# Finding basis of kernel of a linear transformation matrix representation not in standard basis

this question from final exam.

Let $$\Gamma=(\vec v_1 ,\vec v_2, \vec v_3)$$ (that not known), and $$\Lambda = (\vec u_1, \vec u_2, \vec u_3)$$ two basis for $$V=\Re^3$$,

That $$\vec u_1=\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}, \vec u_2=\begin{bmatrix} 1\\ -1\\ 0 \end{bmatrix}, \vec u_3=\begin{bmatrix} 0\\ 0\\ 2 \end{bmatrix}$$,

also givin $$[I_V]^{\Lambda}_{\Gamma}=\begin{bmatrix} 1&1&1\\ 0&1&-1\\ 1&0&0 \end{bmatrix}$$

let $$T:\Re^3\to\Re^2$$ linaer transformation, the function not giving as it, but givin the matrix $$[T]^{\Gamma}_{\Phi}=\begin{bmatrix} 1&2&-1\\ 3&6&-3 \end{bmatrix}$$

givin that $$\Phi=(\vec w_1, \vec w_2)$$ basis for $$\Re^2$$, also $$\vec w_1=\begin{bmatrix} 5\\ 1 \end{bmatrix}, \vec w_2=\begin{bmatrix} 1\\ -5 \end{bmatrix}.$$

the question: Find basis for $$Ker(T)$$.

My Solution:

let $$\vec v_0\in\Re^3$$ therefore there is $$a_1,a_2,a_3\in\Re$$ such that $$[\vec v_0]_{\Lambda}=\begin{bmatrix} a_1\\a_2\\a_3 \end{bmatrix}$$, therefore:

$$[\vec v_0]_{\Gamma}=[I_V]^{\Lambda}_{\Gamma}[\vec v_0]_{\Lambda}$$

$$[\vec v_0]_{\Gamma}=\begin{bmatrix} 1&1&1\\ 0&1&-1\\ 1&0&0 \end{bmatrix}\begin{bmatrix} a_1\\a_2\\a_3 \end{bmatrix}=\begin{bmatrix} a_1+a_2+a_3\\a_2-a_3\\a_1 \end{bmatrix}$$

Then we got $$[T(\vec v_0)]_{\Phi}=[T]^{\Gamma}_{\Phi}[\vec v_0]_{\Gamma}$$

$$[T(\vec v_0)]_{\Phi}=\begin{bmatrix} 1&2&-1\\ 3&6&-3 \end{bmatrix}\begin{bmatrix} a_1+a_2+a_3\\a_2-a_3\\a_1 \end{bmatrix}=\begin{bmatrix} (a_1+a_2+a_3)+2(a_2-a_3)-a_1\\3(a_1+a_2+a_3)+6(a_2-a_3)-3a_1 \end{bmatrix}=\begin{bmatrix} 3a_2-a_3\\9a_2-3a_3 \end{bmatrix}$$

the following claim is based on my claim and not sure about Claim: $$[T(\vec v_0)]_{\Phi}=\vec 0 \iff T(\vec v_0)=\vec 0$$

So $$\vec v_0\in Ker(T) \iff T(\vec v_0)=\vec 0$$

$$T(\vec v_0)=(3a_2-a_3)\vec w_1+(9a_2-3a_3)\vec w_2=\vec 0 \iff 3a_2-a_3=0$$ and $$9a_2-3a_3=0 \iff 3a_2-a_3=0 \iff a_2=a_3/3$$

let $$a_1=t_1\in\Re$$ and $$a_3=t_2\in\Re$$ therefore $$a_2=t_2/3$$

So $$[\vec v_0]_{\Lambda}=\begin{bmatrix} t_1\\t_2/3\\t_2 \end{bmatrix}$$

$$\vec v_0=t_1\vec u_1+t_2/3\vec u_2+t_2\vec u_3=t_1\begin{bmatrix} 1\\2\\0 \end{bmatrix}+t_2/3\begin{bmatrix} 1\\-1\\0 \end{bmatrix}+t_2\begin{bmatrix} 0\\0\\2 \end{bmatrix}=\begin{bmatrix} t_1+t_2/3\\t_1-t_2/3\\2t_2 \end{bmatrix}=t_1\begin{bmatrix} 1\\1\\0 \end{bmatrix}+t_2\begin{bmatrix} 1/3\\-1/3\\2 \end{bmatrix}$$

let $$\vec k_1=\begin{bmatrix} 1\\1\\0 \end{bmatrix}, \vec k_2=\begin{bmatrix} 1/3\\-1/3\\2 \end{bmatrix}$$

$$\vec v_0\in Ker(T) \iff sp(\vec k_1, \vec k_2)$$

the professor claim that all that I wrote is incorrect and it's true just for $$[\vec v_0]_{\Lambda}$$, now I am not sure if it's the complexity of the solution or if there is something that I don't know.

maybe what I found is $$sp(\vec k_1, \vec k_2)\subset Ker(T)$$

$$[T]^{\Gamma}_{\Phi}=\begin{bmatrix} 1&2&-1\\ 3&6&-3 \end{bmatrix}$$
It's a rank $$1$$ matrix, $$\dim(\ker T) =2$$
Row $$2-3×$$ Row$$1$$ :
$$\begin{bmatrix} 1&2&-1\\ 0&0&0\end{bmatrix}$$
$$\big\{\begin{bmatrix} -2\\1\\0\end{bmatrix}$$, $$\begin{bmatrix} 1\\0\\1\end{bmatrix}\big\}$$