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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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  • $\begingroup$ Does diagonalizable have any relevance? If it has an inverse, its rank is 8. So it has 8 eigenvectors...I think? math.stackexchange.com/questions/146927/… $\endgroup$
    – BCLC
    May 14, 2015 at 17:44
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    $\begingroup$ @BCLC No: invertible $n\times n$ matrices need not have $n$ linearly independent eigenvectors. It is also possible to have a singular $n\times n$ matrix with $n$ linearly independent eigenvectors (indeed: every non-zero vector is an eigenvector for the $0$ matrix with eigenvalue $0$). $\endgroup$ May 14, 2015 at 18:26
  • $\begingroup$ I think if the rank of the matrix is n, then there will be n linearly independent eigen vectors. It doesn't matter whether matrix is invertible or not. Although if a matrix is invertible then it means it is full rank i.e rank is n. $\endgroup$ Dec 17, 2020 at 15:23
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    $\begingroup$ @AbhishekPG No, it is not accurate. An n $\times$ n matrix need not have n linearly independent eigen vectors. Eg: consider the 2 $\times$ 2 matrix with first row (3, 1) and second row (0, 3). This matrix has only one linearly independent eigen vector. $\endgroup$ Mar 26, 2021 at 13:12

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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.

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  • $\begingroup$ The OP does not say linearly independent eigenvectors for the max min part XD math.stackexchange.com/a/1282339/140308 $\endgroup$
    – BCLC
    May 14, 2015 at 18:02
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    $\begingroup$ "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.", Could you prove this, please? $\endgroup$
    – user153245
    Oct 21, 2015 at 7:04
  • $\begingroup$ What if each basis vector doubled then there exits an infinite number of independent eigenvectors. $\endgroup$ Nov 16, 2019 at 16:57
  • $\begingroup$ @Rupanshu Yadav That is not true because you can write any vector in the vector space as a linear combination of the basis vectors, which means that all vectors in space (other than the basis vectors) can be represented using the basis vectors. So all vectors other than the basis vectors would become linearly dependent. $\endgroup$ Dec 17, 2020 at 15:14
  • $\begingroup$ if the matrix has rank $r$ would it not have $r$ eigenvectors (and be diagonalizable via those $r$ but have a $n, r$ first matrix and $[r, r]$ diagonal matrix? $\endgroup$ Oct 19, 2021 at 21:21
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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.

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    $\begingroup$ I know its pedantic it just... it gets to me. $\endgroup$
    – user24142
    May 14, 2015 at 17:50
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    $\begingroup$ @user24142 If you really want to be pedantic, you need to specify that you're not working over a finite field where there are only finitely many vectors, so a priori only finitely many eigenvectors. $\endgroup$ May 14, 2015 at 18:29
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    $\begingroup$ For a simple rotation in a 2x2 matrix, you have zero eigenvectors, so the minimum number isn't always infinite. $\endgroup$ May 2, 2019 at 8:11
  • $\begingroup$ The problem statement includes the existence of eigenvalues. If there are eigenvalues, there are eigenvectors. Also, splitting field of the matrix. $\endgroup$
    – user24142
    May 3, 2019 at 0:57
  • $\begingroup$ @Anindya Mahajan that depends on how you specify the rotation. For example if the rotation is by 90 degrees then you are right there are no real eigen values ( but there are two complex eigen values). But if the rotation is by 180 degrees there will be two real linearly independent eigen values for a 2*2 matrix. In fact ,practically complex numbers as eigen value is not of much use but they do exist atleast mathematically. $\endgroup$ Dec 18, 2020 at 16:21
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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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  • $\begingroup$ The OP does not say linearly independent eigenvectors for the max min part XD math.stackexchange.com/a/1282339/140308 $\endgroup$
    – BCLC
    May 14, 2015 at 18:02
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    $\begingroup$ @BLCL yes, so it seems... $\endgroup$ May 14, 2015 at 18:20
  • $\begingroup$ if the matrix has rank $r$ would it not have $r$ eigenvectors (and be diagonalizable via those $r$ but have a $n, r$ first matrix and $[r, r]$ diagonal matrix? $\endgroup$ Oct 19, 2021 at 21:21
  • $\begingroup$ @CharlieParker No, just because a matrix is invertible does not mean it is diagonalizable. Take e.g. $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ which has an eigenvalue $2$ with geometric multiplicity $1$. $\endgroup$ Oct 20, 2021 at 10:43

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