Finding values for which a linear transformation $L: \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ satisfies $v_1, v_2 \in ker(L)$ and $L(v_3) = (1,1,1)$

Question

I am working on an old exam question as preparation for my exam this coming week. The question states that $v_1 = (1, a, a^2), v_2 = (a^2, 1, a), v_3 = (a, a^2, 1)$.

Part A asked to find the values of $a$ for which $\{v_1, v_2, v_3 \}$ is a basis of $\mathbb{R}$. I did this by setting up $v_1, v_2, v_3$ in a matrix, finding the determinant and seeing for which values of $a$ it is equal to zero. I found that $\{v_1, v_2, v_3 \}$ is a basis of $\mathbb{R}$ if $a \neq 1$.

Part B asks: for which values of $a$ is $L: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ a linear map so that $$v_1, v_2 \in ker(L) \text{ and } L(v_3) = (1, 1, 1)$$ I do not really have an idea of where to get started with this problem. I know that $L(v_1) = 0$ and $L(v_2) =0$, but I'm not sure where to go with the problem from there.

Part C asks for the eigenvalues and eigenvectors of $L$, for the values of $a$ given in part B. I assume this is straightforward if I know what the linear map is.

Answer 1

If $a\neq1$, then $\{v_1,v_2,v_3\}$ is a basis, and therefore there is one and only one linear map $L\colon\mathbb{R}^3\longrightarrow\mathbb{R}^3$ such that $L(v_1)=(0,0,0)$, that $L(v_2)=(0,0,0)$, and that $L(v_3)=(1,1,1)$.

Otherwise, $v_1=v_2=v_3=(1,1,1)$. So, no such linear map $L$ exists, since $L(1,1,1)$ cannot be equal to $(0,0,0)$ and to $(1,1,1)$ simultaneously.

Answer 2

Part B : The information you provided allows you to explicitly write the Matrix for the linear transformation L in the basis $B=[v_1, v_2, v_3].$ $$ [L]_B =\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix} $$

Part C: You can remember that if a non zero vector is in the kernel of a linear map that also means that such vector is an eigenvector for the eigenvalue 0. So you already have two eigenvector $ (v_1,v_2) $ which are linearly independent. Once you have made these observations you can see if there are other eigenvalues and finish the exercise.