Problem-18

symmetric matrices
eigenvalues
eigenvectors

Question

Select all true statements.

  1. \(A^TA\) is positive semi-definite.

  2. \(A^TA\) and \(AA^T\) have the same eigenvectors.

  3. \(A^TA\) is symmetric if and only if \(A\) is symmetric.

  4. \(A^TA\) and \(AA^T\) have the same non-zero eigenvalues.

Solution

Options (a) and (d) are correct.

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

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

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

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

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

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