I came across this theorem that says:

Let T be a linear operator and let $\lambda_1, \lambda_2, ... \lambda_k$ be distinct eigenvalues of T. For each i = 1, 2, ..., k, let $v_i \in E_{\lambda_i}$, the eigenspace corresponding to $\lambda_i$. If

$$v_1 + v_2 +... + v_k = 0$$

then $v_i = 0$ for all i.

I don't quite understand the theorem. I thought vectors belonging to eigenspace are eigenvectors and eigenvectors are linearly independent? If that is the case $v_1 + v_2 +... + v_k = 0$ should not hold in the first place. What am I misunderstanding here?

  • 1
    $\begingroup$ Since eigenvectors belonging to distinct eigenvalues are linearly independent. So if there are nonzero vectors in $(v_i)_{i=1}^k$, then these nonzero vectors are linearly dependent. This is a contradiction. So the conclusion of theorem holds. $\endgroup$ – Shine Sep 26 '14 at 10:35

The equation $v_1+v_2+\dots+v_k=0$ can obviously hold if all $v_i=0$.

That it cannot hold for nonzero vectors is a direct consequence of the fact that $E_{\lambda_i}$ are linearly independent:

Say $w_1,\dots, w_n$ are all nonzero vectors in the set $\{v_1,\dots, v_k\}$. Now, since $$1\cdot w_1 + \cdots + 1\cdot w_n =w_1+\dots w_n=0$$

it must follow that, since $w_i$ are independent, that $1=0$ which is false, meaning no such $w_i$ exists


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