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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, 2017 at 2:55

3 Answers 3

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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. $\endgroup$ Apr 12, 2017 at 3:16
  • $\begingroup$ @AmitaiYuval the OP is talking about real matrices here: Euclidean norm, symmetric matrices. $\endgroup$ Apr 12, 2017 at 3:17
  • $\begingroup$ Yes, but every real number is also a complex number. And every real matrix is in particular complex. $\endgroup$ Apr 12, 2017 at 3:19
  • $\begingroup$ And I have a feeling the statement holds for every real $2\times2$ matrix. $\endgroup$ Apr 12, 2017 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. $\endgroup$ Apr 12, 2017 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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