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

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

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The matrix $A^TA^{-1}$ is generally self similar... – DVD Jun 12 '15 at 3:42
One of the themes of Gilbert Strang's books is the ubiquity of $A^T A$ and $A^T CA$ (with $C$ positive semidefinite) in mathematics. For example, the adjoint of the gradient operator is the negative divergence operator, and the Laplacian is the divergence of the gradient. In one book Strang states, "I have learned to look for $A^T C A$". – littleO Jan 4 at 19:57
up vote 69 down vote accepted

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


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 '14 at 20:21
Regular means here the same as invertible. – Simon Markett Mar 11 '14 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



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

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The product $A^TA$ appears in a key role in the normal equations $A^TAx=A^T b$ for solving linear least squares problems.

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What is this key role? – user25959 Apr 16 at 0:07

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 $\langle v,w \rangle = \langle Av, Aw \rangle$ holds for any $v,w \in V$

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

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