Suppose we have two transformations $L$ and $T$. The associated matrices are $A_{5 \times 8}$ and $B_{8 \times 5}$ respectively.

I'm asked whether the matrices $AB$ and $BA$ are defined and are invertible.

My intuition is:

Both are defined ($AB_{5 \times 5}$ and $BA_{8 \times 8}$) but neither is invertible. $AB$ and $BA$ represent the composition of $T$ and $L$ and $L$ and $T$ respectively.

$T$ is not injective and $L$ is not surjective, therefore neither is a bijection and their composition is not a bijection. A matrix is only invertible if the associated linear transformation is a bijection and so neither of those 2 matrices is invertible.

Is my intuition correct?


2 Answers 2


I would invite you to consider the smaller $2 \times 1$ case:

\begin{align*} \mathbf A &= \begin{pmatrix} 1 & 0 \end{pmatrix} \\ \mathbf B &= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{align*}

Do you agree that the map represented by $\mathbf A \mathbf B$ from $\mathbb R \to \mathbb R$ is bijective?

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    $\begingroup$ Thank you for the answer! So to make sure I understand correctly, that map is clearly bijective, it just takes any number in $\mathbb{R}$ and outputs the same number. Is it fair to say then that the composition of an injection and then a surjection could be a bijection? $BA$ in your example would not be bijective and it's basically a projection of $\mathbb{R}^2$ into $\mathbb{R}$? $\endgroup$ Nov 26, 2019 at 19:19
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    $\begingroup$ Yes, a composition of an injective non-surjective function and a surjective non-injective function can be bijective, basically because the codomain of the composition is not the same as the codomain of the first function, so the non-surjectivity isn't a "dealbreaker". You would be right in saying that $\mathbf B \mathbf A$ certainly isn't bijective (from $\mathbb R^2 \to \mathbb R^2$) in my example. $\endgroup$ Nov 26, 2019 at 19:27
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    $\begingroup$ Understood, that was a great example. Thank you! $\endgroup$ Nov 26, 2019 at 19:28
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    $\begingroup$ You're welcome : ) $\endgroup$ Nov 26, 2019 at 19:28
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    $\begingroup$ What a lovely example of a civil, friendly, exchange of ideas <3 $\endgroup$
    – Simon
    Nov 26, 2019 at 21:20

First of all, matrix multiplications $A_{m\times n}B_{n\times m}$ are always defined.

As for the invertibility,

  • $B_{8\times 5}A_{5\times 8}$ is an $8\times 8$ matrix of rank at most 5, therefore it is definitely NOT invertible.
  • $A_{5\times 8}B_{8\times 5}$ may have full rank, and consequently may be invertible. You can construct examples and counter-examples by playing with the identity matrix and the row-reversed identity matrix.

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