# Eigenvectors are linearly independent?

Theorem: Eigenvectors corresponding to distinct eigenvalues are linearly independent.

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

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

## 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 $p \le j$).

• 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? Nov 25, 2020 at 15:03
• @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. Nov 26, 2020 at 12:51

$Av_1=\lambda_1v_1$, $Av_2=\lambda_2v_2$, $\lambda_1\neq\lambda_2$, $v_1,v_2\neq0$. 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 $c(\lambda_2-\lambda_1)=0\implies c=0\implies v_2=0$.

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