Lower bound on quadratic form

Suppose I have a non-symmetric matrix $A$ and I can prove that $x^T A x = x^T \left(\frac{A+A^T}{2}\right) x>0$ for any $x \ne 0$? Can I then say that $x^T A x \ge \lambda_{\text{min}}(A) \|x\|^2 > 0$?

Solution 1:

Let quadratic form $f$ be defined by $f (x) = x^T A x$, where $A \in \mathbb{R}^{n \times n}$. Since $x^T A x$ is a scalar, then $(x^T A x)^T = x^T A x$, i.e., $x^T A^T x = x^T A x$. There are infinitely many matrix representations of $f$. We take affine combinations of $A$ and $A^T$, and any such combination yields $f$

$$x^T (\gamma A + (1-\gamma) A^T) x = f (x)$$

where $\gamma \in \mathbb{R}$. We choose $\gamma = \frac{1}{2}$, which yields the symmetric matrix $\frac{A+A^T}{2}$, which is diagonalizable, has real eigenvalues and orthogonal eigenvectors. Hence, it has the eigendecomposition

$$\frac{A+A^T}{2} = Q \Lambda Q^T$$

Thus,

$$x^T \left(\frac{A+A^T}{2}\right) x = x^T Q \Lambda Q^T x$$

If the eigenvalues are nonnegative, then we can take their square roots

$$x^T \left(\frac{A+A^T}{2}\right) x = x^T Q \Lambda Q^T x = \|\Lambda^{\frac{1}{2}} Q^T x\|_2^2 = \|\Lambda^{\frac{1}{2}} y\|_2^2 \geq 0$$

where $y = Q^T x$. We conclude that $f$ is positive semidefinite. If all the eigenvalues are positive, then $f$ is positive definite. Note that

$$\begin{array}{rl} \|\Lambda^{\frac{1}{2}} y\|_2^2 &= \displaystyle\sum_i \lambda_i \left(\frac{A+A^T}{2}\right) y_i^2\\ &\geq \displaystyle\sum_i \lambda_{\text{min}} \left(\frac{A+A^T}{2}\right) y_i^2 = \lambda_{\text{min}} \left(\frac{A+A^T}{2}\right) \|y\|_2^2\end{array}$$

Since the eigenvectors are orthogonal, $Q^T Q = I_n$, then

$$\|y\|_2^2 = \|Q x\|_2^2 = x^T Q^T Q x = \|x\|_2^2$$

We thus obtain

$$x^T A x = x^T \left(\frac{A+A^T}{2}\right) x \geq \lambda_{\text{min}} \left(\frac{A+A^T}{2}\right) \|x\|_2^2$$