My question is:
Let $M = \mathbb{R}^{5\times 5}$ and let $T:M \to M$ denote the operator $$T (A)=\frac12(A-A^T)$$ where superscript $T$ is the matrix transpose. How can I find kernel, range, nullity, rank, real eigenvalues and eigenvectors?
Firstly, I checked that $T$ preserves sums and multiplication by scalars, so it is a linear map. $T(A+B)=T(A)+T(B) T(c.A)=c.T(A)$. If this matrix was $2\times 2$, I could have let $A= \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$ and then calculated $$T(A)=\frac12(A-A^T)$$ to find kernel, range, nullity, rank.
But this matrix is $(5\times 5)$. So I can't draw like $2\times 2$ matrix and could not calculate $$T (A)=\frac12(A-A^T)$$ Can you suggest another way for me to calculate these values(kernel, range, nullity, rank, real eigenvalues and eigenvectors)?
Some hints:
(1) I leave it to you to argue that the kernel of $T$ is the subspace of symmetric matrices. It is not difficult to show that the space of symmetric $n \times n$ matrices has dimension $\frac{n(n+1)}{2}$. Since $n=5$, $\text{dim(ker}(T)) = \frac{5 \cdot 6}{2} = 15$.
(2) $T(A)$ must be a matrix with zeros along its diagonal (why?). That tells you something about the range. Also, by the Rank-Nullity Theorem, $\text{rank}(T) = \text{dim}(M) - \text{dim(ker}(T)) = 25 - 15 = 10$. I will also leave it to you to give a more explicit description of the range based on these facts.
(3) You can find the eigenvalues and eigenvectors by considering the equation $T(A) = \lambda A$ for $A \neq \mathbf{0}$, that is
$$\frac{1}{2}(A- A^\top) = \lambda A \iff (\frac{1}{2} - \lambda)A = \frac{1}{2} A^\top$$.
Since the diagonal elements of $A$ and $A^\top$ are the same, we must have $(\frac{1}{2} - \lambda) a_{ii} = \frac{1}{2}a_{ii} \implies -\lambda a_{ii} = 0$ for all $i$. This means that $\lambda = 0$ is the only possible eigenvalue, whose eigenspace is the kernel.