Linear Algebra - Diagonalizable matrix

Question

It's a new topic we learn during the linear algebra class and I need a bit help understanding. Lets say, for example, that I have this matrix: \begin{pmatrix}2&1\\x&8\end{pmatrix} and x ∈ R

Can one tell me what may be x in order this matrix will be diagonalizable?

Thanks for helping.

Answer 1

The characteristic polynomial of this matrix (denoted $A$) is $$\chi_A(\lambda)=\det(A-\lambda I)=(2-\lambda)(8-\lambda)-x=\lambda^2-10\lambda+16-x$$ and the reduced discriminant of $\chi_A(\lambda)$ is $$\Delta'=9+x$$ hence

• if $x>-9$ then there's two distinct real eigenvalues of $A$ and then it's diagonalizable over $\Bbb R$
• if $x<-9$ then there's two distinct complex eigenvalues of $A$ and then it's diagonalizable over $\Bbb C$ but not over $\Bbb R$.
• if $x=-9$ then the only eigenvalue is $\lambda=5$ but since $A\ne5 I$ then $A$ isn't diagonalizable.

Answer 2

Hint: a real $2\times 2$ matrix is not diagonalizable $\iff$ it has no eigenvalues, or only one eigenvalue and is not a multiple of the identity.

Answer 3

A matrix is diagonalizable iff it has distinct eigenvalues.

For any $2\times 2$ matrix of the form \begin{bmatrix}a&b\\c&d\end{bmatrix} ($a,b,c,d\in\mathbb{R}$), the characteristic equation is $\det(\lambda I-A)=\lambda^2-(a+d)\lambda+ad-bc=0$. For the equation to have 2 distinct roots, the discriminant has to be greater than $0$. That is, $(a+d)^2-4(ad-bc)>0$. Simplification yields $-4bc<(a-d)^2$.

In your case, $a=2$, $b=1$, $c=x$ and $d=8$. So, $-4bc<(a-d)^2$ implies $-4x<36$. That is, $x>-9$.

Note: If you are considering $\mathbb{C}$, then $A$ is diagonalizable when $-4bc>(a-d)^2$.