I have a question about positive semidefinite matrices that are non diagonalizable. Example: \begin{equation} A= \left(\begin{array}{cc} 2 & 1\\ 0 & 2\\ \end{array}\right) \end{equation} Clearly the (real part of the) eigenvalues of $A$ are non-negative.

But how do I prove in general that the real part of the Eigenvalues of a positive semi-definite real matrix are non-negative? (I have seen the proof where they use diagonalization of the matrix ($B=T^{-1}DT$) but this is not possible for all positive semi-definite real matrices.)

  • 1
    $\begingroup$ Hint: use eigenvectors. $\endgroup$ – TZakrevskiy Jan 30 '14 at 19:53
  • 3
    $\begingroup$ Ok. I guess you mean the following: Let v be an Eigenvector and l the corresponding Eigenvalue of A: Since x'Ax>=0 has to hold for all x it also has to hold for the Eigenvectors v. Hence v'Av=v'lv=l*v'*v>=0 <=> l>=0. Is this correct like this? $\endgroup$ – user137589 Jan 31 '14 at 22:45
  • 1
    $\begingroup$ Yes, your reasoning is correct. $\endgroup$ – TZakrevskiy Feb 1 '14 at 8:28
  • $\begingroup$ @user137589 Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. $\endgroup$ – Julian Kuelshammer Jun 2 '14 at 12:20
  • $\begingroup$ Finally added my comment as an answer. $\endgroup$ – user137589 Aug 16 '15 at 9:32

Of course, we assume that $A$ is a real matrix. The matrix $A$ may have non-real eigenvalues as this one $A=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$. The correct result is:

If $A$ is real and for every $x\in\mathbb{R}^n$, $x^TAx\geq 0$, then the eigenvalues of $A$ have a non-negative real part.

cf. Does non-symmetric positive definite matrix have positive eigenvalues?

Now, I think that to call such a matrix $A$, a positive semi-definite matrix, is a very bad idea. Note that this condition is equivalent to $A+A^T$ is $\geq 0$ in the usual sense. Moreover we encounter questions about this subject on MSE and often the OP does not even report that the studied matrix is not assumed to be symmetric!!

  • $\begingroup$ What do you want to say with this answer (I guess you want to say that something is wrong but I dont get the point)? Should I state the title more precicely (PSD but non-diagonalizable real matrix - proof that real parts of eigenvalues are non-negative)? Or what would be the correct formulation and the corresponding proof? $\endgroup$ – user137589 Aug 18 '15 at 11:23
  • 1
    $\begingroup$ @ user137589 , if you do not see the difference between $\lambda\geq 0$ and $Re(\lambda)\geq 0$, then I cannot help you. $\endgroup$ – loup blanc Aug 18 '15 at 17:27
  • $\begingroup$ Yes I see that. I changed the question. Please write down the correct and complete proof which answers my question. I would appreciate it. Thanks. $\endgroup$ – user137589 Aug 20 '15 at 13:00
  • 1
    $\begingroup$ See my reference above and, inside, the Hui's answer. $\endgroup$ – loup blanc Aug 20 '15 at 13:11

The following answer is copied and modified from achille hui's answer which can be found following the link in loup blanc's answer (Does non-symmetric positive definite matrix have positive eigenvalues?).

For the sake of completeness, I copied it here:

Let $A \in M_{n}(\mathbb{R})$ be any (non-symmetric) real $n\times n$ matrix but "positive semi-definite" in the sense that:

$$\forall x \in \mathbb{R}^n, x \ne 0 \implies x^T A x \geq 0$$ The eigenvalues of $A$ need not be positive. For an example:


has eigenvalue $1 \pm i$. However, the real part of any eigenvalue $\lambda$ of $A$ is always positive.

Let $\mathbb{C} \in \lambda = \mu + \nu i$ where $\mu, \nu \in \mathbb{R}$ be an eigenvalue of $A$. Let $z \in \mathbb{C}^n$ be a right eigenvector associated with $\lambda$. Decompose $z$ as $x + iy$ where $x, y \in \mathbb{R}^n$.

$$(A - \lambda) z = 0 \implies \left((A - \mu) - i\nu\right)(x + iy) = 0 \implies \begin{cases}(A-\mu) x + \nu y = 0\\(A - \mu) y - \nu x = 0\end{cases}$$ This implies

$$x^T(A-\mu)x + y^T(A-\mu)y = \nu (y^T x - x^T y) = 0$$

and hence $$\mu = \frac{x^TA x + y^TAy}{x^Tx + y^Ty} \geq 0$$

In particular, this means that all real parts of all eigenvalues $\lambda$ of $A$ are non-negative.

  • $\begingroup$ I did not downvote your post. Yet it is false; cf. my answer. $\endgroup$ – loup blanc Aug 17 '15 at 17:42
  • $\begingroup$ It is ok. I just wanted a correct answer to my question (which was not clearly stated at the beginning). Thanks for your answer which finally seems correct to me! $\endgroup$ – user137589 Aug 20 '15 at 14:35

I guess PSD matrices are symmetric and a symmetric matrix is orthogonally diagonalizable.We allways consider symmetry because otherwise eigen values can be complex and then it loses the essence of psd.

  • $\begingroup$ No that is not true. PSD matrices are not necessarily symmetric. Example: A= \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} $\endgroup$ – user137589 Jan 31 '14 at 22:36

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.