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I don't want the solution. Please don't post the full solution. I just need a starting clue on how to do this.

  1. Suppose $A$ and $B$ are matrices such that $AB$ and $BA$ are defined.

a) Show that $AB$ and $BA$ are both square matrices.

I actually have no idea where to start. But I note that a square matrix commutes if say:

A was defined as:

$\begin{pmatrix} a & b\\ b & a \end{pmatrix}$

B was defined as:

$\begin{pmatrix} c & d\\ d & c \end{pmatrix}$

Then $AB=BA$

I suppose I need to include this in my argument somehow but for a more general case?

Bobby
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  • Hint: Do you know any relations between the dimensions of A and B? – Jacob Akkerboom May 09 '13 at 15:36
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    When is the product $AB$ of two matrices $A$ and $B$ defined? What conditions on the number of rows and columns must hold? – k.stm May 09 '13 at 15:37
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    Suppose $A$ is an $m\times n$ matrix and $B$ is a $p \times q$ matrix. The fact that $AB$ is defined means what in terms of $n$ and $p$. What happens when you consider $BA$? – Suugaku May 09 '13 at 15:37
  • What about this guys?

    Well I observe that multiplication is only possible with $A$ having a dimension $m$ x $n$ and $B$ must have $n$ x $p$. Hence $AB$ has $m$ x $p$.

    If the reverse is true with $B$ having dimensions $n$ x $p$, $A$ has dimensions $m$ x $n.$ $BA$ now has dimensions $n$ x $n$. Which would imply $p = m.$ Hence I substitute that into dimensions of $AB$. Which $AB$ has $m$ x $m.$ Thus proving that they both must be square matrices?

    But then how would I prove that they are of same size?

     – Bobby May 09 '13 at 15:43
  • @Bobby $AB$ and $BA$ are not necessary of same size. –  May 09 '13 at 15:47
  • @ Sami. Oh oops, I assumed AB=BA, (though its in the next part of the question) – Bobby May 09 '13 at 15:50
  • @Bobby: Don't forget to accept an answer (you can accept one answer per question, by clicking on the $\checkmark$ to the left of the answer you'd like to accept). You can also now upvote as many helpful answers as you'd like! (Click on the "upwards arrow" above an answer's vote-count.) – amWhy May 25 '13 at 04:20

3 Answers3

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Commutativity is not needed. By definition, matrix multiplication of $A$ and $B$ can only be defined if the number of columns of $A$ matches the number of rows of $B$. In that case, if $A$ is an $n \times m$ matrix, and $B$ is an $m \times p$ matrix, then $AB$ will be an $n \times p$ matrix. The way I explain it to my linear algebra students is that you can multiply if the "inner" dimensions match, and the product has dimensions equal to the "outer" dimensions.

$$ \underbrace{A}_{n \times m} \cdot \underbrace{B}_{m \times p} = \underbrace{C}_{n \times p}. $$

What can you say about the dimensions of $A$ and $B$ if both products $AB$ and $BA$ exist?

Shaun Ault
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  • @ Shaun.

    So continuing from your argument,

    $$ \underbrace{B}{m \times p} \cdot \underbrace{A}{n \times m} = \underbrace{D}_{m \times m}. $$

    If $n=p$

    So substituting into your notation for rows and columns

    $$ \underbrace{C}_{n \times n}. $$

    Therefore we have C as a square matrix and so is D.

     – Bobby May 09 '13 at 15:55
  • Correct. From this you get both $AB$ and $BA$ square --- though not necessarily the same size of square. – Shaun Ault May 09 '13 at 15:56
  • Could I ask if your underlining with a curly braces is a valid notation (universally accepted?) – Bobby May 09 '13 at 15:57
  • It's not really a notation... but a way for me to highlight the dimensions of these matrices. In fact, it is standard to simply write $A$, $B$, $C$, .... for matrices without notating their respective dimensions at all. – Shaun Ault May 09 '13 at 15:59
  • Thanks this notation is much clearer than the way I was taught to write it.

    So that would mean they would be both equal in dimensions (size) if $AB=BA$ because we get:

    $$ \underbrace{A}{n \times m} \cdot \underbrace{B}{m \times p} = \underbrace{C}_{n \times n}. $$

    $$ \underbrace{B}{m \times p} \cdot \underbrace{B}{n \times m} = \underbrace{D}_{m \times m}. $$

    If $AB=BA$ implies $C=D$, therefore $m = n$ by implication? Hence both square and equal in size?

     – Bobby May 09 '13 at 16:03
  • All you must show is that both $AB$ and $BA$ are square right? We cannot prove that $C=D$; indeed, $C$ may be a 10-by-10 matrix, and $D$ may be a 1000-by-1000 matrix. But all we need to prove is that they are both square. – Shaun Ault May 09 '13 at 16:05
  • Sorry there was a second part to this:

    If $AB=BA$, show that $A$ and $B$ are both square and of same size

     – Bobby May 09 '13 at 16:06
  • let us continue this discussion in chat – Bobby May 09 '13 at 16:13
  • Kudos to you, Bobby, for asking explicitly for hints rather than a complete solution. You will learn so much more this way! – Ted Shifrin May 09 '13 at 20:53
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Hint To multiply a matrix $A\in\mathcal{M}_{n,p}(\mathbb{F})$ by a matrix $B\in\mathcal{M}_{q,m}(\mathbb{F})$: and get $AB\in\mathcal{M}_{n,m}(\mathbb{F})$ we must have $p=q$ so...

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If A is a (m×n) matrix, for AB to be defined, B must have its rows equal to the coloumns of A, B can be a (n×p) matrix AB = (m×n)•(n×p)=(m×p) But for BA to be defined, BA=(n×p)•(m×n), since BA is defined, p=m Therefore B is a (n×m) matrix. AB=(m×n)•(n×m)=(m×m) thus AB is a (m×m) matrix BA=(n×m)•(m×n)=(n×n) thus BA is a (n×n) matrix Therefore AB and BA are square matrices

SSENYONGA DERRICK
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