2
$\begingroup$

For which values of $a$ the matrix $\left(\begin{array}{ccc} 2 & 0 & 0 \\ 2 & 2 & a \\ 2 & 2 & 2 \end{array}\right)$ is diagonalizable:

  1. above $\mathbb{R}$
  2. above $\mathbb{C}$

We need to look at the characteristic polynomial which is $(x-2)^3-2a(x-2)=x^3-6x^2+2x(6-a)+4(a-2)$ , How do I find the eigenvalues one is $2$ and the other? and how so I continue?

$\endgroup$
4
  • $\begingroup$ Try to factorize the characteristic polynomial. $\endgroup$
    – GBQT
    Aug 26, 2015 at 14:18
  • $\begingroup$ @GBQT done, but the eigenvalues should be the x's for which the polynomial is 0? $\endgroup$
    – gbox
    Aug 26, 2015 at 14:32
  • $\begingroup$ Yes, that's it. Now you just need to find how many eigenvalues there is, and if they are real or not. $\endgroup$
    – GBQT
    Aug 26, 2015 at 14:35
  • $\begingroup$ Similar question-math.stackexchange.com/questions/1314164/… $\endgroup$
    – John_dydx
    Aug 26, 2015 at 14:46

1 Answer 1

0
$\begingroup$

$x^3-6x^2+2x(6-a)+4(a-2)$. Clearly it shows that $x=2$ is a zero of the polynomial. Now ,

$x^3-6x^2+2x(6-a)+4(a-2)=0$

$\implies (x-2)(x^2-4x+4-2a)=0$

$\implies x=2 , x=\frac{4\pm\sqrt{16-4(4-2a)}}{2}=2\pm\sqrt{2a}.$

If $a=0$ then , all roots are $2$ and then the matrix is NOT diagonalizable(check!)

If $a\not=2$ then the matrix is diagonalizable.

$\endgroup$
3
  • 1
    $\begingroup$ If $a=0$ all the roots of the characteristic polynomial are $2$, but that does not imply that the matrix is not diagonalizable. Take twice the identity matrix, for instance. $\endgroup$ Aug 26, 2015 at 14:47
  • 1
    $\begingroup$ If $a=0$, one just need to check whether $x-2$ is the minimal polynomial of the matrix, and that's pretty obvious it is not. $\endgroup$
    – GBQT
    Aug 26, 2015 at 14:49
  • $\begingroup$ What do you mean by "clearly it shows that $x = 2$ is a zero of the polynomial"? $\endgroup$ Aug 26, 2015 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .