Eigenvector proof for repeated eigenvalues

Question

I am stuck trying to solve the following problem:

In diagonalizing a symmetric matrix $S$, we find that two of the eigenvalues ($\lambda_1$ and $\lambda_2$) are equal but the third ($\lambda_3$) is different. Show that any vector which is normal to $\hat{n}_3$ (which is the eigenvector corresponding to $\lambda_3$) is then an eigenvector of $S$ with eigenvalue equal to $\lambda_1$

Can anyone offer hints on, or an outline of, the solution?

Thank you.

Answer 1

The inspiration from this solution came from Elliot G saying that the eigenvectors form an orthonormal basis; I don't agree that $\vec{A}$ has to be parallel to either $\hat{n}_1$ or $\hat{n}_2$ however.

Since $\hat{n}_1$, $\hat{n}_2$, and $\hat{n}_3$ form an orthonormal basis for $\mathbb{R}^3$, any vector $\vec{A}$ can be expressed as

$\vec{A} = \alpha \hat{n}_1 + \beta \hat{n}_2 + \gamma \hat{n}_3$.

Further, since $\vec{A}\cdot\hat{n}_3 = 0$, we can say that $\gamma = 0$, thus

$\vec{A} = \alpha \hat{n}_1 + \beta \hat{n}_2 $.

It follows that

$S\vec{A} = S\alpha \hat{n}_1 + S\beta \hat{n}_2 \\ S\vec{A} = \alpha S\hat{n}_1 + \beta S\hat{n}_2 \\ S\vec{A} = \alpha \lambda_1\hat{n}_1 + \beta \lambda_1 \hat{n}_2 \\ S\vec{A} = \lambda_1 (\alpha \hat{n}_1 + \beta \hat{n}_2) \\ \therefore S\vec{A} = \lambda_1 \vec{A} \\ $.

Answer 2

Proof sketch:

Since $\lambda= \lambda_1=\lambda_2$ and $S$ is symmetric, the geometric multiplicity of $\lambda$ is $2$. Now, if we denote $E_{\mu}$ the eigenspace of $S$ associated to the eigenvalue $\mu$, we have $V=E_{\lambda}\oplus E_{\lambda_3}$ where $S:V\to V$ and $\oplus$ denotes the direct sum. Finally, we know that $V= E_{\lambda_3}\oplus E_{\lambda_3}^{\perp}$ and by identification we get $E_{\lambda_3}^{\perp}=E_{\lambda}$.

Answer 3

Hint: your three eigenvectors form an orthonormal basis for $\Bbb R^3$, so it follows that if a vector is normal to $\hat n_3$, then it is parallel to either $\hat n_1$ or $\hat n_2$

Answer 4

If $A=A^T$ and $\lambda_1, \lambda_2$ are distinct eigenvalues and $\mathbf x, \mathbf y$ corresponding eigenvectors, we have: $$ \lambda_1\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda_1\mathbf{x},\mathbf{y}\rangle = \langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle = \langle\mathbf{x},A\mathbf{y}\rangle = \langle\mathbf{x},\lambda_2\mathbf{y}\rangle = \lambda_2\langle\mathbf{x},\mathbf{y}\rangle \Rightarrow $$ $$ (\lambda_1-\lambda_2)\langle\mathbf{x},\mathbf{y}\rangle=0 $$ and, since $\lambda_1 \ne \lambda_2$, this means that $\mathbf x, \mathbf y$ are orthogonal. In other words:

the eigenspaces corresponding to different eigenvalues are orthogonal.

Since $A$ is diagonalizable then we can chose orthogonal eigenvectors in each eigenspace such that all these vectors are an orthogonal basis for $\mathbb{R}^n$. And this means that a vector orthogonal to an eigenspace is an eigenvector of some other eigenvalue.