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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 $A$.

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2  
Are you familiar with the determinant? –  Alex Becker Sep 20 '12 at 22:43

6 Answers 6

up vote 8 down vote accepted

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 $B B^T$.

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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. –  diimension Sep 20 '12 at 22:58
    
@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. –  M Turgeon Sep 20 '12 at 23:00
    
Okay, now I understand thank you very much! –  diimension Sep 20 '12 at 23:04

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?

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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? –  diimension Sep 20 '12 at 22:52
    
@diimension. Hint: If $A$ and $B$ are invertible then $AB$ is invertible. –  Clive Newstead Sep 20 '12 at 22:54
    
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 ? –  diimension Sep 20 '12 at 23:06
    
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.) –  Clive Newstead Sep 20 '12 at 23:07
    
Great! I understand now! Thank you very much! –  diimension Sep 20 '12 at 23:13

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

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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.

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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.

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

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Not so, $\;A A^T\;$ is its own transpose: \begin{align} & (A A^T)^T \\ = & \;\;\;\;\;\text{"$\;(PQ)^T = Q^T P^T\;$"} \\ & (A^T)^T A^T \\ = & \;\;\;\;\;\text{"simplify: $\;(P^T)^T = P\;$"} \\ & A A^T \\ \end{align} –  Marnix Klooster Nov 19 '13 at 8:18
1  
Edited accordingly. Thanks –  Jack Nov 19 '13 at 13:47

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