3
$\begingroup$

Here's a question:

$A$ is a $3\times 2$ matrix, and $B$ is a $2\times 3$ matrix, so $AB$ is $3\times 3$ matrix.

The problem given to me was to show that the inverse of $AB$ does not exist.

I was able to verify it using actual values, but could not find a way to show it in general. Can anyone help?

Thanks.

$\endgroup$
1
  • 9
    $\begingroup$ Because the rank of $AB$ is capped at 2 already. The max possible rank of $A$ and $B$ is $2$. For $AB$ to be invertible, it needs rank of $3$. $\endgroup$ – xenon Sep 22 '11 at 15:59
6
$\begingroup$

If you think about matrices as coefficients of systems of linear equations, $A$ are the coefficients of a system of $3$ linear equations in two unknowns, while $B$ corresponds to a system of 2 linear equations in 3 unknowns.

That means that the system corresponding to $B$ is under-determined, and as such, the homogeneous system $B\mathbf{x}=\mathbf{0}$ always has nontrivial solutions. So you can find $\mathbf{a}\neq\mathbf{0}$ such that $B\mathbf{a}=\mathbf{0}$. That means that $$(AB)\mathbf{a}=A(B\mathbf{a}) = A\mathbf{0} = \mathbf{0},$$ so the system of 3 linear equations with 3 unknowns given by $AB$ has nontrivial solutions.

But if the coefficient matrix $M$ of a system with the same number of equations as unknowns is invertible then the only solution to $M\mathbf{x}=\mathbf{0}$ is the trivial solution: for multiplying by $M^{-1}$ we get $\mathbf{x}=M^{-1}M\mathbf{x} = M^{-1}\mathbf{0}=\mathbf{0}$.

Since the system of 3 equations in 3 unknowns $(AB)\mathbf{x}=\mathbf{0}$ has a nontrivial solution, then $AB$ cannot be invertible.

(If you know the rank-nullity theorem, and that a square matrix is invertible if and only if its nullity is $0$, you can do the argument above using nullities, noting that $\mathbf{N}(B)\subseteq \mathbf{N}(A)B$, so $\mathrm{nullity}(AB)\geq \mathrm{nullity}(B)$, and that $3 = \mathrm{rank}(B)+\mathrm{nullity}(B)$ and $\mathrm{rank}(B)\leq 2$.)

$\endgroup$
0
$\begingroup$

Think of $B$ as a linear map from $k^3$ to $k^2$. Then it must have a nullspace since you're mapping to a lower dimensional space. So some $v\in k^3$ which isn't zero also maps to zero. You're not going to be able to invert that operation since it's not injective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.