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Suppose I have a data set of $m$ vectors in $\mathbb{R}^d$, $D = \{x_1,\ldots,x_m\}$.

Let $S = \sum_{i=1}^{m}x_ix_i^T$ be the scatter matrix.

My question is: thus the eigenvectors of $S$ form a base to $\mathbb{R}^d$?

I know that if I have $d$ eigenvectors of $S$ with distinct eigenvalues, than those vectors are orthogonal to each other because $S$ is symmetric. Then, I know orthogonal vectors are independent of each other. Thus, if $S$ has $d$ eigenvectors with distinct eigenvalues, then those vectors are $d$ (= dimension of the space) independent vectors and thus form a base.

But, I don't sure that the eigenvectors of $S$ necessarily form a base because maybe there are eigenvectors with same eigenvalues and thus not necessarily orthogonal to each other. Also, I don't sure if $S$ necessarily has $d$ eigenvectors, maybe can be less.

Can $S$ has non distinct eigenvalues?

Can $S$ has less than $d$ eigenvectors?

Elaboration on when the eigenvectors of $S$ form a base (maybe always or maybe under some conditions) will be helpful.

George
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2 Answers2

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Note that $S = \sum_{i=1}^m x_i x_i^T$ is symmetric. A fundamental property of symmetric matrices is the property of (orthogonal) diagonalizability which means that the eigenvectors of $S$ span $\mathbb{R}^d$. For reference see:

why symmetric matrices are diagonalizable?

I assume that $m < d$, so eigenvalue $0$ must have a multiplicity of at least $d - m$. This means that $S$ can have non-distinct eigenvalues (not only $\lambda = 0$ can be degenerate).

You are right that eigenvectors with the same eigenvalues are not necessarily orthogonal to each other but the eigenspaces belonging to certain eigenvalues all have a orthogonal basis that can be acquired, for instance, through the Gram-Schmidt-process (meaning that the linearly independent eigenvectors belonging to the same eigenvalue can be chosen orthogonal).

So, to summarize, $S$ is diagonalizable, therefore it has $d$ linearly independent eigenvectors which do not necessarily have to belong to different eigenspaces (there can be non-distinct eigenvalues).

$S$ cannot have less than $d$ eigenvectors.

Community
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hickslebummbumm
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Question $1$ : Does the eigenvectors of a symmetric matrix form a basis for $\mathbb{R}^d$ ? Yes.

Suppose you have two eigenvectors $v$ and $w$ corresponding to $\lambda$. Every vector in the span of $v$ and $w$ is also an eigenvector. $S(\alpha v + \beta w)=S(\alpha v ) + S(\beta w)=\alpha S(v) + \beta S(w)=\lambda (\alpha v + \beta w)$. Thus we can choose any two orthogonal vectors in the eigenspace corresponding to $\lambda$ and they will be eigenvectors.

Question $2$ : Can $S$ have non-distinct eigenvalues ? Yes.

Question $3$ : Can $S$ have less than $d$ eigenvectors ? No.

Every $d$ x $d$ real symmetric matrix, has $d$ real eigenvalues. Since symmetric matrices are diagonalizable, for a given eigenvalue $\lambda$, the algebraic multiplicity (number of times it divides the characteristic polynomial) and geometric multiplicity (dimension of the space spanned by the corresponding eigenvectors) are equal, which implies that the sum of dimensions of eigenspaces of $S$ is $d$.

Srinivas K
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