I am wondering if this true: $\lambda_{\min}(A) \ge \lambda_{\min}(\frac{A+A^T}{2})$, given that $A$ is non-symmetric but with real eigenvalues. I came across this inequality in one of the math-stackexchange posts but wonder why it is true? I did MATLAB simulations with for many rand(2,2) matrices and it seems to hold up but is not sufficient to be taken as a fact. Please let me know.


Fact. If $S$ is a real symmetric matrix (of size $n$), then $$\forall X\in\mathbb{R}^n,\ X^T\,S X\geq\lambda_{\min}\|X\|^2,$$ where $\lambda_{\min}$ is the minimum eigenvalue of $S$.

Proof. We know that a real symmetric matrix is diagonalizable in an orthonormal basis. Let $(X_1,\ldots,X_n)$ be an orthonormal basis of eigenvectors of $S$, with $X_k$ associated with the eigenvalue $\lambda_k$. Now let $X\in\mathbb{R}^n$ and decompose it on the basis: there exists $x_1,\ldots,x_n\in\mathbb{R}^n$ such that $X=x_1X_1+\cdots+x_nX_n$. Then $$X^T SX=\lambda_1x_1^2\|X_1\|^2+\cdots+\lambda_nx_n^2\|X_n\|^2\geq\lambda_{\min}\|X\|^2.$$

Now let $A$ be a square real matrix with real coefficients. Let $\lambda$ be a real eigenvalue of $A$ and let $X_\lambda$ be an associated eigenvector. Then $$X_\lambda^T AX_\lambda=\lambda\|X_\lambda\|^2.$$ Now, transposing this one-by-one matrix yields $$X_\lambda^TA^TX_\lambda=\lambda\|X_\lambda\|^2$$ too, hence $$X_\lambda^T\left(\frac{A+A^T}2\right)X_\lambda=\lambda\|X_\lambda\|^2.$$ Hence, from the preliminary fact, and since $S=\dfrac{A+A^T}2$ is a real symmetric matrix, we must have $$\lambda\|X_\lambda\|^2\geq\lambda_{\min}\|X_\lambda\|^2$$ where $\lambda_{\min}$ is the minimal eigenvalue of $S$. Since $X_\lambda\neq0$ we conclude that $\lambda\geq\lambda_{\min}$ i.e., that:

every real eigenvalue of $A$ is non-less than $\lambda_{\min}$.

You can generalize it slightly with the non-real eigenvalues of $A$ too: let $\lambda\in\mathbb{C}$ be an eigenvalue of $A$ and let $X_\lambda\in\mathbb{C}^n$ be an eigenvector of $A$ associated with $\lambda$. Then: $$\overline{X_\lambda^T}AX_\lambda=\lambda\|X_\lambda\|^2,$$ and also (transpose and take the conjugate, using the fact that $A$ has real coefficients): $$\overline{X_\lambda^T}A^TX_\lambda=\overline{\lambda}\|X_\lambda\|^2.$$ Hence $$\overline{X_\lambda^T}\left(\frac{A+A^T}2\right)X_\lambda=\Re(\lambda)\|X_\lambda\|^2.$$ Extending the preliminary fact to complex vectors (and taking the associated hermitian product) yields $\Re(\lambda)\geq\lambda_{\min}$, i.e.,

the real part of every (complex) eigenvalue of $A$ is non-less than $\lambda_{\min}$.

  • $\begingroup$ Thanks for your answer. I will go through it carefully. By the way, by similar argument, can we also say $\lambda_{\max}(A) \le \lambda_{\max}(\frac{A+A^T}{2})$ $\endgroup$ – ems Dec 24 '17 at 19:50
  • $\begingroup$ @ems: Yes of course! $\endgroup$ – gniourf_gniourf Dec 24 '17 at 20:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.