2
$\begingroup$

$A$ 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?


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


(2) $A$ is diagonalizable.


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


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

How can I find the true statement(s)?

$\endgroup$
  • $\begingroup$ What is your attempt? $\endgroup$ – iamvegan Sep 3 '16 at 11:43
  • $\begingroup$ @iamvegan When I posted the question I really didn't know ho to continue. Now, thanks to Li Chun Min I've provided this answer, but I don't know if it is right. $\endgroup$ – ᴜsᴇʀ Sep 4 '16 at 9:53
3
$\begingroup$

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 $\alpha \neq \beta$.


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


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.


Option (4) is false because:

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


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

$\endgroup$
  • 1
    $\begingroup$ If $A$ is symmetric then diagonalizable. So $\mathbf{(2)}$ is also true. Your work seems to be okay. $\endgroup$ – iamvegan Sep 4 '16 at 14:40
2
$\begingroup$

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 $V_5$.)

$\endgroup$
  • $\begingroup$ Thanks for your help: from your answer I've understand how to start. I've write this answer, but I really don't know if it's right. $\endgroup$ – ᴜsᴇʀ Sep 4 '16 at 9:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.