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Let matrix $A= \begin{bmatrix} 1 & -1 & 1\\ 2 & 1 & 2 \end{bmatrix}$. Find a two dimensional subspace $S^*$ s.t.

$$\underset{{x \in S^*,\|x\|_2 =1}} \min \|Ax\|_2=\underset{{\dim S=2, x \in S},} \max \underset{{\|x\|_2 =1}} \min \|Ax\|_2.$$ I'm not sure how to solve this kind of minimax characterization problem. Thanks!

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    You could try using the squared $l_2-$ norm. This will probably make computation easier. – 2017-01-02
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    @YukiJ You mean to find the singular value of $\|Ax\|_2$? Since we usually define $\|A\|_2 = \max(x^{H}A^{H}Ax)^{1/2}=\sigma_1 (A)$. Not sure whether this is what you meant. – 2017-01-02
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    Would you mind giving a little bit of more clarification. To my understanding $$\underset{{\|x\|_2 =1}} \min \|Ax\|_2 = K$$ is a constant. This means $$\underset{{\dim S=2, x \in S}} \max \underset{{\|x\|_2 =1}} \min \|Ax\|_2 = \underset{{\dim S=2, x \in S}} \max K = K$$ is the same constant. Is my reasoning sound? – 2017-01-02
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    @SlavenGlumac Ok? But how do you find the subspace? – 2017-01-04
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    At https://en.wikipedia.org/wiki/Singular_value_decomposition#Applications_of_the_SVD there is a section titled Total least squares. Your problem is very similar except you have to take two least singular values. – 2017-01-04

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