Is a matrix multiplied with its transpose something special?

In my math lectures, we talked about the Gram-Determinant where a matrix times its transpose are multiplied together.

Is $A A^T$ something special for any matrix $A$?

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The main thing is presumably that $AA^T$ is symmetric. Indeed $(AA^T)^T=(A^T)^TA^T=AA^T$. For symmetric matrices one has the Spectral Theorem which says that we have a basis of eigenvectors and every eigenvalue is real.

Moreover if $A$ is regular, then $AA^T$ is also positive definite, since $$x^TAA^Tx=(A^Tx)^T(A^Tx)> 0$$

Then we have: A matrix is positive definite if and only if it's the Gram matrix of a linear independent set of vectors.

Last but not least if one is interested in how much the linear map respresented by $A$ changes the norm of a vector one can compute

$$\sqrt{\left<Ax,Ax\right>}=\sqrt{\left<A^TAx,x\right>}$$

which simplifies for eigenvectors $x$ to the eigenvalue $\lambda$ to

$$\sqrt{\left<Ax,Ax\right>}=\sqrt \lambda\sqrt{\left<x,x\right>},$$

The determinant is just the product of these eigenvalues.

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May I ask, in addition, what happens if $A$ is just a column vector? Does $A\times A^{T}$ have additional special properties? –  digital-Ink Jul 24 '12 at 21:12
Then you can write $\mathbb R^n\cong A\bot V$. What is $AA^TA?$ and what is $AA^Tv$ for $v\in V$? How does $AA^T$ hence look like? –  Simon Markett Jul 25 '12 at 7:57
Or in other words: can you see that the rank of $AA^T$ is $1$? What does that mean for the dimension of the eigenspace to the eigenvector $0$? –  Simon Markett Jul 25 '12 at 9:55
What "if $A$ is regular" exactly mean? There seem to be several interpretations on Wikipedia. –  qazwsx Mar 11 at 20:21
Regular means here the same as invertible. –  Simon Markett Mar 11 at 20:57

$AA^T$ is positive semi-definite, and in a case in which $A$ is a column matrix, it will be a rank 1 matrix and have only one non-zero eigenvalue which equal to $A^TA$ and its corresponding eigenvector is $A$. The rest of the eigenvectors are the null space of $A$ i.e. $\lambda^TA = 0$.

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One could name some properties, like if $B=AA^T$ then

$$B^T=(AA^T)^T=(A^T)^TA^T=AA^T=B,$$

so

$$\langle v,Bw\rangle=\langle Bv,w\rangle=\langle A^Tv,A^Tw\rangle.$$

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If you have a real vectorspace equipped with a scalar product, and an Orthogonal matrix $A$ then $AA^T=I$ holds. An Matrix is orthogonal if for the scalarproduct $<v,w> = <Av, Aw>$ holds for any $v,w \in V$

However I don't see a direkt link to the Gram-Determinant.

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