prove $2\times 2$ real matrix has an eigenvalue = $\max_{x\neq0}\frac{\|Ax\|}{\|x\|}$ Let $A$ be $2 \times 2$ real matrix and set 
$r(A) = \max_{x\neq0}\frac{\|Ax\|}{\|x\|}$
where$\left\|\cdot\right\|$ is the Euclidean norm. Prove the matrix $A$ always has an eigenvalue $\lambda$
with $ |\lambda| = r(A).$
I know that if a matrix is symmetric, then $\lambda_{\max} = \max_{x\neq0} <Ax,x>$. But A is not always symmetric, how can I solve this problem?
 A: For a general matrix, the statement is no longer true. Consider a rotation matrix
$$A=\left(\begin{array}{rr}\cos a & \sin a\\ -\sin a & \cos a\end{array}\right)$$
Then $$\max_{x\neq0} \frac{\|Ax\|}{\|x\|} = 1,$$ yet $A$ has no eigenvalues.
You would need the fact that $A$ has eigenvalues for the statement to hold. If $A$ does have at least an eigenvalue, then we can prove the statement by looking at two cases:


*

*if $A$ has a double eigenvalue (i.e. $\lambda_1=\lambda_2$).

*if $A$ has two different eigenvalues $\lambda_1>\lambda_2$.

A: Hint: Note that $$||Ax||^2=x^TA^TAx$$ Now try and convince yourself that $$\max_{||x||=1}||Ax||=\sqrt{\lambda_{max}(A^TA)}$$Now try and find the relationship between eigenvalues of $A$ and $A^TA$. Note that the steps so far doesn't make use of the fact that $A$ is a $2\times 2$ matrix.
A: I think this might not be true in general. For example, consider the matrix:
\begin{equation*}
A = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}
\end{equation*}
$A$ has repeated eigenvalue $\lambda_1 = \lambda_2 = 0$; while $r(A) = 1$ since $x = \left[0, 1 \right]^T$ maximizes $\frac{||Ax||}{||x||}$
