When is the given matrix not diagonizable?

Question

Given the following matrix $$A = \begin{bmatrix} 0 & 1\\ bk+b & -bk\end{bmatrix}$$ where $b \neq 0$ and $k \neq 0$, find when the given matrix is not diagonizable.

My initial thought would be when the matrix doesn't have $2$ linearly independent eigenvectors. In this case it would be when the eigenvalue has a multiplicity of 2. I.e. $\lambda_1=\lambda_2 $.

I find the eigenvalues:

$ \lambda_1=\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\ \lambda_2=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2} $

I then let them equal each other:

$ (1)\qquad\frac{-bk+\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}=\frac{-bk-\sqrt{(bk)^2-4\cdot(-bk-b)}}{2}\\ \quad\\ \quad\\ (2)\qquad\sqrt{(bk)^2-4\cdot(-bk-b)}=-\sqrt{(bk)^2-4\cdot(-bk-b)} $

Here my initial thought would be that the only case this is true if the term under the square root is equal zero.

$ b^2k^2+4bk+4b=0\\ b(bk^2+4k+4)=0 $

Since $b\neq0$ I need the term inside the parantheses to be equal 0.

$ bk^2+4k+4=0 $

I'm a bit lost from here.

I know the solution is $ -bk-k=(\frac{-bk}{2})^2 $.

The solution was provided by someone else here on Stack Exchange, he used the determinant and the trace of the matrix to come to that conclusion, which I can't seem to replicate.

Answer 1

We have that

• $Tr(A)=\lambda_1+\lambda_2=-bk$

• $\det(A)=\lambda_1\lambda_2=-(bk+b)$

and $\lambda_1=\lambda_2=\lambda$ implies

• $\lambda=-\frac{bk}2$

• $\lambda^2=-(bk+b)$

and therefore $\lambda_1=\lambda_2$ when

$$\left(-\frac{bk}2\right)^2=-(bk+b)$$

Finally recall that for $\lambda_1\neq \lambda_2$ $A$ is diagonalizable but for $\lambda_1=\lambda_2$ we need to check directly whether or not $A$ is diagonalizable.

Answer 2

$$b\neq 0, k\neq 0, A=\begin{pmatrix} 0 & 1\\ bk+b & -bk \end{pmatrix}\Rightarrow \det (A-xI)=\begin{vmatrix} -x & 1\\ bk+b & -bk-x \end{vmatrix}=0\Rightarrow$$ $$\Rightarrow x^2+bkx-(bk+b)=0\Rightarrow x=\dfrac{-bk\pm \sqrt{b^2k^2+4bk+4b}}{2}$$ $$x_1=x_2\Rightarrow b^2k^2+4bk+4b=0\Rightarrow b^2k^2=-4(bk+b)\Rightarrow \left( \dfrac{bk}{2}\right)^2=-bk-b$$ Let us now look at the eigenvectors. $$x=-\dfrac{bk}{2}\Rightarrow A-xI=\begin{pmatrix} \frac{bk}{2} & 1\\ -\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2} \end{pmatrix}$$ $$\begin{pmatrix} \frac{bk}{2} & 1\\ -\left( \frac{bk}{2}\right)^2 & -\frac{bk}{2} \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 0\\ 0 \end{pmatrix}\Rightarrow \left\{ \begin{array}{lcc} \dfrac{bk}{2}x+y=0 \\ \\ -\left( \dfrac{bk}{2}\right)^2x-\dfrac{bk}{2}y=0 \end{array} \right.\Rightarrow bkx+2y=0\Rightarrow$$ $$E_{-\frac{bk}{2}}=\langle (-2,bk)\rangle \Rightarrow \dim \left( E_{-\frac{bk}{2}}\right) =1$$

Not diagonalizable.

Answer 3

The characteristic polynomial is $$ \det\begin{bmatrix} -X & 1\\ bk+b & -bk-X\end{bmatrix}= X^2+bkX-(bk+b) $$ The discriminant is $$ (bk)^2+4(bk+b)=b^2k^2+4bk+4b $$ If the discriminant is nonzero, then the matrix is diagonalizable (over the complex numbers), because it has two distinct eigenvalues.

If the discriminant is zero, then the matrix is not diagonalizable: a $2\times2$ matrix with a unique eigenvalue (of multiplicity $2$) is diagonalizable if and only if it is already diagonal. But it's not necessary to believe to the preceding statement: just note that the unique eigenvalue is $-bk/2$ so we need to compute the rank of $A+(bk/2)I$: $$ A+\frac{bk}{2}I= \begin{bmatrix} bk/2 & 1\\ bk+b & -bk/2\end{bmatrix} $$ Such a matrix surely has rank less than $2$, but it cannot have rank $0$, which would be required for diagonalizability.

If you want diagonalizability with a real matrix, the eigenvalues must so be distinct and real; this is the same as $$ b^2k^2+4bk+4b>0 $$

Thus the answer is:

1. $A$ is not diagonalizable over the complex numbers if and only if $b^2k^2+4bk+4b=0$.
2. $A$ is not diagonalizable over the real numbers if and only if $b^2k^2+4bk+4b\le0$.