I am given the eigenvalues of a square, 8x8, matrix. They are all non-zero. I have determined that the matrix is diagonalizable and has an inverse. In one part of the problem, I am asked to find the maximum and minimum number of eigenvectors that the matrix could possibly have?
Since A is diagonalizable does that mean it will have n linearly independent eigenvectors. So, is the max and min number of eigenvectors is 8?
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$\begingroup$Correct, an $n\times n$ matrix which is diagonalizable must have a set of $n$ linearly independent eigenvectors -- the columns of the diagonalizing matrix are such a set.
In general, if an $n\times n$ matrix has $k$ distinct eigenvalues, then there may in general be anywhere between $k$ and $n$ linearly independent eigenvectors.
For any of this, it doesn't matter whether or not the eigenvalues are non-zero.
$\endgroup$ 5 $\begingroup$If $\vec v$ is an eigenvector, then so is $t \vec v$ for all real $t$. If they're asking about linearly independent eigenvectors, then you're right, but if they're just asking about eigenvectors, I would say the min and max is always infinite.
$\endgroup$ 7 $\begingroup$Yes: If $A$ is diagonalizable, then there is a basis $v_1,\dots,v_8$, s.t. $D=U^{-1}AU$ is diagonal, where $U$ has $v_1,\dots, v_8$ as columns. In that case every $v_i$ is an eigenvector to the $i$-th diagonal element of $D$.
EDIT: Of course every matrix with at least one eigenvalue $\lambda$ has infinitely many eigenvectors (as pointed out in the comments), since the eigenspace corresponding to $\lambda$ is at least one-dimensional.
Well, to be more precise: It depends on the underlying field...
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