dimension of the subspace $W$ of $10\times 10$ matrices , $W=\{[a_{ij}]: a_{ij}=0\text{ if i is even }\}$ Could any one tell me how to find the dimension of the subspace $W$ of $10\times 10$ matrices , $W=\{[a_{ij}]: a_{ij}=0\text{ if i is even }\}$
from the given data I see that row  $2,4,6,8,10$ are $0$, and then  it becomes a $5\times 10$ matrix, so dimension $50$?
am I right? 
 A: The difficulty in this question is to write one thing that is clear so let's do it even so. We know that the standard basis of $\mathcal M_{10}(\mathbb R)$ is $\mathcal B=(E_{ij})_{1\leq i,j\leq10}$ where $E_{ij}=(e_{pq})_{1\leq p,q\leq10}$ is the matrix defined by the Kronecker symbol $\delta$
$$e_{pq}=\delta_{ip}\delta_{jq}$$
and for all matrix $A=(a_{ij})\in \mathcal M_{10}(\mathbb R)$ we have
$$A=\sum_{i=1}^{10}\sum_{j=1}^{10}a_{ij}E_{ij} $$
Now if $A\in W$ then
$$A=\sum_{\substack{i=1\\ i\ \text{ odd}}}^{10}\sum_{j=1}^{10}a_{ij}E_{ij} $$
so $\mathcal B'=(E_{ij})_{\substack{i\in\{1,3,5,7,9\}\\1\leq j\leq10}}$ is basis for $W$ so $\dim W=50$.
A: It does remain a $10 \times 10$ matrix, but because of all the zero rows, you can get away with $50$ independent "vectors" (which in this case are all matrices) to form a basis.  You have to be able to construct any given matrix from a linear combination of the basis vectors.  As suggested in the comments below, the basis should consist of 10x10 matrices each of which has a 1 in a cell which is or can be non-zero; and zeros elsewhere.
