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I have a question about eigenvectors corresponding to eigenvalues with an algebraic multiplicity $>1$. I am relatively new to linear algebra and my resources as well as my mathematical background are quite limited as I am a biologist.

The question arose because of an exercise I stumbled across, for which I unfortunately do not have the solutions. The exercise is to find eigenvalues and eigenvectors of a symmetric matrix $\boldsymbol{E}$ and to show that the matrix in question is diagonalizable by proving that $\boldsymbol{P^T}\boldsymbol{E}\boldsymbol{P} = \boldsymbol{\Lambda}$, where $\boldsymbol{P}$ is an orthogonal matrix and $\boldsymbol{\Lambda}$ is a diagonal matrix of eigenvalues.

$$ E = \begin{bmatrix} -1 & -2 & 1 \\ -2 & 2 & -2 \\ 1 & -2 & -1\end{bmatrix}$$

The eigenvalues are $det(\boldsymbol{E}-\lambda \boldsymbol{I})=0$, which gives us $-(\lambda + 2)^2(\lambda-4)$ and thus we get $\lambda_1 = -2$ with multiplicity $m_1 = 2$ and $\lambda_2 = 4$ with $m_2 = 1$.

$\boldsymbol{E} - \lambda_1 \boldsymbol{I} = \begin{bmatrix}1 &-2 & 1 \\ -2 & 4 & -2 \\ 1 & -2 & -1\end{bmatrix}$. Since $rank(\boldsymbol{E} - \lambda_1 \boldsymbol{I}) = 1$, there should be 2 linearly independent eigenvectors for $\lambda_1$. To find the eigenvectors, I solved $(\boldsymbol{E}-(-2)\boldsymbol{I})\vec{u}=\vec{0}$ using the reduced matrix :$\begin{bmatrix}1 & -2 & 1 \\ 0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}u_1 \\ u_2 \\ u_3\end{bmatrix} = \begin{bmatrix}0 \\ 0\\ 0 \end{bmatrix}$.

Since the matrix is of rank $1$, the values of $2$ of the $3$ elements of the eigenvector can be chosen freely. E.g. $u_1,u_2=1$ and thus $u_3=1$, yielding $\vec{x_1}\begin{bmatrix}1\\1\\1\end{bmatrix}$. Next, I chose a vector $\vec{x_2}$ such that $\vec{x_2} \neq c\vec{x_1}$. One such vector is $\vec{x_2}=\begin{bmatrix}1 \\ -2 \\ -5\end{bmatrix}$.

For $\lambda_2$, we have : $\boldsymbol{E}- \lambda\boldsymbol{I}= \begin{bmatrix} -5 & -2 & 1 \\ -2&-2&-2\\1&-2&-5\end{bmatrix}$ which can be reduced to $\begin{bmatrix}1 & 0 & -1 \\ 0&1&2\\0&0&0\end{bmatrix}$.

Thus, $u_1 = u_3$ and $u_2=-2u_3$ and the resulting vector should have the form $\vec{x_3}=\begin{bmatrix}u_3\\-2u_3\\u_3\end{bmatrix}$. For $u_3=1$ I get a vector $\vec{x_3}$ that is orthogonal to $\vec{x_1}$ and $\vec{x_2}$. However, $\vec{x_1}$ and $\vec{x_2}$ are not orthogonal.

So this is where I start to get stuck. How can I find three orthogonal eigenvectors such that I can demonstrate that $\boldsymbol{P^T}\boldsymbol{E}\boldsymbol{P} = \boldsymbol{\Lambda}$.

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  • $\begingroup$ "According to my linear algebra book, eigenvectors of symmetric matrices should be orthogonal, regardless of whether they correspond to different eigenvalues or to an eigenvalue with multiplicity $>1$": this statement is incorrect and is unlikely to be what your book said. The correct statement is that for a symmetric matrix, there exists an orthogonal basis of eigenvectors $\endgroup$ May 19, 2021 at 13:36
  • $\begingroup$ The first step where you go wrong is in selecting $\vec x_2$. It is not sufficient to have $\vec x_1^T \vec x_2 = 0$, we must also have $(E - (-2)I) \vec x_2 = 0$. $\endgroup$ May 19, 2021 at 13:40
  • $\begingroup$ Thank you for the hint, I missed that. The vector should have been $\vec{x_2}=\begin{bmatrix}1 &-2&-5\end{bmatrix}$, which is then not orthogonal to $\vec{x_1}$. Is there a particularly good way to find $\vec{x_1}$, $\vec{x_2}$, and $\vec{x_3}$ such that they are orthogonal? $\endgroup$ May 19, 2021 at 17:00
  • $\begingroup$ The paragraph in my book I was referring to is as follows: "Symmetric matrices have eigenvectors that are orthogonal to one another. We establish this in two cases: (1) eigenvectors corresponding to different eigenvalues, and (2) eigenvectors corresponding to a multiple eigenvalue." But that does not imply what I wrote above, I will edit it. $\endgroup$ May 19, 2021 at 17:04
  • $\begingroup$ One approach is to apply the Gram-Schmidt process to the set $\{\vec x_1,\vec x_2 \}$ in order to get an orthogonal basis of the eigenspace $\endgroup$ May 19, 2021 at 19:04

