A real symmetric matrix with less than $n$ distinct eigenvalue

Question

If $A$ is a real symmetric matrix with less than $n$ distinct eigenvalues, is it true that for any $\vec{v}$, $\{\vec{v},A\vec{v},...,A^{n-1}\vec{v} \}$ is always linearly dependent?

Answer 1

Yes, this is always the case (assuming $A$ is $n\times n$).

Since $A$ is symmetric, it is diagonalizable. Since it has fewer than $n$ distinct eigenvalues, at least one eigenspace of $A$ must have at least two dimensions. Which is to say, there is an eigenbasis for $A$ (being careful to choose the basis vectors from the multidimensional eigenspace) such that $v$ expressed in that basis has one component equal to $0$. And this will also be true of all $A^kv$, and any linear combination of them. Since we have $n$ vectors that do not span all of $n$-dimensional space, they must be linearly dependent.

Answer 2

Since $A$ is a real symmetric matrix, it is diagonalizable. Therefore it satisfies $p(A) =0$ where $p(\lambda) = \prod_i (\lambda- \lambda_i)$, $\lambda_i$ the distinct eigenvalues. This has degree $\le n-1$, so $ 0 = p(A) v$ is a nontrivial linear combination of $v, Av, \ldots, A^{n-1} v$.