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Let $a$ be a real number and consider the square matrix of order 3,

$$ N=\left[ \begin{array}{ccc} a&1&1\\ 1&a&1\\ 1&1&a \end{array} \right] $$

Determine all values of $a$ for which:

$(a)$ The matrix $N$ is invertible.

$(b)$ For every $v \in \mathbb R^3$, $v^TNv\ge0.$

please check if the solution is ok, if there is an error please correct, thanks

Solution :

$a)$ We show for what values of $a$ the determinant of N is zero,

$$\begin{vmatrix} a&1&1\\ 1&a&1\\ 1&1&a \end{vmatrix}=0 $$

$\Rightarrow \quad$ $(a-1)^2(a+2)=0$ $\quad \Rightarrow \quad$ $a=1\quad $ or $\quad a=-2$.

therefore $N$ is invertible for all $a\in \mathbb R -\{1,-2\} $

$b)$ we show for what values of $a$, the matrix N is positive semi-definite.

$i)$ $a\ge0$ $\quad ($in particular for $\quad v = e_i, \quad i=1,2,3 \quad \Rightarrow $, ${e_i}^TNe_i\ge0.$)

$ii) $all the eigenvalues of N are greater than or equal to zero.

$\quad $the characteristic polynomial of N is given by,

$\quad p(\lambda)=det(N-\lambda I)$

$$\begin{vmatrix} a-\lambda&1&1\\ 1&a-\lambda&1\\ 1&1&a-\lambda \end{vmatrix}=0 $$

$\quad \Rightarrow \quad $ $(a-\lambda-1)^2(a-\lambda+2)=0$ $\quad \Rightarrow \quad$ $a-\lambda-1=0\quad $ or $\quad a-\lambda+2=0$.

$\quad \Rightarrow \quad $ $\lambda =a-1\quad $ or $\quad \lambda = a+2$

as the eigenvalues must be non-negative so that the matrix N is positive semi-definite.

$\quad \Rightarrow \quad $ $a\ge 1 \quad$ or $\quad a\ge -2 $

then of $(i)$ and $(ii)$we have

$a\ge 0$, $a\ge 1$ and $a\ge -2$.

therefore

$a\ge 1$.

please check if the solution is ok, if there is an error please correct, thanks

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  • $\begingroup$ Seems fine. For (b) the first part is not needed. It suffices to find the values of $a$ for which all eigenvalues are non-negative. $\endgroup$ – Test123 Nov 29 '17 at 17:51
  • $\begingroup$ very minor quibble "the eigenvalues must be non-negative so that the matrix N is positive semi-definite." $\implies a\ge 1$ and $a\ge -2$ and I would drop $i)$ from your discussion. $\endgroup$ – Doug M Nov 29 '17 at 18:09
  • $\begingroup$ It does look ok (I trust you for calculations though). As stated above, i) is unnecessary $\endgroup$ – tommy1996q Nov 29 '17 at 18:15
  • $\begingroup$ You could’ve saved yourself a little work by computing the eigenvalues right off the bat instead of computing the determinant separately, since you’re going to need the eigenvalues later, anyway. The determinant of the matrix is their product. $\endgroup$ – amd Nov 30 '17 at 2:31
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Your solution seems fine. As mentioned in one comment though, the first part of your writings at $(b)$ isn't needed, as all you need to do is find the cases for which the eigenvalues are non-negative.

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