# eigenvectors and eigenvalues of a symmetric positive semi-definite matrix

Given a symmetric, positive semi-definite matrix $M$ with $p$ dimensions, and its eigenvalues are $\lambda_1=\lambda_2>...>\lambda_p$, how to show that the corresponding eigenvectors $u_1$ and $u_2$ are not unique and span 2-dimensional subspace?

Tried:

let $U = (u_1, u_2,...,u_p)$, $D =$diag$(\lambda_1,...,\lambda_p)$, then we have $M = UDU^\intercal$, and due to orthogonality of $U$, we have $D = U^\intercal MU$, and satisfies $u_1^\intercal Mu_1=u_2^\intercal Mu_2$.

Also, we know that $u_1^\intercal u_1=u_2^\intercal u_2 = 0$, as well as $u_1^\intercal u_1 = u_2^\intercal u_2 =0$. I just don't know how to make use of these relations... Any ideas??

You know that $M=U D U^T$, with $D$ diagonal and $U$ unitary. Suppose $D$ is ordered so that $D= \operatorname{diag}(\lambda_1,\lambda_1,\lambda_3,...,\lambda_p)$.

Then $(M-\lambda_1 I)x = 0$ iff $(D-\lambda_1 I) U^T x = 0$.

Note that $D-\lambda_1 = \operatorname{diag}(0,0,\lambda_3-\lambda_1,...,\lambda_p-\lambda_1)$ and let $y_1 = U e_1, y_2 = U e_2$, then $y_1,y_2$ are linearly independent and $(D-\lambda_1 I) U^T y_k = (D-\lambda_1 I) e_k = 0$ (for $k=1,2$).

Hence $\operatorname{sp} \{ y_1,y_2\} \subset \ker (M-\lambda_1 I)$.

Since $\operatorname{sp} \{ y_1,y_2\} = \operatorname{sp} \{ y_1-y_2, y_1+y_2 \}$, we see that the eigenvectors are not 'unique'.

As an aside, eigenvectors are never unique. Even if normalized, if $u$ is an eigenvector, so is $-u$.

• for your last remark, if we only consider normalized 'positive' eigenvectors, would they be unique? In other words, suppose that all eigenvalues are distinct, would $u_i$ be unique under normalization? Feb 15, 2015 at 22:15
• Well, what does positive mean for an eigenvector such as $(1,-1)^T$? Generally it is better to think in terms of eigenspaces, as there is no uniqueness issue to deal with. Feb 15, 2015 at 22:49

There exists a basis of the space consisting of eigenvectors of $M$ ( $M$ is diagonalizable, being real symmetric) : $\ M \cdot u_i = \lambda_i \cdot u_i$, and $u_1$, $\ldots$, $u_n$ a basis. The extra assumption is $\lambda_1= \lambda_2 = \lambda$. $u_1$, $u_2$, being part of a basis, are linearly independent, and span the subspace of vectors $u$ such that $Mu = \lambda\cdot u$.

A symmetric positive definite matrix $A$ is in particular diagonalizable. It's a general fact about diagonalizable matrices that, for each eigenvalue $\lambda$, the algebraic multiplicity equals the geometric multiplicity.

• The algebraic multiplicity of $\lambda$ is the maximum exponent $m$ such that $(\lambda-X)^m$ divides the characteristic polynomial $\det(A-\lambda X)$.

• The geometric multiplicity of $\lambda$ is the dimension of the eigenspace $E_A(\lambda)=\{v:Av=\lambda v\}$.

In your case, since the eigenvalue $\lambda_1$ has (algebraic) multiplicity $2$, also the eigenspace has dimension $2$, so you find two linearly independent eigenvectors $u_1$ and $u_2$.