# Is following matrix diagonalizable?

$$A:= \begin{pmatrix} a & c \\ b & d \\ \end{pmatrix}$$

Determine the conditions on $𝑎,𝑏,𝑐,𝑑$ so that $𝐴$ is diagonalizable i.e. $𝐴$ has two linearly independent eigenvectors.

I think for $A$ to be diagonalizable, it should be a symmetric matrix but Im not sure whether the answer should be: $b = c$ (the case for symmetric matrix).

• Which is your base field? $\mathbb R$? $\mathbb C$? – José Carlos Santos Sep 14 '17 at 8:02
• Have you written out the characteristic polynomial in terms of $a,b,c,d$? – Tobias Kildetoft Sep 14 '17 at 8:02

No, it is not necessary for $A$ to be symmetric, in order to be diagonalizable. Consider the non-symmetric matrix $$A=\begin{pmatrix} 1 & 2 \cr 3 & 4 \end{pmatrix},$$ which is diagonalizable. In general, one just has to follow the steps with the characteristic polynomial and eigenvalues as in this duplicate, but for general $a,b,c,d$.

$A$ doesn't have to be symmetric. "Almost all" matrices are diagonalizable.

First, if there are two distinct eigenvalues, then the matrix is diagonalizable. Since you can compute the two eigenvalues directly, this is straightforward; this sufficient condition for $A$ to be diagonalizable is $(a-d)^2 + 4bc \neq 0$.

Even if the matrix has only one eigenvalue (which then has algebraic multiplicity 2) the matrix may still be diagonalizable. An example is the identity matrix. However, there may be only one eigenvecor, an example is

A = [0,1; 0 0]

which is not diagonalizable.

So, for the case of a single eigenvalue $\lambda$, to determine if $A$ can be diagonalizable, compute the reduced matrix

$\lambda I - A$

Since $\lambda$ is an eigenvalue, we know that this matrix is singular, and a null vector of this matrix is an eigenvector of $A$. However, if this matrix is the zero matrix, then there are two linear independent eigenvectors. The condition for this to happen is clearly $a=d$, $b=c=0$.

In summary: $A$ can be diagonalized if

1) $(a-d)^2 + 4bc \neq0$ (two distinct eigenvalues each with an eigenvector)

or

2) $a=d$, $b=c=0$ (a single eigenvalue with two independent eigenvectors)

• I am not sure what you mean that example to be. It is certainly not diagonalizable. – Tobias Kildetoft Sep 14 '17 at 8:45
• @Tobias: I assume you refer to the example A=[0,1;0 0]. Together with the example A=I, it shows that when there is only one eigenvalue, more analysis is needed to check if A can be diagonalized. I edited to make that more clear. Hope it helps. – Uffe_Thygesen Sep 14 '17 at 9:02
• Ahh, ok. I see. – Tobias Kildetoft Sep 14 '17 at 9:03

So a standard vector $\vec{v} = [\begin{smallmatrix}x \\ y \end{smallmatrix}]$ is really just a linear combination of the basis vectors such that $\vec{v} = x \boldsymbol{ \hat{i}} + y \boldsymbol{\hat{j}}$ where $\boldsymbol{\hat{i}} = [\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}]$ and $\boldsymbol{\hat{j}} = [\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}]$. So what a matrix $A = [\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}]$ means is, where would be vector end up in our traditional basis if I decided to replace $\boldsymbol{\hat{i}}$ and $\boldsymbol{\hat{j}}$ with $[\begin{smallmatrix} a \\ c \end{smallmatrix}]$ and $[\begin{smallmatrix} b \\ d \end{smallmatrix}]$ respectively, meaning $A\vec{v} = x[\begin{smallmatrix} a \\ c \end{smallmatrix}] +y[\begin{smallmatrix} b \\ d \end{smallmatrix}]$.

In the case where $[\begin{smallmatrix} a \\ c \end{smallmatrix}]$ and $[\begin{smallmatrix} b \\ d \end{smallmatrix}]$ are linearly dependent, that means that $A$ projects all of 2-space onto a single line. So here's a question, what would that transformation do to the area of a 2D figure by reducing it to a line? Right, it would reduce the area to $0$. Therefore, if $\det(A) =0$, then $A$ is not diagonizable.

The other instance is when there is only one eigenvector span, e.g. $[\begin{smallmatrix} 1 & 1 \\0 & 1 \end{smallmatrix}]$. That one gets into the eigenvector definition, which actually covers both cases:

$$A\vec{v} = \lambda I \vec{v}$$ $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \vec{v} = \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \vec{v}$$ $$\begin{bmatrix} a & b \\ c & d \end{bmatrix} \vec{v} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} \vec{v} = \vec0$$ $$\bigg(\begin{bmatrix} a & b \\ c & d \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}\bigg) \vec{v} = \vec0$$ $$\begin{vmatrix} a - \lambda & b \\ c & d-\lambda \end{vmatrix} = 0$$ $$(a - \lambda)(d-\lambda ) - bc = 0$$

So what you now have is a quadratic equation to determine your eigenvalues. Using factoring (or the quadratic formula if you're super desperate), you can find if you have no real solutions (no eigenvalues - bad), one solution, or rather two identical solutions (one eigenvalue - bad), or two real solutions (two eigenvalues - possible).

If you have two real solutions, you'll want to check your eigenvalues by using them to solve a linear system of equations, to wit:

$$(a - \lambda) x + by = x$$ $$cx + (d - \lambda) y = y$$

If those solutions are nontrivial, (i.e. $x \ne 0$ or $y \ne 0$), then congratulations, you have a diagonizable linear transformation matrix in 2-space.

tl;dr

$$(\det(A) \ne 0) \land (N(\lambda)=2) \land (\lambda = p + 0i | p \in \mathbb{R}) \land (\vec{v}_{\lambda} \ne \vec0) \implies A \text{ is diagonizable}$$

• There are certainly diagonalizable non-invertible matrices. – Tobias Kildetoft Sep 14 '17 at 10:16
• To wit? I'm trying to think of how a non-invertible matrix can be diagonalized. – Adam Gluntz Sep 14 '17 at 10:23
• Just take a diagonal matrix with one of the diagonal entries equal to $0$. – Tobias Kildetoft Sep 14 '17 at 10:32
• I did not understand that to be a proper "diagonal" matrix. I was thinking of a matrix that has an eigenbasis. – Adam Gluntz Sep 14 '17 at 10:34
• Jeez, I thought I was being all cool and now here I am with egg all over my face. – Adam Gluntz Sep 14 '17 at 10:40