Proving a matrix is diagonalizable given the characteristic polynomial.

Question

If we are given a real symmetric 2x2 matrix how can we deduce that it is diagonalizable given that we know the characteristic polynomial?

Answer 1

I think this is a bit of a trick problem, but oh well. We know a $\text{2 x 2}$ matrix is diagonalizable if the eigenvalues are distinct, so it suffices to show a real symmetric $\text{2 x 2}$ matrix with repeated eigenvalues is already diagonal. (I take for granted the standard quadratic form argument that eigenvalues are necessarily real.)

We can assume WLOG that $A$ has 2 positive eigenvalues. If this wasn't the case, then instead the argument would be on $A' := A + \delta I$ for large enough $\delta$ so $A' \succ 0$, and $A$ is diagonalizable iff $A'$ is. (More simplistically, if we prove $A'$ has all off-diagonal components equal to zero then $A'-\delta I =A$ does too.)

Then in the case of repeated eigenvalues, examining the characteristic polynomial we can see:

$0\lt \lambda_1\cdot\lambda_2 = \lambda_1^2 =\det\big(A\big)=a_{1,1}\cdot a_{2,2} -a_{1,2}\cdot a_{2,1}= a_{1,1}\cdot a_{2,2}-a_{1,2}^2$
$0\lt \lambda_1+\lambda_2 = 2\cdot\lambda_1 = \text{trace}\big(A\big) = a_{1,1}+a_{2,2}$

but we also have
$ \lambda_1 = \big(\lambda_1^2\big)^\frac{1}{2}=\big(a_{1,1}\cdot a_{2,2}-a_{1,2}^2\big)^\frac{1}{2} \leq \big(a_{1,1}\cdot a_{2,2}\big)^\frac{1}{2} \leq \frac{1}{2}\big(a_{1,1}+a_{2,2}\big) = \frac{1}{2}\big(2\lambda_1\big) = \lambda_1$

the first inequality follows because
$a_{1,1}\cdot a_{2,2}-a_{1,2}^2 \leq a_{1,1}\cdot a_{2,2}$
i.e. since we are working in reals we know $a_{1,2}^2\geq 0$.

Since this upper bound is met with equality, we see that repeated eigenvalues for a $\text{2 x 2}$ real symmetric matrix implies $0=a_{1,2}=a_{2,1}$ -- i.e. the matrix is already diagonal. (The second inequality is of course $\text{GM}\leq \text{AM}$ and since it is met with equality this also implies $a_{1,1}=a_{2,2}$, i.e. $A\propto I$.)