I have a bit of an issue with determining the kernel of a function when the function is represented by matrices in two different bases. I give below an example to illustrate my problem.
A basis of $\mathbb{R}^3$ is given as $B=\lbrace\vec{b}_1,\vec{b}_2,\vec{b}_3 \rbrace$ with $\vec{b}_1 = [1,0,1]$, $\vec{b}_2 = [1,1,1]$, $\vec{b}_3 = [0,1,1]$. A function $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ is given as $f(\vec{b}_1)=[2,4,5]$, $f(\vec{b}_2)=[0,-2,-4]$ and $f(\vec{b}_3)=[2,0,-3]$. The questions are as follows:
(a) Give the matrix representation of $f$ $A^f_{\mathcal{B}\mathcal{B}_0^3}$ where $\mathcal{B}_0^3$ refers to the standard basis in $\mathbb{R}^3$.
(b) Give $A^f_{\mathcal{B}_0^3\mathcal{B}_0^3}$.
(c) Give $A^f_{\mathcal{B}\mathcal{B}}$.
(d) Is $f$ injective, surjective, bijective?
Here are my attempts at answering each question:
(a) $A^f_{\mathcal{B}\mathcal{B}_0^3} = \begin{bmatrix} 2 & 0 & 2 \\ 4 & -2 & 0 \\ 5 & -4 & -3 \end{bmatrix}$
(b) $A^f_{\mathcal{B}_0^3\mathcal{B}_0^3} = A^f_{\mathcal{B}\mathcal{B}_0^3} \left( T_{\mathcal{B}\mathcal{B}_0^3} \right)^{-1}$ where $\left( T_{\mathcal{B}\mathcal{B}_0^3} \right)^{-1}$ is the inverse of the matrix $T_{\mathcal{B}\mathcal{B}_0^3}=\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$.
That results in $A^f_{\mathcal{B}_0^3\mathcal{B}_0^3} = \begin{bmatrix} -2 & -2 & 4 \\ -2 & -6 & 6 \\ -1 & -9 & 6 \end{bmatrix}$
(c) $A^f_{\mathcal{B}\mathcal{B}} = T_{\mathcal{B}_0^3\mathcal{B}} A^f_{\mathcal{B}\mathcal{B}_0^3} = \left( T_{\mathcal{B}\mathcal{B}_0^3} \right)^{-1} A^f_{\mathcal{B}\mathcal{B}_0^3} = \begin{bmatrix} 1 & -2 & -3 \\ 1 & 2 & 5 \\ 3 & -4 & -5 \end{bmatrix}$
(d) Now here comes the problem. If $f$ is injective, then $\text{Ker } f = \lbrace 0 \rbrace$. So I want to determine the kernel of $f$. I would think that the kernel would be independent of the basis, since the function is still the same, however maybe I'm wrong since I get the two following results for $A^f_{\mathcal{B}\mathcal{B}}$ and $A^f_{\mathcal{B}_0^3\mathcal{B}_0^3}$:
$\text{Ker } A^f_{\mathcal{B}\mathcal{B}} = \left\lbrace \mathbb{R} \cdot \begin{bmatrix} -1 \\ -2 \\ 1 \end{bmatrix} \right\rbrace$
$\text{Ker } A^f_{\mathcal{B}_0^3\mathcal{B}_0^3} = \left\lbrace \mathbb{R} \cdot \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} \right\rbrace$
Both kernels indicate that the function is not injective, but they have different basis. Did I make a mistake somewhere? Is that normal? The next question asks to give a basis of the kernel, so I think something went wrong here.
Thank you in advance for your suggestions.
Julien.
Yes the kernel is independent of the basis.
I didn't check you computations, but what you got was that $\ker f$ is spanned by the vector whose coordinates with respect to the basis $\mathcal B$ are $-1$, $-2$, and $1$ and also by the vector whose coordinates with respect to the basis $\mathcal{B}_0^3$ are $3$, $1$, and $2$. These are not two distinct answers. These are the ways of describing the same line with respect to two distinct basis.