Find the number of distinct real values of $c$ such that $A^2x=cAx$

Asked May 25 '15 at 14:52

Active May 25 '15 at 16:08

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Let $$A= \begin{pmatrix} 5 & -3 & 0 \\ -3 & 5 & 0 \\ 0 & 0 & 2 \end{pmatrix}$$ and $c$ be a real no. such that $A^2x=cAx$ for some non-zero vector $x$. Then the number of distinct real values of $c$ is

a) $0$ b) $1$ c) $2$ a) $3$

$A^2x=cAx$ $\to$ $A.Ax=c.Ax$ $\to $ $c$ is eigenvalue of $A$.

Eigenvalue of $A$ are $8,2,2$. So c) becomes true...

Is this correct?

edited May 25 '15 at 16:08

Martin Sleziak

53,687

asked May 25 '15 at 14:52

Tani

1,026

3

You have to say why $Ax\neq 0$. – Bernard May 25 '15 at 15:00
2

Your statement about $c$ being an eigenvalue of $A$ holds, but only because $A$ is non singular! – b00n heT May 25 '15 at 15:00

Find the number of distinct real values of $c$ such that $A^2x=cAx$

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