I want to know the dimension of the space of all symmetric matrices with trace $0$ and $a_{11}=0$,
I can show that the dimension of space of all symmetric matrices $S$ is $n(n+1)/2$, now I give a linear map $T\colon S\rightarrow\mathbb{R}^2$ by $T(A)=(a_{11},\operatorname{trace}(A))$ so the kernel is exactly the space I want,
so now it is enough to show the map is surjective so that I can apply rank nullity theorem. am I in right path?
in that case $\dim\ker(T)=n(n+1)/2 -2$.
Best Answer
As Paul already noted, you are on the right path (although you have to assume that $n\geq 2$ as of course the zero matrix is the only matrix with $a_{11}=0$).
Now, how to show that this map is surjective if $n\geq 2$: Let $(r,s)\in \mathbb{R}^2$. Then define the symmetric matrix $A$ by $a_{11}=r$ and $a_{nn}=s-r$ and all other entries equal to zero. Then $T(A)=(r,s)$, so $T$ is surjective. Now you can apply rank-nullity and get the result you stated.
To comment on the solution that Daniel Rust gave in the comments: To define a symmetric matrix you have to give a value to each diagonal entry and to each entry above the diagonal (the values below the diagonal are then determined by $a_{ij}=a_{ji}$. There are $n$ entries on the diagonal. But $a_{11}=0$ by assumption, so there is no choice and also say, for $a_{nn}$ there is no choice as $0=\operatorname{trace}(A)=\sum a_{ii}$. Thus $a_{nn}=-\sum_{i\neq n}a_{ii}$. So you are left with $n-2$ choices.
For the part above the diagonal note that there are $n^2$ entries of a matrix in total. Of these all but $n$ don't lie on the diagonal, that's $n^2-n$. Exactly half of them lie above the diagonal, so that's $\frac{1}{2}(n^2-n)$. For all of them you have no constraints from your conditions. Adding all choices up, you have $(n-2)+\frac{1}{2}(n^2-n)=\frac{1}{2}n(n+1)-2$.