The maximum value of "a" such that the matrix pmatrix -3 & 0 & -2 \\ 1 & -1 & 0 \\ 0 & a & -2 pmatrix has three linearly independent real eigenvectors is

$\frac{1 + 2\sqrt{3}}{3\sqrt{3}}$

$\frac{1 + \sqrt{3}}{3\sqrt{3}}$

GATE 2015 EE – Question 36

Engineering Mathematics · Linear Algebra: Eigen values, Eigen vectors · 2 marks · Multiple choice

The maximum value of "$a$" such that the matrix $\begin{pmatrix} -3 & 0 & -2 \\ 1 & -1 & 0 \\ 0 & a & -2 \end{pmatrix}$ has three linearly independent real eigenvectors is

$\frac{2}{3\sqrt{3}}$
$\frac{1}{3\sqrt{3}}$
$\frac{1 + 2\sqrt{3}}{3\sqrt{3}}$
$\frac{1 + \sqrt{3}}{3\sqrt{3}}$

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Correct answer: (B) $\frac{1}{3\sqrt{3}}$

Explanation

The characteristic polynomial is $-[(\lambda + 1)(\lambda + 2)(\lambda + 3) + 2a]$. Three independent real eigenvectors need three real eigenvalues, so the cubic $(\lambda + 1)(\lambda + 2)(\lambda + 3) = -2a$ must have three real roots. This cubic has a local extreme value of magnitude $\frac{2}{3\sqrt{3}}$, so we need $2a$ no larger than that. So the maximum value of $a$ is $\frac{1}{3\sqrt{3}}$.