If $A$ is a positive semi definite matrix, is $\left[ \begin{matrix}c_1A & c_2A \\ c_3A & c_4A\end{matrix} \right]$ positive semi definite? ($c_1, c_2, c_3, c_4 > 0)$

In general, what about $\left[ \begin{matrix}c_{1,1}A & c_{1,2}A & \ldots & c_{1,n} A\\ \vdots & \vdots & \ddots & \vdots\\c_{n,1}A & c_{n,2}A & \ldots & c_{n,n}A\end{matrix} \right]$ ? ($c_{i,j} > 0$)

If the above block matrices are not semi-definite in general, is it possible to gain constraints on the constants, under which they will be semi definite?


Not necessarily. Certainly, this fails if $c_2 \neq c_3$.

Even if $c_2 = c_3$, take $A = 1$ and consider $c_1 = 1, c_2 = c_3 = 10, c_4 = 1$.

The general matrix you describe can be written in the form $$ \left[ \begin{matrix}c_{1,1}A & c_{1,2}A & \ldots & c_{1,n} A\\ \vdots & \vdots & \ddots & \vdots\\c_{n,1}A & c_{n,2}A & \ldots & c_{n,n}A\end{matrix} \right] = C \otimes A $$ Where $C$ is the matrix with entries $c_{ij}$ and $\otimes$ denotes the Kronecker product. For vectors $u,v$ of appropriate size, we have $$ (u \otimes v)^*(C \otimes A)(u \otimes v) = (u^*Cu)(v^*Av) $$ So, at the very least, $C$ must be positive semidefinite as well.

In fact, this condition is sufficient, as can be deduced from the properties of the Kronecker product given in the link. Namely: if $A,C$ are both symmetric, then so is $C \otimes A$, and if both have only non-negative eigenvalues, then so does $C \otimes A$.

  • $\begingroup$ (+1) I have added some more question to the original post. Please do give it a read as well. $\endgroup$ – TenaliRaman Sep 15 '14 at 22:09
  • $\begingroup$ @TenaliRaman see my update. Your bigger matrix will be PSD if and only if $C$ is PSD as well. $\endgroup$ – Omnomnomnom Sep 15 '14 at 22:11
  • $\begingroup$ This is an excellent piece of information! Thank you so very much! $\endgroup$ – TenaliRaman Sep 15 '14 at 22:12

Assume $A$ is diagonal. Making $R_2 := R_2 - \frac{c_3}{c_1}R_1$, we get: $$\begin{bmatrix} c_1A & c_2 A \\ \mathbf{0} & \left(c_4 - \frac{c_2c_3}{c_1}\right)A\end{bmatrix}$$

From here, making $R_1:= R_1 - \frac{c_2}{\left(c_4 - \frac{c_2c_3}{c_1}\right)}R_2$ gives: $$\begin{bmatrix} c_1A & \mathbf{0} \\ \mathbf{0} & \left(c_4 - \frac{c_2c_3}{c_1}\right)A\end{bmatrix}$$ Since $c_1 > 0$, we need only worry about $\left(c_4 - \frac{c_2c_3}{c_1}\right)$. But: $$\left(c_4 - \frac{c_2c_3}{c_1}\right) < 0 \implies c_4 < \frac{c_2c_3}{c_1}$$ If you take $c_1 = c_2 = c_3 = 1$ and $c_4 = 0.5$, the matrix will be indefinite.

  • $\begingroup$ (+1) I added some more questions to the original. Please, do give it a read as well. $\endgroup$ – TenaliRaman Sep 15 '14 at 22:08
  • $\begingroup$ I think the same reasoning will do. The expressions will get uglier, though. In fact, maybe it is possible to make $c_{ij} = 1$, for all $(i,j) \neq (n,n)$, before doing the calculations, and then see what value of $c_{nn}$ will make the matrix be indefinite. $\endgroup$ – Ivo Terek Sep 15 '14 at 22:12
  • $\begingroup$ True, the general matrix will also continue to be indefinite. @Omnomnomnom has given necessary and sufficiency conditions as well. Thank you so very much to both of you for your answers. $\endgroup$ – TenaliRaman Sep 15 '14 at 22:14
  • $\begingroup$ Don't mind it. Glad to help. $\endgroup$ – Ivo Terek Sep 15 '14 at 22:16

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