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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.

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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 $3$. –  xenon Sep 22 '11 at 15:59
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2 Answers

up vote 5 down vote accepted

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 $\mathrm{rank}(B)\leq 2$.)

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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.

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