Prove these vectors are Linearly Independent.

Question

Given Matrix $A_{n\times n}$. Suppose it has $k$ distinct eigenvalues : $u_1, u_2, \cdots u_k$.

$a_1, a_2,\cdots, a_k$ : Represents Algebraic Multiplicities of that $k$ distinct eigenvalues.

$$a_1+ a_2 +\cdots + a_k = n$$

Assume Geometric Multiplicity = Algebric Multiplicity, for all eigenvalues, respectively.(GM=AM)

Now we know that: Any $k$ eigenvectors corresponding to those $k$ distinct eigenvalues are Linearly Independent.

We will have $n$ eigenvectors in total(because of condition : GM=AM) : $a_1$ vectors corresponding to $u_1$, $a_2$ vectors corresponding to $u_2$, and so on...

How can I prove that, these $n$ vectors are Linearly Independent?

Answer 1

Let
$\alpha_{11},\dots,\alpha_{1a_1}$ be linear independent eigenvectors of $u_1$,
$\alpha_{21},\dots,\alpha_{2a_2}$ be linear independent eigenvectors of $u_2$, ...,
$\alpha_{k1},\dots,\alpha_{ka_k}$ be linear independent elgenvectors of $u_k$.

If they are linear dependent, then we have $$ \lambda_{11}\alpha_{11}+\dots\lambda_{ka_k}\alpha_{ka_k}=0 $$ and not all $\lambda_{ij}=0$. Let $\beta_i=\lambda_{11}\alpha_{11}+\dots\lambda_{1a_1}\alpha_{1a_i}$ for $i=1,\dots,k$. Not all of $\beta_i$'s are zero(if all of them are zero, then some eigenvectors of the same eigenvalue would be linear dependent). Denote these nonzeros as $\beta_{i_1},\dots,\beta_{i_s}$, so we get $$ \beta_{i_1}+\dots+\beta_{i_s}=0. $$ So $\beta_{i1},\dots,\beta_{is}$ are linear dependent. But they are eigenvectors of different eigenvalues, they can not be linear dependent.