6
$\begingroup$

I want to know how to prove this question:

Prove that if $A$ is invertible matrix then $AA^T$ and $A^T A$ are also invertible.

My attempt:

Since $A$ is invertible we have that $AA^{-1} = I$ and if we denote $B = A^{-1}$, we have that $AB=I$ so if we take the transpose of both sides then we have $(AB)^T = I^T = I$, but this is where I get my problem since the transpose, inverse of $B$ isn't equal to $A$.

$\endgroup$
1
  • 2
    $\begingroup$ Are you familiar with the determinant? $\endgroup$ Sep 20, 2012 at 22:43

6 Answers 6

10
$\begingroup$

Let B be a matrix, so that $AB = I$ and $BA = I$. Then $A^T B^T = (BA)^T = I^T = I$ and $B^T A^T = (AB)^T = I^T = I$. So the inverse for $A^T$ is $B^T$. Now we get :

$(A A^T)(B^T B) = A(A^T B^T) B = A(I)B = AB = I$ and $(A^T A)(B B^T) = I$. So the inverse for $AA^T$ is $B^TB$ and for $A^T A$ we have $B B^T$.

$\endgroup$
3
  • $\begingroup$ Thank you very much!! but how do you that the inverse of A^T is B^T? Is that already implied? And also how did you get (AA^T)(B^T)(B) = A(A^T)(B^T)B? Forgive my ignorance, this is only my second week dealing with proofs. $\endgroup$
    – diimension
    Sep 20, 2012 at 22:58
  • 1
    $\begingroup$ @diimension The first fact follows from the fact that inverses are unique and from the computation in the first paragraph. For your second question, this is simply associativity. $\endgroup$
    – M Turgeon
    Sep 20, 2012 at 23:00
  • $\begingroup$ Okay, now I understand thank you very much! $\endgroup$
    – diimension
    Sep 20, 2012 at 23:04
3
$\begingroup$

Using your notation, you've shown that $(AB)^T = I$, so far so good; but also $(AB)^T = B^TA^T = (A^{-1})^TA^T$, and this tells you that $A^T$ is invertible and $(A^T)^{-1}=(A^{-1})^T$. Can you see where to go with this?

$\endgroup$
5
  • $\begingroup$ I kind of see where you are going. Okay, since we got the A^T is invertible we must show some how that (A^-1)^T = A, the only thing I can think of is that if I take the transpose again but that I wont work since it will just start me back were I started, but if I premultiply will that work? $\endgroup$
    – diimension
    Sep 20, 2012 at 22:52
  • $\begingroup$ @diimension. Hint: If $A$ and $B$ are invertible then $AB$ is invertible. $\endgroup$ Sep 20, 2012 at 22:54
  • $\begingroup$ Okay, Turgeon gave me the answer in another way but I want to know how to also proof it doing your way, but can you give a more specific hint ? $\endgroup$
    – diimension
    Sep 20, 2012 at 23:06
  • $\begingroup$ Well, if you know that $A$ is invertible (with inverse $A^{-1}$) and $B$ is invertible (with inverse $B^{-1}$), then $ABB^{-1}A^{-1}=AIA^{-1}=AA^{-1}=I$, and so $AB$ has inverse $B^{-1}A^{-1}$. But here $B=A^T$, which we've proved to be invertible. (And so on.) $\endgroup$ Sep 20, 2012 at 23:07
  • $\begingroup$ Great! I understand now! Thank you very much! $\endgroup$
    – diimension
    Sep 20, 2012 at 23:13
3
$\begingroup$

Recall a matrix is invertible if and only if its determinant is non-zero.

Now what can you say about

A) $\text{det}(A^T)$

B) $\text{det}(AB)$?

Work these out, and it becomes clear

$\endgroup$
1
$\begingroup$

last line $(A^TA)^T=A^TA$.

$AA^T$ is equivalent to $(A^T)^T(A^T)$, so the rest follows when $A^T$ is treated as A.

$\endgroup$
1
$\begingroup$

An alternative approach:

To show that $A^TA$ is invertible, think about the linear system $A^TAx=0$. It suffices to show that $x=0$ is the unique solution. But $A^TAx=0$ implies that $\langle Ax,Ax\rangle= x^TA^TAx=0$. Thus $Ax=0$. Now use the fact that $A$ is invertible.

For $AA^T$, note that $AA^T=(A^T)^T(A^T)$

$\endgroup$
1
  • $\begingroup$ A cute approach! But please stop deleting your old posts. The process has generated a flag already. Soon I have to start undeleting, and I don't really cherish the prospect. Has something gone terribly wrong? $\endgroup$ Jul 29, 2017 at 20:04
0
$\begingroup$

You need to find an $\;X\;$ such that $\;A A^T X = I\;$, given that $\;A\;$ is invertible.

Let's calculate what this means, by repeatedly left-multiplying by a matrix that cancels out: \begin{align} (*) \;\;\; \phantom{\equiv} & A A^T X = I \\ \Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;A^{-1}\;$; use $\;A^{-1} A = I\;$"} \\ & A^T X = A^{-1} \\ \Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;(A^{-1})^T\;$"} \\ & (A^{-1})^T A^T X = (A^{-1})^T A^{-1} \\ \equiv & \;\;\;\;\;\text{"simplify using $\;P^T Q^T = (Q P)^T\;$ and $\;A A^{-1} = I\;$"} \\ (**) \;\;\; \phantom{\equiv} & X = (A^{-1})^T A^{-1} \\ \Rightarrow & \;\;\;\;\;\text{"left-multiply by $\;A A^T\;$ -- working our way back to $(*)$"} \\ & A A^T X = A A^T (A^{-1})^T A^{-1} \\ \equiv & \;\;\;\;\;\text{"simplify using $\;P^T Q^T = (Q P)^T\;$, $\;A^{-1} A = I\;$, and $\;A A^{-1} = I\;$"} \\ & A A^T X = I \\ \end{align}

This 'ping pong' argument shows that $(*)$ and $(**)$ are equivalent, so $\;(A^{-1})^T A^{-1}\;$ is the inverse of $\;A A^T\;$.

The other half of the question can be solved in the same way.

$\endgroup$

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.