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what is the dimension of the set of $n\times n$ matrices whose sum of first row and diagonal entries are each $0$, I know that set of all trace $0$ matrices is of dimension $n^2-1$ by using rank-nullity theorem to this linear map $A\mapsto trace(A)$, so here another condition is added $x_{11}+\dots +x_{1n}=0$ along with $x_{11}+\dots+x_{nn}=0$ so if I define the linear map $A\mapsto \sum_{i=1}^{n} x_{1i}$ then clearly this is also a linear functional hence this set of matrices has the dimension $n^2-1$, so when both simultaneously happen, the dimension will be $n^2-2$? please help.

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    yes kernel is precisely those matrices I want2012-10-27

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Another way to do this question is to plug your subspaces into this formula: $\dim(S \cap T)=\dim(S)+\dim(T)-\dim(S+T).$ It is not too hard to see that if $n \ne 1$, $S+T$ is just the space of all $n \times n$ matrices, so you end up with $\dim(S \cap T)=n^2-1+n^2-1-n^2=n^2-2.$ Note that if $n=1$, the dimension is not $-1$!

Edit: If we are given a $3 \times 3$ matrix $\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix},$ we can split it up as follows: $\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix} = \begin{bmatrix}0&0&0\\d&e&f\\g&h&a+i\end{bmatrix} + \begin{bmatrix}a&b&c\\0&0&0\\0&0&-a\end{bmatrix}.$ It should be clear why this works for any $n \times n$ matrix except for the case $n=1$.

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    could you tell me why $S+T=M_n(\mathbb{R})$?2012-10-27