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Thanks to Ben Grossmann I found the solution. The set of vectors $\vec{x_1}=\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix},\vec{x_2}\begin{bmatrix}1 \\ -2 \\ -5 \end{bmatrix}$, and $\vec{x_3}\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$ form a basis of the eigenspace. An orthonormal basis can be found by applying the Gram-Schmidt process.

First, let's start in a 1-dimensional subspace $V_1$, which is spanned by $\vec{x_1}$. The vector $\vec{x_1}$ can be normalized, $\frac{\vec{x_1}}{||\vec{x_1}||} = \frac{1}{\sqrt{3}}\vec{x_1}$. Hence, $V_1=span(\vec{n_1})$ as $\vec{n_1}$ is a linear combination of $\vec{x_1}$.

By adding the next vector, we get subspace $V_2=span(\vec{n_1}, \vec{x_2})$. The vector $\vec{x_2}$ can be created as a linear combination of a multiple of $\vec{n_1}$ ($c \vec{n_1}=\vec{v_2}$) and another vector ($\vec{w_2}$) that is orthogonal to it. The vector $\vec{v_2}$ is the projection of $\vec{x_2}$ onto $V_1$ ($\text{proj}_{V_1}(\vec{x_2})$). The projection can be described as follows: $\text{proj}_{V_1}(\vec{x_2})=(\vec{x_2}\cdot\vec{n_1})\vec{n_1}$.

Hence, $\vec{w_2} = \vec{x_2}-(\vec{x_2}\cdot\vec{n_1})\vec{n_1} = \begin{bmatrix}1 \\ -2 \\ -5\end{bmatrix}-\left(\begin{bmatrix}1 \\ -2 \\ -5\end{bmatrix}\cdot \frac{1}{\sqrt{3}}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}\right)\frac{1}{\sqrt{3}}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} = \begin{bmatrix}3 \\ 0 \\ -3\end{bmatrix}$. Normalize the new vector, $\vec{n_2}= \frac{\vec{w_2}}{||\vec{w_2}||}=\frac{1}{\sqrt{18}}\vec{w_2}$.

The last step is now to find the last orthogonal basis vector in the subspace $V_3$. $\vec{w_3} = \vec{x_3}-(\vec{x_3}\cdot\vec{n_1})\vec{n_1}--(\vec{x_3}\cdot\vec{n_2})\vec{n_2} = \begin{bmatrix}1 \\ -2 \\ 1\end{bmatrix}-\left(0\right)\frac{1}{\sqrt{3}}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix} - \left(0\right)\frac{1}{\sqrt{18}}\begin{bmatrix}3 \\ 0 \\ -3\end{bmatrix}= \begin{bmatrix}1 \\ -2 \\ 1\end{bmatrix}$. And normalize, $\vec{n_3}=\frac{1}{\sqrt{6}}\vec{w_3}$. The orthogonal basis of the eigenspace is $\Bigg\{\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix},\frac{1}{\sqrt{18}}\begin{bmatrix} 3 \\ 0 \\ -3 \end{bmatrix},\frac{1}{\sqrt{6}}\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}\Bigg\}$. By constructing $\boldsymbol{P}$ as $\begin{bmatrix}\vec{n_1},\vec{n_2},\vec{n_3}\end{bmatrix}$, we get $\boldsymbol{P^T}\boldsymbol{E}\boldsymbol{P}=\begin{bmatrix}-2 & 0 & 0\\ 0 & -2 & 0 \\ 0& 0 &4\end{bmatrix}$.

If my choice of words is inappropriate, please correct me.

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