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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 $x^T A x \ge \lambda_{\text{min}}(A) \|x\|^2 > 0$?

Rodrigo de Azevedo
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Controller
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    For non symmetric (real) matrices eigenvalues need not be real. It is not clear what $\lambda_\min(A)$ should be defined to be. Of course you can define it to be $\lambda_\min(\frac{A+A^T}2)$, and then this question evaporates. – Marc van Leeuwen May 23 '16 at 07:41

2 Answers2

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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

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

Rodrigo de Azevedo
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  • I do not understand why

    $$ |\Lambda^{\frac{1}{2}} y|_2^2 = \displaystyle\sum_i \lambda_i \left(\frac{A+A^T}{2}\right) y_i^2 $$

     – Faroq AL-Tam Jun 27 '18 at 13:01
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    @WesamF. $$ |\Lambda^{\frac{1}{2}} y|2^2 = y^T \Lambda ,y = \displaystyle\sum_i \Lambda{ii} y_i^2 = \displaystyle\sum_i \lambda_i \left(\frac{A+A^T}{2}\right) y_i^2 $$ – Rodrigo de Azevedo Jun 27 '18 at 13:16
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    @ Rodrigo de Azevedo Thank you very much,

    If we have this term $$x^\top A x + v^\top x$$, where $A$ is both positive definite and symmetric, what do I need to do to proof that it is bounded below as: $$x^\top A x \geq \lambda_{\min} |x|_2^2 + |v| + |x|$$, can you give me a hint please.

     – Faroq AL-Tam Jun 27 '18 at 13:37
  • sorry it is - $$x^\top A x \geq \lambda_{\min} |x|_2^2- |v| |x|$$ – Faroq AL-Tam Jun 27 '18 at 13:52
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    @WesamF. $$\frac 12 x^\top A x - b^\top x$$ is only bounded below if $A$ is positive semidefinite and $b$ is in the column space of $A$. In that case, one can try to rewrite it in the form $$\frac 12 (x-c)^\top A (x-c) + \text{constant}$$ – Rodrigo de Azevedo Jun 27 '18 at 22:18
  • Thank you very much, is there a book or that explains the proof of these two conditions. – Faroq AL-Tam Jun 28 '18 at 00:05
  • @WesamF. All the material is here on MSE. It is just not that easy to find. – Rodrigo de Azevedo Jun 28 '18 at 07:31
  • how can $|\Lambda^{\frac{1}{2}} y|_2^2 = \displaystyle\sum_i \lambda_i \left(\frac{A+A^T}{2}\right) y_i^2$ be true. If $\left(\frac{A+A^T}{2}\right)$ is a matrix. – Dylan Dijk Aug 22 '23 at 07:20
  • @DylanDijk $\displaystyle\lambda_i \left(\frac{A+A^T}{2}\right)$ is a scalar, not a matrix. – Rodrigo de Azevedo Nov 03 '23 at 16:48
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Hint: If $A$ is positive definite you can diagonalize the Matrix

$$A = V^{-1}\Lambda V$$, where $V$ is the matrix of normalized eigenvectors and $\Lambda = \operatorname{diag} (\lambda_1,\cdots,\lambda_n)$ is a diagonal matrix of the eigenvalues.

Rodrigo de Azevedo
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MrYouMath
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  • The thing is that since $A$ is non symmetric, I can only prove that $\frac {A+A^T}{2}$ is positive definite. Thus, $x^T \frac {A+A^T}{2}x\ge\lambda_{min}(\frac {A+A^T}{2})|x|^2$ and as far as I know $\lambda_{min}(A)\ge\lambda_{min}(\frac {A+A^T}{2})$ – Controller May 22 '16 at 16:59
  • Then you should change your statement that $x^TAx=x^T\frac{A+A^T}{2}x>0, as this implies A symmetric. – MrYouMath May 22 '16 at 17:12
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    Why? This relation is valid for any matrix $A$ – Controller May 22 '16 at 17:16
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    @Controller Do you have a proof that says $\lambda_{\min}(A) \ge \lambda_{\min}(\frac{A+A^T}{2})$, given that $A$ is non-symmetric but with real eigenvalues? If so, please let me know. – ems Dec 24 '17 at 18:32