Question-44

eigenvalue and eigenvector
quadratic form

Let \(\displaystyle A\) be a positive definite matrix.

Which of these options is correct?

  • Spectral theorem
  • Eigenvalue equation

Thanks to Vivek, Sherry and Aniruddha for this solution.

This solution uses the result that a matrix is positive definite if and only if all its eigenvalues are positive. Since \(\displaystyle A\) is positive definite, let \(\displaystyle ( \lambda _{i} ,v_{i})\) be the \(i^{th}\) eigenpair of \(\displaystyle A\) with \(\lambda_i > 0\), for \(\displaystyle i=1,\cdots ,n\). Pre-multiplying both sides by \(A\): \[ \begin{equation*} \begin{aligned} Av_{i} & =\lambda v_{i}\\ A^{2} v_{i} & =\lambda _{i}^{2} v_{i} \end{aligned} \end{equation*} \]

We see that \(\displaystyle \left( \lambda _{i}^{2} ,v_{i}\right)\) is an eigenpair of \(\displaystyle A^{2}\) for \(\displaystyle i=1,\cdots ,n\). It follows that \(\displaystyle \left\{\lambda _{1}^{2} ,\cdots ,\lambda _{n}^{2}\right\}\) is the set of eigenvalues of \(\displaystyle A^{2}\), where each entry is strictly positive. Therefore, \(\displaystyle A^{2}\) is also positive definite.

Since \(\displaystyle A\) has \(\displaystyle n\) positive eigenvalues, its rank is \(\displaystyle n\) and it is invertible. Pre-multiplying both sides by \(A^{-1}\) and dividing both sides by \(\lambda_i\) (we can do this since \(\lambda_i > 0\):

\[ \begin{equation*} \begin{aligned} Av_{i} & =\lambda v_{i}\\ A^{-1} v_{i} & =\frac{1}{\lambda _{i}} v_{i} \end{aligned} \end{equation*} \]

We see that \(\displaystyle \left(\frac{1}{\lambda _{i}} ,v_{i}\right)\) is an eigenpair of \(\displaystyle A^{-1}\). The set of eigenpairs of \(\displaystyle A^{-1}\) is given by \(\displaystyle \left\{\frac{1}{\lambda _{1}} ,\cdots ,\frac{1}{\lambda _{n}}\right\}\). It is clear that all these are positive. Therefore, \(\displaystyle A^{-1}\) is also positive definite.

Since \(\displaystyle A\) is p.d., we can decompose \(\displaystyle A\) as:

\[ \begin{equation*} A=Q\Sigma Q^{T} \end{equation*} \]

where \(\displaystyle Q\) is an orthogonal matrix and \(\Sigma\) is a diagonal matrix with positive entries on the diagonal. It follows that:

\[ \begin{equation*} \begin{aligned} A^{2} & =Q\Sigma ^{2} Q^{T}\\ A^{-1} & =Q\Sigma ^{-1} Q^{T} \end{aligned} \end{equation*} \]

We see that both these matrices are positive definite.