Consider Linear transformation Given the linear transformation $ f: \mathbb{R}^{4\times 4} \to  \mathbb{R}^{4\times 4}$ defined by $A \to f(A)= A + A^T$

How do you determine the dimensions of $\text{Ker}(f)$ and $\text{Img}(f)$?
First, you need to prove that the image of $f$ is precisely the space of symmetric matrices: the matrices $B$ for which $B=B^T$, or in other words those of the form
$$\begin{bmatrix}b_1&amp;b_2&amp;b_3&amp;b_4\\b_2&amp;b_5&amp;b_6&amp;b_7\\b_3&amp;b_5&amp;b_8&amp;b_9\\b_4&amp;b_6&amp;b_9&amp;b_{10}\end{bmatrix}$$

The dimension of the space of symmetric $4\times 4$ matrices is $10$: They are uniquely determined by the upper triangular part (including the diagonal), which contains 10 entries ($b_1,\ldots,b_{10}$ above). To find a basis of this space, for each $i=1,\ldots,10$, set $b_i=1$ and all $b_j=0$, for $j\neq i$ in the matrix above.

In other words, $\dim(im(f))=10$.

By the rank-nullity theorem $\dim(\ker(f))=\dim(\mathbb{R}^{4\times 4})-\dim(im(f))=16-10-6$.



Alternatively, you can notice that $A\in\ker(f)$ iff $A=-A^T$, so $A$ has the form
$$\begin{bmatrix}0&amp;a_1&amp;a_2&amp;a_3\\-a_1&amp;0&amp;a_4&amp;a_5\\-a_2&amp;-a_4&amp;0&amp;a_6\\-a_3&amp;-a_5&amp;-a_6&amp;0\end{bmatrix}.$$

So the dimension of $\ker(f)$ is $6$. Again by the rank-nullity theorem, $\dim(im(f))=10$.
Clearly $Ker(f)=\{A\in M(4,4,\mathbb{R})\,|\, A^T=-A\}$, the space of antisymmetric matrix that has dimension $6$(Skew-symmetric matrix). The image is clearly contained in the space of symmetric matrix, that has dimension $10$, and so by a famous theorem(Dimension Theorem) we can conclude by dimension argument