# For Which Value The Matrix is Diagonalizable?

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?

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

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

• 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. Aug 26, 2015 at 14:47
• 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.
– GBQT
Aug 26, 2015 at 14:49
• What do you mean by "clearly it shows that $x = 2$ is a zero of the polynomial"? Aug 26, 2015 at 15:53