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I have read these two properties for eigen values and eigen vectors:

1) $n \times n$ Matrix A is guaranteed to have $n$ independent eigenvectors if all its eigenvalues are different.

2)If A has repeated eigenvalues, it may or may not have $n$ independent eigenvectors.

I can see how 2) might be true:
The nullspace of $(A-\lambda I )$ might have dimension $>1$, thus leading to more than one independent eigen vector $X$ when we solve $(A- \lambda I)X=0$.

Now I want to know:

If $A$ has repeated eigen values, with $m<n$ distinct eigen values, Is $A$ guaranteed to have atleast $m$ independent eigen vectors?


Equivalently, I'm trying to ask:

Is the nullspace of $(A - \lambda I)$ surely different for different values of $\lambda$?

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if $v$, non-zero satisfies $(A-\lambda I)v=0$ and $(A-\mu I)v=0$, then by subtracting you get $(\lambda - \mu)Iv=0$, which implies $\lambda = \mu$.

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To different eigen-values (even just two of them) correspond linearly independent eigen-vectors, so the answer is YES.

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