Suppose that the 2x2 real matrix B = satisfies ad - bc = 1 and a+d = 2.For which values of the entries a,b,c and d will B be diagonalizable by a real matrix?Explain.
Hello,
A matrix 2x2 is diagonalizable if its eigenvalues are distinct (or that the sum of the dimensions of its eigenspaces is 2)
So the characteristic polynomial (remember that the eigenvalues are the root of this polynomial) of the matrix is defined as :
$\displaystyle \chi_B (\lambda)=\begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix}=(a-\lambda)(d-\lambda)-bc$
$\displaystyle (a-\lambda)(d-\lambda)-bc=ad+\lambda^2-a \lambda - d \lambda - bc=\underbrace{ad-bc}_{1}-(\underbrace{a+d}_{2}) \lambda + \lambda^2$
$\displaystyle \chi_B (\lambda)=\lambda^2-2\lambda+1=(\lambda-1)^2$
So 1 has to be an eigenvalue and generate an eigenspace of dimension 2.
Let's study the eigenspaces generated by 1.
We have to find $\displaystyle x_1, \ x_2$ such as :
$\displaystyle \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 1 \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$
So it's equivalent to solving this system of equations :
$\displaystyle \begin{bmatrix} ax_1+bx_2=x_1 \\ cx_1+dx_2=x_2 \end{bmatrix}$
First of all, we know that $\displaystyle a+d=2$, that is to say $\displaystyle d=2-a$.
So the system is now :
$\displaystyle \begin{bmatrix} ax_1+bx_2=x_1 \longleftrightarrow (a-1)x_1+bx_2=0 \\ cx_1+(2-a)x_2=x_2 \longleftrightarrow cx_1+(1-a)x_2=0\end{bmatrix}$
And from here, i can't really know... If a=1, nothing tells us that b and c are different from 0, which is a condition that could help because it can let us say that a has to be different from 1 (or else, the eigenspace will be the null vector).
Can you go further though ?