- #1
ybhan23
- 2
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Is it true that if an n by n matrix A has n-linearly independent eigenvectors, then it must also be invertible because these n-eigenvectors span n-space. But does this reasoning work the other way around: that is if A is invertible, does that imply n-linearly independent eigenvectors can be found? More generally, is there some connection between the column space and eigenspace?
On a side note, here is an unrelated but interesting question: why is it that if the square of a matrix is the negative identity matrix (-I), then it implies that the matrix has an even number of columns and rows?
Thanks in advance
On a side note, here is an unrelated but interesting question: why is it that if the square of a matrix is the negative identity matrix (-I), then it implies that the matrix has an even number of columns and rows?
Thanks in advance