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 $sp(\vec k_1, \vec k_2)\subset Ker(T)$


1 Answer 1


$[T]^{\Gamma}_{\Phi}=\begin{bmatrix} 1&2&-1\\ 3&6&-3 \end{bmatrix}$

It's a rank $1$ matrix, $\dim(\ker T) =2$

Row $2-3×$ Row$1$ :

$\begin{bmatrix} 1&2&-1\\ 0&0&0\end{bmatrix}$

Hence the basis for the null space :

$\big\{\begin{bmatrix} -2\\1\\0\end{bmatrix}$, $\begin{bmatrix} 1\\0\\1\end{bmatrix}\big\}$


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