0 & 9\end{bmatrix}$. [ hint:if $AB=I$ then we have $BA=I$]

I imagine that the spectral theorem has to play a role here as $AB$ is real symmetric... But I don't know how to play with it. — mathcounterexamples.net, Feb 27 '16 at 21:26
@mathcounterexamples.net It's true, for example eigenvalues of the matrix $3\times 3$ are 9;9;0. — echzhen, Feb 27 '16 at 21:27
I computed it directly. Not very elegant, but it works. Just solving polynomial equations with elimination method. — Dietrich Burde, Feb 27 '16 at 21:30

Anurag A · Accepted Answer · 2016-02-28T03:43:54.247

5

Note that rank$(AB)=2$. But rank$(BA) \geq \text{rank}(AB)^2$ because $(AB)^2=(AB)(AB)=A(BA)B$. We compute $$(AB)^2=\begin{bmatrix}72 & 18 & -18\\18&45&36\\-18&36&45\end{bmatrix}=9(AB).$$ But rank$(AB)^2=2$(follows from above matrix equality), so rank$(BA) \geq 2$. Since $BA$ is $2 \times 2$, this means $BA$ is invertible. Now we have \begin{align*} (BA)^3 & = (BA)(BA)(BA)\\ &=B(AB)^2A\\ &=B(9AB)A\\ &=9(BA)(BA)\\ &=9(BA)^2\\ BA&=9I. \end{align*}

edited Feb 28 '16 at 03:43

answered Feb 27 '16 at 22:46

Anurag A

41,067

you've mentioned that because $(AB)^2=A(BA)B$ so $rank(BA) \geq rank(AB)^2 $. I can't understand why. Is there any special theorem which proves this claim? – F.K Feb 29 '16 at 08:18
1

@C.B I am just using the fact that rank$(MN) \leq \min(\text{rank}(M), \text{rank}(N))$. In the proof above, $BA$ is one of the components in the product describing $(AB)^2$ so.... – Anurag A Feb 29 '16 at 09:59
I got it. thank you so much – F.K Feb 29 '16 at 10:39

score 3 · Answer 2 · edited Apr 13 '17 at 12:19

Probably there are shorter arguments, but maybe something like this:

$AB$ is symmetric and has an ON basis of eigenvectors $v_1$, $v_2$ and $v_3$ with corresponding eigenvalues $9$, $9$ and $0$.
$AB$ and $BA$ have the same non-zero eigenvalues (see for example here).
Multiplying by $B$ from the left, we find that $Bv_1$ and $Bv_2$ are eigenvectors corresponding to $BA$ with eigenvalues $9$. (Note that these vectors are non-zero.)
The vectors $Bv_1$ and $Bv_2$ are linearly independent, since, multiplying by $A$, we find that $Bv_1=kBv_2$ implies $9v_1=k9v_2$ which cannot hold since $v_1$ and $v_2$ are orthogonal.
This implies that $BA$ is a diagonalizable $2\times2$-matrix with double eigenvalue $9$. We conclude that $BA=9I$, since, we can write $BA=SDS^{-1}$ with $D$ diagonal with nines on the diagonal.

To the down voter: Is something wrong with this solution? – mickep Feb 28 '16 at 08:50 — mickep, Feb 28 '16 at 08:50

prove that $BA=\begin{bmatrix}9 & 0\\ 0 & 9\end{bmatrix}$

2 Answers2