Determine a basis for $\operatorname{Im}f$ and $\operatorname{Ker}f$

Question

Let $B=(\underline{e_1}, \underline{e_2}, \underline{e_3}, \underline{e_4})$ a basis for a vector space $V$ on $\mathbb{R}$. Given the map $$ f:V\longrightarrow V,\qquad f(\underline x)=-\underline x, $$ determine a basis for $\operatorname{Im}f$ and $\operatorname{Ker}f$.

My attempt. I compute the matrix associated with $f$ and it easily results that $$ M^{B, B}(f)=\begin{pmatrix}-1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}. $$ So, a basis for $\operatorname{span}((-1, 0, 0, 0), (0, -1, 0, 0), (0, 0, -1, 0), (0, 0, 0, -1))$ is $((-1, 0, 0, 0), (0, -1, 0, 0), (0, 0, -1, 0), (0, 0, 0, -1))$ and hence a basis for $\operatorname{Im}f$ is $$ B_{\operatorname{Im}f}=(-\underline e_1, -\underline e_2, -\underline e_3, -\underline e_4). $$ For the kernel, it easily results that $\operatorname{Ker}f=\{0\}$. Is my attempt correct?

Thank You

Answer 1

It is correct. Also you could see that since $f$ is an automorphism a basis for $Im(f)$ is the normal $\{e_1,e_2,e_3,e_4\}$.

Answer 2

Note that $\text{im} \;f = V$, and $\dim \text{im}\; f + \dim\text{ker} \; f = \dim V = 4$. So indeed $\dim \text{ker} \; f = 0$, which means $\ker \; f = \{0\}$.