Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A$?

share|cite|improve this question
    
The matrix $A^TA^{-1}$ is generally self similar... – DVD Jun 12 '15 at 3:42
2  
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 67 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

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

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.

share|cite|improve this answer
    
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
1  
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
2  
What "if $A$ is regular" exactly mean? There seem to be several interpretations on Wikipedia. – qazwsx Mar 11 '14 at 20:21
2  
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. $\lambda^TA = 0$.

share|cite|improve this answer

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

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

so

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

share|cite|improve this answer

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

share|cite|improve this answer
    
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.

share|cite|improve this answer

protected by Community Jul 5 '15 at 7:55

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.