Why A is diagonalizable?

Question

This is a part of a question.
Let $A$ be a real $2\times 2$ matrix such that $\,\det A=1,\operatorname{tr}A>2$. Why is $A$ diagonalizable?

The original question asks to prove that $A$ and $B$ have a common eigenvector if $\det B=1,\operatorname{tr}B>2,\operatorname{tr}\left(ABA^{-1}B^{-1}\right)=2$.

Answer 1

If $\lambda_1, \lambda)_2$ are the two (complex) eigenvalues of a real matrix $A$ (size $2\times 2$), then we always have $\det(A) = \lambda_1 \lambda_2$ and $\operatorname{tr} A = \lambda_1 + \lambda_2$.

If the eigenvalues would be equal the eigenvalues would be $1,1$ or $-1,-1$ and thus give trace $-2$ or $2$ which is not the case.

So, the eigenvalues are different, and we now have a base of eigenvectors: the two linearly independent (because they have different eigenvalues) eigenvectors of $A$ must form a base of $\mathbb{R}^2$.

Answer 2

MooS' first & prompt comment brings up the principal reason: $A$ is diagonalisable because the eigenvalues must be different under these givens.

$A\text{ certainly has two}\:\mathbb C\text{omplex eigenvalues}$ $$\lambda_1 = x+iy\quad\text{and}\quad\lambda_2 = \frac 1{\lambda_1} = {x-iy\over x^2+y^2}$$ owing to the condition $\,\det A=1$, in particular both eigenvalues are non-zero. If the imaginary part $\,y\,$ were to be non-zero, then the condition $\:\operatorname{tr}A>2\:$ could not be satisfied. Hence both eigenvalues are real and even greater than $0\,$.

The trace-condition can be read as AM-GM inequality $$\operatorname{tr}A>2\quad\iff\quad \frac 12\left(\lambda_1 + \frac1{\lambda_1}\right)\:>\: \lambda_1\cdot\frac1{\lambda_1} = 1$$ which is strict, and strictness implies $\lambda_1\ne\frac1{\lambda_1}=\lambda_2\,$.

Answer 3

This is because the characteristic polynomial of $A$ is $$\chi_A(\lambda)=\lambda^2-(\operatorname{Tr} A) \lambda+\det A,$$ and that it has two distinct real roots since its discriminant is $$\Delta=(\operatorname{Tr} A)^2-4\det A>4-4=0$$ by the hypotheses.