1. ## Symmetric matrix proof

Donīt know if the descriptive title is misleading but here goes.

Show that for every $m/n$matrix $A$ with a range $r$ there exists a matrix $AA^T$ which is symmetric and has a range $r$.

This is what I have:

The matrix $AA^T$ is a square $m/m$ matrix and also symmetric because $(AA^T)^T=A^{TT}A^T=AA^T$.

The part with range is still confusing. Now this should be true

$dimR(A)=dimR(A^T)=r$

but how should I pursue $dimR(AA^T)=r$? Im thinking about Orthogonal matrices. Because if $AA^T$ is symmetric which it is, then there exists an orthogonal matrix $Q$ such that $Q^TAA^TQ=D$ where $D$ is a diagonal matrix with the eigenvalues of $AA^T$ in the diagonal.

any help appreciated

2. Originally Posted by wilbursmith
Donīt know if the descriptive title is misleading but here goes.

Show that for every $m/n$matrix $A$ with a range $r$ there exists a matrix $AA^T$ which is symmetric and has a range $r$.

This is what I have:

The matrix $AA^T$ is a square $m/m$ matrix and also symmetric because $(AA^T)^T=A^{TT}A^T=AA^T$.

The part with range is still confusing. Now this should be true

$dimR(A)=dimR(A^T)=r$

but how should I pursue $dimR(AA^T)=r$?
Use the rank + nullity theorem. If $A^{\textsc{t}}x = 0$ then clearly $AA^{\textsc{t}}x = 0$. Conversely, if $AA^{\textsc{t}}x = 0$ then $0 = x^{\textsc{t}}AA^{\textsc{t}}x = (A^{\textsc{t}}x)^{\textsc{t}}(A^{\textsc{t}}x)$, which implies that $A^{\textsc{t}}x=0$. So $AA^{\textsc{t}}$ and $A^{\textsc{t}}$ have the same kernel, hence the same nullity, hence the same rank.