# If $A$ and $B$ are symmetric matrices, so is $A+B$

This question looks pretty easy and hard at the same time. Here's how it goes:

Let A and B be symmmetric nxn matrices. Show that A + B is also symmetric.

To me this sounds like a pretty obvious fact, well, if its not obvious I still have a good idea of why this is true, but I'm new to university level maths and I don't really know how I would go about writing a formal proof for this. Any help would be much appreciated. =D

• FFR, it's best to echo the question in the title, not the tags. Sep 6, 2014 at 11:07
• I assume your difficulty that the solution is so totally obvious that you are unclear what you are expected to write? This kind of question is simply bad. It is a lazy way of trying to find out if you have grasped what two definitions mean (symmetric and matrix addition). So you just put down the minimum to show that you have grasped them. Ukhrir's answer is fine. But something shorter than that would also be fine. Sep 6, 2014 at 11:10

This is how I would write a proof of this statement:

Let $A=(a_{ij})_{i,j=1}^n$,$B=(b_{ij})_{i,j=1}^n$ be symmetric matrices, then it holds that $a_{ij}=a_{ji}$ and correspondingly for $B$. Then consider the sum $C=A+B$, then $c_{ij}=a_{ij}+b_{ij}$, and $c_{ji}=a_{ji}+b_{ji}$. Then since $A$, $B$ are both symmetric $a_{ji}+b_{ji}=a_{ij}+b_{ij}$ and thus $c_{ji}=c_{ij}$ and therefore $C$ must be symmetric.

Hint: Prove that the transpose of the sum of two matrices is the sum of their transposes.

If $C=A+B$ then $C_{ij}=A_{ij}+B_{ij}=A_{ji}+B_{ji}=C_{ji}$. So $C$ is symmetric.

Hint: write down the symmetric matrices explicitly, that is $$A= \begin{pmatrix}a_{11}&a_{12}&a_{13}&\cdots&a_{1n}\\a_{12}&a_{22}&a_{23}&\cdots&a_{2n}\\a_{13}&a_{23}&a_{33}&\cdots&a_{3n}\\\vdots&\vdots&\vdots&\ddots&\vdots\\a_{1n}&a_{2n}&a_{3n}&\cdots&a_{nn}\end{pmatrix},\qquad B=\begin{pmatrix}b_{11}&b_{12}&b_{13}&\cdots&b_{1n}\\b_{12}&b_{22}&b_{23}&\cdots&b_{2n}\\b_{13}&b_{23}&b_{33}&\cdots&b_{3n}\\\vdots&\vdots&\vdots&\ddots&\vdots\\b_{1n}&b_{2n}&b_{3n}&\cdots&b_{nn}\end{pmatrix}$$

and compute $A+B$ to see what is going on. For the formalities, if you have troubles, look at the answer of Uhkrir.

a) Since (A+B)^T = A^T + B^T, A = A^T and B = B^T, and you have that (A+B)^T = (A+B) That is, A + B is symmetric.

I thought of doing it by writing down the matrices out in full but I thought that would be messy. This looks like a good answer.

Thanks for replies tho!!