Question-20

If \(A\) is a real matrix, select all true statements.

• \(A^TA\) is positive semi-definite.
• \(A^TA\) and \(AA^T\) have the same eigenvectors.
• \(A^TA\) is symmetric if and only if \(A\) is symmetric.
• \(A^TA\) and \(AA^T\) have the same non-zero eigenvalues.

Answer

• \(A^TA\) is positive semi-definite.
• \(A^TA\) and \(AA^T\) have the same eigenvectors.
• \(A^TA\) is symmetric if and only if \(A\) is symmetric.
• \(A^TA\) and \(AA^T\) have the same non-zero eigenvalues.

Solution

Let \(\displaystyle ( \lambda ,v)\) be an eigenpair of of \(\displaystyle A^{T} A\) with \(\displaystyle \lambda \neq 0\):

\[ \begin{aligned} \left( A^{T} A\right) v & =\lambda v\\ \left( AA^{T}\right) (Av) & =\lambda ( Av)\\ \left( AA^{T}\right)( Av) & =\lambda ( Av) \end{aligned} \]

This is of the form \(\displaystyle \left( AA^{T}\right) u=\lambda u\). For \(\displaystyle ( \lambda ,u)\) to be an eigenpair of \(\displaystyle AA^{T}\), \(\displaystyle u=Av\) has to be non-zero. To show this, do the following. If \(\displaystyle u=Av=0\), then \(\displaystyle A^{T} Av=0\Longrightarrow \lambda v=0\). Since \(\displaystyle \lambda \neq 0\), \(\displaystyle v=0\) which contradicts the fact that \(\displaystyle ( \lambda ,v)\) is an eigenpair of \(\displaystyle A^{T} A\).

---

Let \(\displaystyle ( \lambda ,v)\) be an eigenpair of \(\displaystyle A^{T} A\). Then:

\[ \begin{aligned} A^{T} Av & =\lambda v\\ v^{T} A^{T} Av & =\lambda \left( v^{T} v\right)\\ ( Av)^{T}( Av) & =\lambda \left( v^{T} v\right)\\ \Longrightarrow \lambda & =\frac{||Av||^{2}}{||v||^{2}} \geqslant 0 \end{aligned} \]

We can divide by \(\displaystyle ||v||^{2}\) as \(\displaystyle v\neq 0\) for it is an eigenvector. Since all the eigenvalues of \(\displaystyle A^{T} A\) are non-negative, \(\displaystyle A^{T} A\) is positive semi definite.