# Positive-definiteness of block diagonal matrix

Given a block diagonal matrix $A$ like so:

$$A= \begin{bmatrix} B & & \\ & C & \\ & & D \\ \end{bmatrix}$$

Given $B$ is symmetric and has real, positive eigenvalues (obviously $B$ is positive-definite), $C$ is

$$C = \begin{bmatrix} 11 & -4 \\ 0 & 2 \end{bmatrix}$$

and $D$ is

$$D = \begin{bmatrix} 11 & -3 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix}$$

I'm asked to determine whether $C$, $D$ and $A$ are positive-definite. Well, $C$ and $D$ are asymmetric, so obviously we need to stretch the rules and use the definition of positive-definiteness that includes asymmetric matrices, which states that an asymmetric matrix is positive-definite iff its symmetric part (the sum of the matrix plus its transpose, divided by $2$ (optionally)) is positive-definite.

So, calculate the symmetric parts of $C$ and $D$, see if they're positive-definite, easy enough. However, not knowing the data of $B$ (besides its properties) doesn't allow me to calculate the symmetric part of $A$, so I need to work by properties only. What am I not seeing here?

• Ouch. Forgot that. Yes, it's the block matrix itself. Will edit now. Nov 26, 2012 at 19:43
• You can apply the symmetric component rule to $A$, because its symmetric part is just the block diagonal made of the symmetric parts of $B$, $C$, and $D$ (do you see why?). Then note that if the blocks of a block-diagonal matrix are p.d., then so is the matrix. Nov 26, 2012 at 19:46
• Yeah, I see why. Answer now seems so obvious. You should post your comment as an answer so that I can pick it. Thanks a lot. Nov 26, 2012 at 19:48

## 3 Answers

Posting as answer by request.

You can apply the symmetric component rule to $A$, because its symmetric part is just the block diagonal made of the symmetric parts of $B$, $C$, and $D$. Then note that if the blocks of a block-diagonal matrix are p.d., then so is the matrix.

You already know that $B$ is p.d., and as user1551 stated, $B$ is itself its symmetric component. It just remains to show that $C$ and $D$ have p.d. symmetric components, but you've done that already.

The symmetric part of $B$ is $B$ itself, because $B$ is symmetric.

• As long as you take the option to divide by $2$, at least. Nov 26, 2012 at 20:04
• The symmetric part by definition requires the division by 2. The optional part was my addition, in order to say that positive scalar multiplication wouldn't affect anything here. Nov 26, 2012 at 23:13
• @CameronBuie Ha ha, you're right. I was too rush to give an answer. Nov 27, 2012 at 9:40

The determinant of a diagonal block matrix is the product of the determinants of the blocks. So, if you apply this to the way you calculate the characteristic polynomial, you will get what you want.

EDIT: A matrix is positive definite if it has positive eigenvalues. You already know that $B$, $C$, and $D$ have positive eigenvalues. But, the eigenvalues of $A$ can be obtained from $B$, $C$, and $D$ (which uses user1551's answer).

• Thanks for the answer, but my limited knowledge doesn't allow me to fully understand it. I'm aware of the property you mention, but I don't know how to apply it to my problem. Nov 26, 2012 at 19:47
• @LefterisAslanoglou I have edited to add an explanation.
– J126
Nov 26, 2012 at 19:52
• Regarding your edit, positive eigenvalues are a criterion for symmetric matrices. So you're saying that if C and D have positive eigenvalues, so will their symmetric parts, and so will the symmetric part of A? Nov 26, 2012 at 19:54