# If $\lambda_1, …, \lambda_m$ are eigenvalues, then the corresponding eigenvectors are linearly independent

Suppose $T:V\rightarrow V$ is a linear transformation, where $V$ is a finite dimensional vector space with dimension $n$.

1. Show that if $\lambda_1,...,\lambda_m$ are the eigenvalues of $T$, then the sequence of eigenvectors comprising bases for the eigenspaces $E_{\lambda_1},...,E_{\lambda_m}$ is linearly independent.

2. If $T$ has eigenvalues $\lambda_1,...,\lambda_m$ and a concatenation of the bases of the eigenspaces $E_{\lambda_1},...,E_{\lambda_m}$ is a basis $B$ for $V$, show that the $B$-matrix of $T$ is diagonal.

It doesn't seem to me that 1. should be true, since the algebraic multiplicity for a certain eigenvalue can be greater than 1, so wouldn't this eigenvalue have multiple eigenspaces that are the same? Nonetheless, if we assume we only take one eigenspace from each distinct eigenvalue, then I don't see why the statement is actually true.

For number 2, we have to find $[T[E_{\lambda_1}...E_{\lambda_m}]]_B$, but how is this diagonal when the coordinate vector is only one column?

• There seems to be an assumption missing that $T$ is a bijection. – ml0105 Dec 9 '14 at 2:20
• @ml0105 what changes if this assumption is made? How would you answer the question with that assumption? – Carley Dec 9 '14 at 2:45
• We can write $T$ as a matrix. So if $T$ is not bijective, $T$ is not of full rank. So there will be at least one eigenvector repeated, and so we won't have an eigenbasis (a basis of eigenvectors). I'm not entirely sure off hand how to answer this. It's been a while since I've done some of this, and my textbook isn't in reach right now. – ml0105 Dec 9 '14 at 2:52
• I will also add that the distinct eigenvectors form the basis of a subspace of $V$, known as the eigenspace. If $T$ is bijective, then the eigenvectors generate $V$. So you need either $T$ is bijective or the eigenvalues $\lambda_{1}, ..., \lambda_{m}$ are distinct (or have equal algebraic and geometric multiplicities). Whichever way, I think more assumptions are needed. Perhaps someone else will have a better idea though. – ml0105 Dec 9 '14 at 2:55
• @Nishant I've proved part (a) under the assumption of distinct eigenvalues. Do you have any ideas for part (b)? – Carley Dec 9 '14 at 6:15