# For which numbers is the matrix diagonalizable?

We have the matrix \begin{equation*}A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}\end{equation*}

I want to find for which real numbers $a,b,c,d$ the matrix is diagonalizable in $\mathbb{R}^{2\times 2}$ and for which in $\mathbb{C}^{2\times 2}$.

The charachteristic polynomial is \begin{align*}\det (A-\lambda I)=\lambda^2-(a+d)\lambda +(ad-cb)\end{align*}

So, the eigenvalues are

We have the following cases:

As an element of $\mathbb{R}^{\times 2}$ we have the following:

• Expression under the root $< 0$: no real eigenvalue, so the matrix is not diagonalizable.
• Expression under the root $> 0$: two different eigenvalues, that means that the matrix is not diagonalizable, or not?
• Expression under the root $= 0$, we have an eigenvalue of multiplicity $2$. What do we have in this case?

As an element of $\mathbb{C}^{\times 2}$ we have the following:

• Expression under the root $\neq 0$: two different eigenvalues , that means that the matrix is not diagonalizable, right?
• Expression under the root $= 0$, we have an eigenvalue of multiplicity $2$. What do we have in this case?
• Two different eigenvalues implies diagonalizable. If you have both the same value for the eigenvalue, it means that $$\chi_{A}\left(X\right)=\left(X-a\right)^2$$ hence if it was diagonalizable you would have $A=PDP^{-1}$ where $D=aI_2$ then $A=aI_2$ that's the only case where it will be diagonalizable. – Atmos Jan 21 '18 at 13:43
• So, can we not get a further condition that has to be satisfied so that we know when the matrix is diagonizable in the case of two same eigenvalues? @Atmos – Mary Star Jan 21 '18 at 13:55
• At condition (2) it says that it has to have at least n real roots, if it has n complex eigenvalues, what does then hold? @Moo – Mary Star Jan 21 '18 at 13:58
• I haven't really understood why in general, if a matrix has complex eigenvalues, it is not diagonalizable. Coudyou explain it further to me? @Moo – Mary Star Jan 21 '18 at 15:35

A matrix is diagonalizable iff its minimum polynomial is a product of simple linear factors.

Also, the distinct roots of the minimum polynomial are the same as the distinct eigenvalues of the matrix.

So for a 2x2 matrix, the only case when it is NOT diagonalizable is when the matrix has 2 repeated eigenvalues.

The matrix has repeated eigenvalues when:

$$det(A-\lambda I) = \lambda^2-\lambda \cdot (a+d) + (ad-bc)$$ has repeated roots....ie: $b^2-4ac=0$, which should simplify to: $$(a-d)^2+4bc=0$$

So all values of a, b, c and d that satisfy $(a-d)^2+4bc\ne0$ make A diagonalizable.

• If the matrix has repeated eigenvalues then its characteristic polynomial is $(x-\alpha )^2$ where $\alpha$ is the eigenvalue. The matrix can still be diagonalizable in this case, but this requires further analysis. – Dave Jan 21 '18 at 14:26
• So my solution doesn't hold only when A is a diagonal matrix and with both its diagonal elements being the same. Am I correct? – Dylan Zammit Jan 21 '18 at 14:34
• Yes, because this is the only case when the matrix with repeated eigenvalue has two dimensional eigenspace. Perhaps a minor special case, but nonetheless a necessary addition. – Dave Jan 21 '18 at 14:40
• I got stuck right now. When a 2x2 matrix has two repeated eigenvalues, the matrix is diagonalizable only when the the eigenvectors are independent? @Dave – Mary Star Jan 21 '18 at 16:00
• In the repeated eigenvalue case: the matrix is diagonalizable if and only if $c=b=0$. – Dave Jan 21 '18 at 16:40