Understanding Eigenvectors: Solving for Eigenvalues and Corresponding Vectors

[squenshl]

Homework Statement

Let $\mathbb{R}^3\mapsto \mathbb{R}^3$ be given by $$T\begin{pmatrix}
x \\
y \\
z
\end{pmatrix} := \begin{pmatrix}
x+y-2z \\
2x-2z \\
y-x
\end{pmatrix}.$$

1. Give a basis of eigenvectors for $\text{ker}(T)$.
2. Give a basis of eigenvectors for $\text{ran}(T)$.

Relevant Equations

None

Okay so I found the eigenvalues to be $\lambda = 0,-1,2$ with corresponding eigenvectors $v = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.
Not sure what to do next. Thanks!

 

[fresh_42]

What are $\ker T$ and $\operatorname{im}T$?

 

[Mark44]

1. Give a basis of eigenvectors for $\text{ker}(T)$.
2. Give a basis of eigenvectors for $\text{ran}(T)$.

I don't think it makes sense to talk about a basis of eigenvectors for $\text{ker}(T)$ and $\text{ran}(T)$.
It's more usual for a problem to ask for a basis for each of these subspaces.

 

[fresh_42]

If $T$ is semisimple (diagonalizable) then we can find a basis of eigenvectors, in any case generalized eigenvectors. He has even calculated them already.

 

[squenshl]

Ok so the kernel of $T$ is $(x,y,z)$ such that $T(x,y,z)=0$ & this only occurs when we have $(1,1,1)$ so I guess that is the basis for the kernel right?

 

[squenshl]

So the basis for the range of $T$ are the other two eigenvectors.

 

[fresh_42]

Yes. But $(1,1,1)$ is not the only vector, all multiples are as well sent to zero. If the kernel is one dimensional, then the range is two dimensional. The other two eigenvectors are in the range. Now do they span the range? Or more generally: Are eigenvectors to distinct eigenvalues always linearly independent, and why?

 

[squenshl]

Right the basis for the kernel is the span of $(1,1,1)$. Yes eigenvectors are linearly independent so they do span the range thanks!

 

[Mark44]

Right the basis for the kernel is the span of (1,1,1).

A basis for the kernel is the vector <1, 1, 1>, not the span of this vector. $\text{Ker} (T)$ is the set of all constant multiples of <1, 1, 1>; i.e., the span of <1, 1, 1>.

 

[WWGD]

Notice that for $x$ the kernel $Ax=0 = \lambda 0$ for any $\lambda$