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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 $T:S\rightarrow\mathbb{R}^2$ by $T(A)=(a_{11},trace(A))$ 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 nulity theorem. am I in right path? in that case $dimKer(T)=n(n+1)/2 -2$

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Yes, I think you are in the right path. Except you may need to consider the trivial case when $A$ is $1\times 1$ matrix. In that case, your $T$ is given by $T(A)=(a_{11},a_{11})$, which is not surjective. – Paul Dec 21 '12 at 5:43
:-o :-o :-o :-o :(, I was given $n\ge 2$ – Taxi Driver Dec 21 '12 at 5:44
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There is always the tried and tested method of counting choices. You have a choice of every element above the diagonal, which is $\frac{1}{2}(n-1)n$ choices, and a choice for every element except two on the diagonal, which is $n-2$ more choices. Giving a total of $\frac{1}{2}(n^2+n-4)=\frac{1}{2}n(n+1)-2$ choices. – Daniel Rust Dec 21 '12 at 15:52

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