Eigenvalues and eigenvectors in a symmetric matrix

Question

$A$ is a 3x3 symmetric matrix. You know that $2$ and $5$ are eigenvalues of $A$, and

$$V_5 = Span \left( \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} ,\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right) $$

is the eigenspace of the eigenvalue $5$. Which of the following statements is true?

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(1) $$V_2 = Span \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} $$

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(2) $A$ is diagonalizable.

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(3) There aren't three eigenvectors of $A$ orthogonal between them.

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(4) $$ V_2 = \left\{ \begin{pmatrix} x \\ y \\ z \end{pmatrix} | x + y + z = 0 \right\} $$

How can I find the true statement(s)?

Answer 1

You know that $A$ is a square symmetric matrix. So, from the Spectral Theorem:

$A$ is symmetric $\iff$ $A$ is orthogonally diagonalizable

From that you have the following:

$A$ is symmetric $\implies$ the eigenspaces of $A$ are pairwise orthogonal

This means that $V_\alpha \bot V_\beta$ for every autovalue $\alpha \neq \beta$.

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From the previous you know that option (3) is false.

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You can find the cartesian equation of $V_5$:

$$2x - y - z = 0$$

and the parametric equation of $V_{2_1}$:

$$ \begin{cases} x = 2 \alpha \\ y = -\alpha \\ z = -\alpha \end{cases} $$

So:

$$n[V_5] = d[V_{2_1}] = \begin{pmatrix}2 \\ -1 \\ -1 \end{pmatrix}$$

This means that $V_{2_1} \bot V_5$, so option (1) is true.

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Option (4) is false because:

$$n[V_5] = \begin{pmatrix}2 \\ -1 \\ -1 \end{pmatrix} \neq d[V_{2_4}] = \begin{pmatrix}1 \\ 1 \\ 1 \end{pmatrix}$$

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Option (2) is true because: $A$ is symmetric $\implies$ $A$ is diagonalizable

Answer 2

You may use the fact that a matrix is orthogonally diagonalizable iff it is symmetric. You may prove this fact if you are interested. Now you can immediately know whether statements 2 and 3 are true or not. Then what remains is to find a basis (consists of 1 vector $\vec v$ only) for $V_2$. (Hint: you know that $\vec v$ is orthogonal to $V_5$.)