I had a question about the relationship between eigenvalues and their corresponding vectors.
So the question that I was working on is as follows:
Suppose $T\in$ $\mathscr L$(V) and dim(range T) = k. Prove that the number of eigenvalues is at most k+1.
Now I think I understand the solution, but I wanted to make sure that I was understanding it correctly.
So if you let $e_1,e_2, …, e_m$, be distinct eigenvalues with $v_1, v_2, …, v_m$ the list of corresponding nonzero vectors. If $e_j\neq0$, then $T\frac {v_j}{e_j}=v_j$ ( this part I understand because a vector is an eigenvalue if $Tv = ev$ and $v\neq$0)
Now the solution says that since at most one of $e_1,…, e_m$ equals zero, there is at least one of $v_1, …, v_m$ that is in the range of T.
So my question is concerning this specifically and this was my thinking around it: if there exists an eigenvalue that equals zero, then its corresponding eigenvector would be in the range of $T$ because this would mean $Tv=0v=0$, and since 0$\in$range$T$ because it is a subspace of $V$, this implies that there is at least one vector in the linearly independent list $v_1,…,v_m$ ( the zero vector) that is in the range of $T$.
Could anyone help me out here by confirming whether my reasoning is correct? And if not, where I got off track?
Please & Thank You!
Best Answer
No, it is not correct. Take$$\begin{array}{rccc}T\colon&\Bbb R^3&\longrightarrow&\Bbb R^3\\&(x,y,z)&\mapsto&(0,z,0).\end{array}$$Then $0$ is an eigenvalue and the eigenvectors corresponding to this eigenvalue are those of the form $(x,y,0)$, with $x,y\in\Bbb R$. But, among these, only those with $x=0$ belong to the range of $T$, since the range of $T$ is $\{(0,x,0)\mid x\in\Bbb R\}$.
You went off track when you wrote that, since $T(v)=0$ and since $0\in\operatorname{range}(t)$, then some eigenvector must be in range of $T$. Why? By the definition of eigenvector, no eigenvector is equal to $0$.