7
$\begingroup$

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 $A$.

  • 2
    Are you familiar with the determinant?2012-09-20

6 Answers 6

9

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 $B B^T$.

  • 0
    Okay, now I understand thank you very much!2012-09-20
3

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?

  • 0
    Great! I understand now! Thank you very much!2012-09-20
3

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

1

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 $AA^T=(A^T)^T(A^T)$

  • 0
    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?2017-07-29
1

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.

0

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.