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

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.

• For part C, you should be able to spot two eigenvectors of $L$ and their associated eigenvalues without constructing $L$ explicitly. – amd Jan 19 at 1:23

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.
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.