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Let $A∈\mathscr{M}_{3×2}(\mathbb{R})$ and $B∈\mathscr{M}_{2\times3}(\mathbb{R})$ be matrices satisfying $AB =\begin{bmatrix} 8 &2 &−2\\ 2 &5 &4\\ −2 &4& 5 \end{bmatrix}$. Calculate $BA$. (Golan, The Linear Algebra a Beginning Graduate Student Ought to Know, Exercise 426.)

Maybe it can be solved by solving a system of equations, but I think there is a shorter way since this problem was in my exam.

Thanks.

Martin Sleziak
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Guadalupe
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2 Answers2

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$(AB)^2=9(AB)\Rightarrow(BA)^3=9(BA)^2\Rightarrow \mu_{BA}(X)\mid X^2(X-9)$

$P_{AB}(X)=X(X-9)^2\Rightarrow P_{BA}(X)=(X-9)^2\Rightarrow\mu_{BA}(X)\mid (X-9)^2$

Conclusion: $\mu_{BA}(X)=X-9$, so $BA=9I_2$.

(Here $\mu_X$, respectively $P_X$ stands for the minimal polynomial, respectively the characteristic polynomial of a square matrix $X$.)

user26857
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  • nice work. you got it with minimal computation(+1). – abel May 07 '15 at 00:35
  • It wouldn't hurt to explain the implication $P_{AB}(X)=X(X-9)^2\Rightarrow P_{BA}(X)=(X-9)^2$ a little. I thought maybe you were using the fact that $AB$ and $BA$ have the same nonzero eigenvalues but, at a glance, I don't see how that's enough. – Mike F May 07 '15 at 00:45
  • @MikeF You can find a proof here or here. – user26857 May 07 '15 at 06:38
  • @MikeF More precisely, $AB$ and $BA$ in general have the same multiset of nonzero eigenvalues. – user1551 May 07 '15 at 09:41
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Hint: I'll consider the same problem assuming, instead, that \begin{align} A\text{ is }3 \times 2 && B \text{ is } 2 \times 3 && AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}. \end{align}

  • Since $AB$ has rank $2$, and $A$ is $2 \times 3$, the rank of $A$ must also be $2$. In fact, the column space of $A$ must equal the column space of $AB$. So, $A$ has the form $A=\begin{bmatrix} * & * \\ * & * \\ 0 & 0 \\ \end{bmatrix}$ for some undetermined $2 \times 2$ invertible matrix.
  • For similar reasons, the null space of $B$ must equal that of $AB$. So, $B$ has the form $B=\begin{bmatrix} * & * & 0 \\ * & * & 0 \\ \end{bmatrix}$ for some undetermined invertible matrix.
  • From the equation $AB = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$, we see that the undetermined $2 \times 2$ matrices are inverses of one another. Given this, one checks that $BA = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$

Remark: For the original unmodified problem, note that, since $AB$ is symmetric, it is diagonalizable (even orthogonally so). Moreover, its diagonal form turns out to be $9$ times projection onto two coordinates. Let $U$ be an invertible $3 \times 3$ matrix with $$UABU^{-1} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}.$$ Apply the above argument to $A_1 = UA$ and $B_2 = BU^{-1}$ in order to compute $B_2A_2 = BA$.

Mike F
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  • A matrix like $A$ of rank two has not necessary a zero row, so you must perform some elementary transforms in order to get there. I wonder now if these elementary transforms don't interfere with the ones used for $B$ in a wrong way, that is, you have multiply $PAQ$ by $RBS$ with $Q\ne R^{-1}$. – user26857 May 07 '15 at 14:11
  • I have raised objections on the argument of your post. This comment has nothing to do with my concern. – user26857 May 07 '15 at 18:15
  • @user26857: OK I have deleted my last comment, which did not address your concern. I didn't think it was necessary to perform any elementary row operations on $A$. Is it to one of these assertions that you are objecting? (1) $col(A) = col(AB)$ so that $col(A) = { (x,y,0) : x,y \in \mathbb{R}}$. (2) Since $col(A) = { (x,y,0) : x,y \in \mathbb{R}}$, $A$ has the form $A=\begin{bmatrix}
    • & * \
    • & * \

    0 & 0 \ \end{bmatrix}$ for some undetermined $2 \times 2$ invertible matrix.

     – Mike F May 07 '15 at 18:35
  • I see it now. It would have been useful to include your first comment into the answer. – user26857 May 07 '15 at 18:41
  • @user26857: OK I have added something to that effect. I was trying to just leave a hint, but I think you may be right that the hint was too vague. – Mike F May 07 '15 at 19:12