linear algebra basis of spac3 and dimensions

Question

For a square matrix A of size n, its trace, denoted by tr A, is the sum of all diagonal elements, i.e. trA = [A]11 + [A]22 + ... + [A]nn. Show that the set of all square matrices of size 5 whose trace is 0 form a vector space. Find a basis of the space and its dimension. I already answered the first part, i just need to find basis and dimension. I think the dimension is 5 but im not sure about the rest

Answer 1

Since $[A]_{5,5}=-([A]_{1,1}+[A]_{2,2}+[A]_{3,3}+[A]_{4,4})$, the subspace has dimension $24$.

A basis is given by the $20$ matrices that have a $1$ in the $i,j$th position, for$i\not=j$, and zeros everywhere else; together with four matrices with a $1$ in the $i,i$th place, $i\not=5$, and a $-1$ in the $5,5$ place, and zeros everywhere else...

Remember the space $M_{5×5}$ has dimension $25$. Your subspace is the kernel of the linear transformation trace, $\text{ tr}:M_{5×5}\to\mathbb R$, which takes a linear transformation to it's trace...The rank of trace is $1$. So the nullity is $24$.