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
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.
I found this here: http://mymathforum.com/linear-algebra/8319-if-b-symmetric-show-b-also-symmetric.html
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!!
