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Is number of linearly independent eigenvector same as number of distinct eigenvalues of a matrix?

I have seen many of my friends using this shortcut to find out number of linearly independent eigen vectors, but does this hold true in each and every case?

Rodrigo de Azevedo
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Turing101
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    No: the largest possible number of linearly independent eigenvectors is at least as many as the number of distinct eigenvalues of a matrix, but it can be bigger. For example, the $n\times n$identity matrix has just one eigenvalue, but you can find $n$ linearly independent eigenvectors. – Arturo Magidin Jan 20 '20 at 05:41
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    https://math.stackexchange.com/questions/29371/how-to-prove-that-eigenvectors-from-different-eigenvalues-are-linearly-independe – TheStudent Jan 20 '20 at 05:42
  • Is there any shortcut by which the total number of linearly indipendent eigen vectors can be found? actual process takes a lot of time tbh – Turing101 Jan 20 '20 at 05:49

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Eigenvectors for different eigenvalues are independent. But eigenvectors for the same eigenvalue can be independent also. This happens when the dimension of the eigenspace is greater than one.

So, we can say that there are at least as many independent eigenvectors as distinct eigenvalues.

  • maximum number of linearly independent vector for an $N*N$ matrix is $N$ right? – Turing101 Jan 20 '20 at 07:05
  • @Turing101 Well, what’s the maximum number of linearly-independent vectors in an $n$-dimensional vector space? – amd Jan 20 '20 at 07:11
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It's true only when all eigenvalues are 1-fold.

For example, consider $A$ matrix as $$ A = \pmatrix{3 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 5}. $$ Then $A$ has one 1-fold eigenvalue 3 and one 2-fold eigenvalue 5, totally two distinct eigenvalues. But $A$ has three orthonormal eigenvectors with respect to eigenvalues 3, 5 and 5 $$ \pmatrix{1\\0\\0}, \pmatrix{0\\ 1 \\ 0}, \pmatrix{0 \\ 0 \\ 1}. $$

UpbtEZ
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