# Symmetric matrix proof

• Apr 30th 2010, 09:41 AM
wilbursmith
Symmetric matrix proof
Donīt know if the descriptive title is misleading but here goes.

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

This is what I have:

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

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

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

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

any help appreciated
• Apr 30th 2010, 10:31 AM
Opalg
Quote:

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

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

This is what I have:

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

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

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

but how should I pursue \$\displaystyle dimR(AA^T)=r\$?

Use the rank + nullity theorem. If \$\displaystyle A^{\textsc{t}}x = 0\$ then clearly \$\displaystyle AA^{\textsc{t}}x = 0\$. Conversely, if \$\displaystyle AA^{\textsc{t}}x = 0\$ then \$\displaystyle 0 = x^{\textsc{t}}AA^{\textsc{t}}x = (A^{\textsc{t}}x)^{\textsc{t}}(A^{\textsc{t}}x)\$, which implies that \$\displaystyle A^{\textsc{t}}x=0\$. So \$\displaystyle AA^{\textsc{t}}\$ and \$\displaystyle A^{\textsc{t}}\$ have the same kernel, hence the same nullity, hence the same rank.