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Theorem: Eigenvectors corresponding to distinct eigenvalues are linearly independent.

Could someone give me a geometric interpretation of the theorem? Thanks!

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  • $\begingroup$ Try to see a matrix as a linear transformation. (rotation, reflection, etc.) And look at what $Av=v$ means. This helped me. $\endgroup$
    – user112167
    Dec 3, 2013 at 1:01
  • $\begingroup$ @user112167 I already thought about it in that sense, but no luck. Could you maybe turn your comment into a full answer? $\endgroup$
    – dfg
    Dec 3, 2013 at 1:09
  • $\begingroup$ Maybe this would help: youtube.com/watch?v=vs2sRvSzA3o $\endgroup$
    – user112167
    Dec 3, 2013 at 1:17
  • $\begingroup$ @user112167 It does help, thank you! $\endgroup$
    – dfg
    Dec 3, 2013 at 1:32
  • $\begingroup$ Is the converse also true? I mean, if eigenvectors are linearly independent, thus their corresponding eigenvalue are distinct? $\endgroup$
    – C. Bishop
    Nov 25, 2020 at 15:01

3 Answers 3

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Two vectors

An eigenvector $v$ of a transformation $A$ is a vector that, when the transformation is applied to it, doesn't change its direction, i.e., $Av$ is colinear to $v$. The only thing that may change is its length. The factor of that change is its eigenvalue $\lambda$.

So, if

$$Av_1 = \lambda_1 v_1, \quad Av_2 = \lambda_2 v_2, \quad \lambda_1 \ne \lambda_2,$$

it means that $A$ stretches vectors $v_1$ and $v_2$ differently. This would, of course, be impossible, had they had the same direction, i.e., if they were colinear, which is the same as being linearly dependent.

More than two vectors

Let us assume that we have eigenvectors $(v_i)_{i=1}^k$ with their respective eigenvalues $(\lambda_i)_{i=1}^k$, where $k > 2$. Assume that the vectors are arranged in a way that $\mathcal{B} := \{v_1,\dots,v_j\}$ is linearly independent (for some $j$, $1 < j < k$), while $\{v_1,\dots,v_j,v_l\}$ is linearly dependent for all $l > j$.

Obviously, $\mathcal{B}$ forms basis of the space spanned by vectors $(v_i)_{i=1}^k$. Geometrically, however, every linear combination of vectors in $\mathcal{B}$ forms a parallelepiped (in $j$ or less dimensions) with the vertex oposite of $0$ being $v_l := \sum_{i=1}^j \alpha_i v_i$. Now, what happens with that parallelepiped when we apply a linear transform?

Since all the eigenvalues $(\lambda_i)_{i=1}^j$ are distinct, each of its edges stretches differently, which means that the diagonal of that parallelepiped will not be colinear with the original. In other words,

$$\not\exists \lambda_l \colon A v_l = \lambda_l v_l.$$

In other words, a vector linearly dependent with the eigenvectors having distinct eigenvalues cannot be an eigenvector itself (unless it's a trivial case $v_l = \alpha_p v_p$ for some $p \le j$).

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  • $\begingroup$ Does the convers hold true? I mean, if eigenvectors are linearly independent, thus they come from distinct eigenvalues? it could be possible that one comed from the null eigenvalue? $\endgroup$
    – C. Bishop
    Nov 25, 2020 at 15:03
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    $\begingroup$ @C.Bishop, consider the identity matrix (or, equivalently, linear operation). Any vector space basis will also be a set of linearly independent eigenvectors, yet there is only one distinct eigenvalue. $\endgroup$ Nov 26, 2020 at 12:51
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$Av_1=\lambda_1v_1$, $Av_2=\lambda_2v_2$, $\lambda_1\neq\lambda_2$, $v_1,v_2\neq0$. Suppose $v_2=cv_1$. Then $Av_2=\lambda_2v_2=c\lambda_2v_1$ and $Av_2=Acv_1=cAv_1=c\lambda_1v_1$, hence $c(\lambda_2-\lambda_1)=0\implies c=0\implies v_2=0$.

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Hint: If some eigenvector $v$ lies on eigenspace corresponding to eigenvalue $\lambda$, then eigenvalue of $v$ must also be $\lambda$.

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