For which numbers is the matrix diagonalizable?

Question

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
\begin{align*}\lambda^2&-(a+d)\lambda +(ad-cb)=0 \\ \Rightarrow &\lambda_{1,2}=\frac{(a+d)\pm \sqrt{(a+d)^2-4(ad-cb)}}{2}=\frac{(a+d)\pm \sqrt{a^2+2ad+d^2-4ad+4cb}}{2}\\ & =\frac{(a+d)\pm \sqrt{a^2-2ad+d^2+4cb}}{2} =\frac{(a+d)\pm \sqrt{(a-d)^2+4cb}}{2}\end{align*}

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?

Answer 1

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.

Answer 2

The matrix can be diagonalised iff it has a complete set of eigenvectors.

To be more explicit for a $2\times2$ matrix we can distinguish a few cases:

1. If the two eigenvalues are distinct, which occurs when $(a-d)^2+4bc\neq0$, then the eigenvectors are linearly independent, and the matrix can be diagonalised.

2. If the eigenvalues are identical (degenerate), one needs to check if the eigenvectors are linearly dependent. The degenerate condition requires $(a-d)^2+4bc = 0$, and the eigenvalues are $\lambda_1 = \lambda_2 = (a-d)/2$. The equations for the eigenvector(s) $\mathbf{x} = (x,y)$ are therefore \begin{align} \begin{cases} \frac{a-d}{2} x + b y = 0 \\ c x - \frac{a-d}{2} y = 0 \end{cases} \end{align} These equations are linearly dependent because of the degeneracy condition, so there are two possibilities:

   • We can determine no relation between $x$ and $y$, and therefore two eigenvectors exist (we can choose any linearly independent pair). This can occur only if all coefficients of the eigenvectors system of equations vanish, i.e. $(a-d) = b = c = 0$. The matrix in this cases reduces to the (already diagonal) form \begin{equation} A = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} \end{equation}

   • As soon as one of the coefficients of the eigenvectors equations does not vanish, we establish a relation between $x$ and $y$, thus choosing one specific direction for the (only) eigenvector. As a consequence the matrix is defective and not diagonalisable. This can manifest in two forms:

     • If $(a-d)=0$ then we must have that either $b=0$ or $c=0$, because we still must satisfy the degenerate condition $(a-d)^2+4bc = 0$. In this case the matrix has the form \begin{equation} A = \begin{bmatrix} a & b \\ 0 & a \end{bmatrix} \mbox{ with } b\neq0 \qquad \mbox{ or } \qquad A = \begin{bmatrix} a & 0 \\ c & a \end{bmatrix} \mbox{ with } c\neq0. \end{equation} which are classic non-diagonalisable Jordan blocks.

     • If $(a-d)\neq0$ then, for $(a-d)^2+4bc = 0$ to hold, we must have that both $b\neq0$ and $c\neq0$. Also in this case there exist only one eigenvector, and the matrix is non-diagonalisable. An example of such a matrix is

\begin{equation} A = \begin{bmatrix} 5 & 1 \\ -4 & 1 \end{bmatrix} \end{equation}