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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?

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  • $\begingroup$ There are three possibilities to consider for the roots of the characteristic polynomial of a real $2\times2$ matrix: distinct real roots, repeated real root, or complex conjugate roots. Seems like it shouldn’t be too much work to examine each case separately. $\endgroup$ – amd Apr 12 '17 at 2:55
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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.

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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:

  1. if $A$ has a double eigenvalue (i.e. $\lambda_1=\lambda_2$).
  2. if $A$ has two different eigenvalues $\lambda_1>\lambda_2$.
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  • $\begingroup$ Of course $A$ has an eigenvalue, like any other complex matrix. $\endgroup$ – Amitai Yuval Apr 12 '17 at 3:16
  • $\begingroup$ @AmitaiYuval the OP is talking about real matrices here: Euclidean norm, symmetric matrices. $\endgroup$ – Quang Hoang Apr 12 '17 at 3:17
  • $\begingroup$ Yes, but every real number is also a complex number. And every real matrix is in particular complex. $\endgroup$ – Amitai Yuval Apr 12 '17 at 3:19
  • $\begingroup$ And I have a feeling the statement holds for every real $2\times2$ matrix. $\endgroup$ – Amitai Yuval Apr 12 '17 at 3:25
  • $\begingroup$ Everybody knows that. But you have to consider the context of the question. And yes, the statement holds for any matrix, if we talks about real matrices as complex matrices. It's not that difficult to modify the first case to suite that context. $\endgroup$ – Quang Hoang Apr 12 '17 at 3:28
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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 $\frac{||Ax||}{||x||}$

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