Say we have a symmetric matrix $A \in R^{n\times n}$ with all non-negative diagonals.

Then, to my understanding, $$\operatorname{trace}(A) = \sum_{i=1}^n A_{ii} = \sum_{i=1}^n \lambda_i \ge 0$$

Because $A$ is symmetric, by the spectral decomposition theorem it can be written as $$A = UDU^T $$ where U is an orthogonal matrix with the eigenvectors of A as columns and $D = \operatorname{diag}(\lambda_1, \lambda_2, ... , \lambda_n)$.

We know that the condition for $A$ to be PSD is $x^TAx\ge0 $ for any $x\in R^n$. Then $$x^TUDU^Tx \ge 0 $$

If we define vector $y=U^Tx$, this becomes $$y^TDy\ge0$$ or $$x^TAx=\sum_{i=1}^n\lambda_iy_i^2$$

Because we established above that $\sum_{i=1}^n\lambda_i\ge0$ and we know that $y_i^2\ge0$ for all $i=1,\ldots,n$, we know that the above expression for $x^TAx$ must be $\ge0$ and thus matrix A is PSD.

What am I missing? Why isn't this sufficient to show that $A$ is PSD?

Thank you so much.

  • $\begingroup$ You only know that the sum of the eigenvalues is non-negative, but not the individual eigenvalues. Take, for example, $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$. The sum of the eigenvalues is $1+1=2$, but the product of the eigenvalues, the determinant, is $-3$. For this matrix, if you take $x=(-1,1)^T$, then $Ax=-x$, then $x^TAx=-\|x\|^2=-2<0$. $\endgroup$
    – user752802
    Feb 25, 2020 at 18:46

1 Answer 1


Suppose that $\lambda_1=-1$ and that $\lambda_2=2$. Then $\lambda_1+\lambda_2=1\geqslant 0$. However,$$\lambda_1\times3^2+\lambda_2\times1^2=-1<0.$$So, no, it doesn't follow from $\lambda_1+\lambda_2+\cdots+\lambda_n\geqslant0$ that you always have$$\lambda_1y_1^{\,2}+\lambda_2y_2^{\,2}+\cdots+\lambda_ny_n^{\,2}\geqslant0$$


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.