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 $A$?
matrices
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 $A$?
Best Answer
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 invertible, 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 represented 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.