I have been doing the following exercise problem:
Let $T: R^3-> R^3$ be a linear transformation so that:
Determine eigenvalues and eigenspace of T.
So, I determined that $0$ and $1/2$ are eigenvalues, with eigenvectors $(1,1,1)$ and $(0,2,0)$ respectively. But the unclear part is as follows: It says in the solutions, apart from this, that:
"Vector $(0,1,1)$ is not an eigenvector. But we see that $2*T([0,2,0]) = T([0,1,0])$, which means that: $2*[0,2,0]-[0,1,0]=[0,3,1]$ is in nullspace of T.
Thus, eigenspace for eigenvalue 0 has the base $([0,1,0],[0,3,1])$. Eigenspace for eigenvalue 1/2 is $([0,2,0])$."
What I do not understand is: Why is $[0,3,1]$ an eigenvector for eigenvalue $0$?
Thanks!
Best Answer
The solutions are wonky, and have some computational errors. I think they meant the vector $(0, 3, -1)$ instead. Here's how it works (remembering $T$ is linear): \begin{align*} T\begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix} &= T\left( 2\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \right) \\ &= 2T\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} - T\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \\ &= 2\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \\ &= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = 0\begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix}. \end{align*}
I know the question gave you two eigenvectors immediately, but beware of concluding from this that there aren't any others! In order to find all the eigenvectors, you may need to find some that aren't listed.
In this case, you were given the subtle clue that the linearly independent list $((1, 1, 1), (0, 2, 0), (0, 1, 1))$ mapped to a very linearly dependent list $((0, 0, 0), (0, 1, 0), (0, 2, 0))$. The first list is a basis for $\mathbb{R}^3$, and in particular, spans all of $\mathbb{R}^3$. The range of $T$ is the span of the second list, being the image of a spanning list. Notice how all three vectors are parallel, and so the range is $1$-dimensional. That is to say, $$\operatorname{Rank}(T) = 1.$$ By the Rank-Nullity theorem, $$\operatorname{Rank}(T) + \operatorname{Null}(T) = 3 \implies \operatorname{Null}(T) = 2.$$ In other words, there should be two linearly independent eigenvectors corresponding to $0$. The other one can be found by finding a non-trivial linear combination such that $$a_1 \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + a_2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + a_3 \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix},$$ (e.g. $a_1 = 0$, $a_2 = 2$ and $a_3 = -1$, as used in the solutions), and then form the same combination of the vectors that mapped to $(0, 1, 0)$ and $(0, 2, 0)$, i.e. $$a_1 \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + a_2\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} + a_3 \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}.$$ By linearity, such a vector must map to $T(0, 0, 0) = (0, 0, 0)$.