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7:39 PM
Is the sum of the dimensions of the eigenspaces of an $n\times n$ matrix necessarily equal to $n$?
at first I thought using the fact that for diagonalizable matrices, it has to be equal to $n$.
but I was advised to simply show that $1\leq\mathrm{dim}(\text{sum of all the eigenspaces})\leq n$ by verifying it for a $2\times 2$ matrix
how can we do that?
 
 
2 hours later…
10:00 PM
We know that eigenvectors to different eigenvalues are linearly independent. So the sum of dimensions cannot be larger than the dimension of $F^n$, which is $n$. (Simply because any subspace of $F^n$ has dimension at most $n$.)
The inequality can be strict, for example, $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has only one eigenvalue, it has eigenspace of dimension one. (You can take any matrix which is not diagonalizable.)
 
 
2 hours later…
11:53 PM
@MartinSleziak Thanks!
 

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