kernel of matrix in given basis Linear transformation $f$ is given in basis ${v_1, v_2, v_3}$ by matrix $M$:
$$\begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix}.$$
Find kernel and image of matrix in terms of basis ${v_1, v_2, v_3}$
Okay so in order to find kernel I transform matrix using elementary row operations and I find out that $v=(1,2,4)$ is a vector such that if I apply matrix on this vector ($Mv$) i get $0$ vector. And every sclar multiple of $v$ has same property of course, exluding them none of other vectors have this property so the kernel is one dimensional.
My concern is about what is kernel in terms of ${v_1, v_2, v_3}$. Is it $span((1,2,4))$ or is it $span((v_1+2v_2+4v_3))$ or maybe its something else.
I know if the matrix was given in standard basis kernel would be $span((1,2,4))$.
Answering my question please explain why.

Consider the matrix representation of linear transformation $f$ in the ordered basis $ \beta=\{v_{1},v_{2},v_{3}\}$ given by $$[f]_{\beta}=\begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix}$$
By definition, the image of $[f]_{\beta}$ can be find as
$${\rm im}([f]_{\beta})=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\in {\bf R}^3: \exists \begin{pmatrix} a\\b\\c\end{pmatrix}\in {\bf R}^{3}: \begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}x\\y\\z\end{pmatrix} \right\}$$
that is, solving the linear system we have
$${\rm im}([f]_{\beta})=\left\{\begin{pmatrix}x\\y\\z\end{pmatrix}\in {\bf R}^{3}: x-2y+z=0 \right\}={\rm span}\left\{\begin{pmatrix}2\\1\\0\end{pmatrix},\begin{pmatrix} -1\\0\\1\end{pmatrix}\right\}$$
Since  $$[f(v)]_{\beta}=[f]_{\beta}\cdot [v]_{\beta}, \quad (*)$$ then using $(*)$
$${\rm im}(f)={\rm span}\left\{2\cdot v_{1}+1\cdot v_{2}+0\cdot v_{3},-1\cdot v_{1}+0\cdot v_{2}+1\cdot v_{3} \right\}$$
$$\boxed{{\rm im}(f)={\rm span}\left\{2v_{1}+v_{2},-v_{1}+v_{3} \right\}}$$
By definition, the kernel of $[f]_{\beta}$ can be find as
$$\ker([f]_{\beta})=\left\{\begin{pmatrix}a\\b\\c\end{pmatrix}\in {\bf R}^{3}: \begin{pmatrix}0 & 2 & -1\\ -2 & 5 & -2\\ -4 & 8 & -3\end{pmatrix} \begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix} \right\}$$
that is, solving the linear system we have
$$\ker([f]_{\beta})=\left\{\begin{pmatrix}a\\b\\c\end{pmatrix}\in {\bf R}^{3}: y=2x, z=4x, x\in  {\bf R}  \right\}={\rm span}\left\{\begin{pmatrix}1\\2\\4\end{pmatrix} \right\}$$
Using $(*)$ we have
$${\rm ker}(f)={\rm span}\left\{1\cdot v_{1}+2\cdot v_{2}+4\cdot v_{3} \right\}$$
$$\boxed{\ker(f)={\rm span}\left\{ v_{1}+2v_{2}+4v_{3}\right\}}$$
