I'm sorry if my question is rather stupid, but I have a brainfreeze right now.

I want to prove that, for every $A\in GL(2,\mathbb{R})$ and for every $v\in \mathbb{R}^2$, $\|v\|=1$, I can find $w\in \mathbb{R}^2$, $\|w\|=1$ such that $\|Av\|=1/\|A^{-1}w\|$.

Writing $v=(v_1,v_2)$ and $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ I have to find a solution $(w_1,w_2)$ for the equation


such that $w_1^2+w_2^2=1$.

Does this solution always exist?

According to my calculations this is equivalent to ask if it is always true

$(-2CD-E^2)^2-4C^2(D^2+E^2)\ge0$, with $D=b^2+a^2-d^2-c^2$, $E=-db-ac$, $C=\frac{(ad-bc)^2}{(av_1+bv_2)^2+(cv_1+dv_2)^2}-(d^2+c^2)$, but I can not go any further in the proof..


Let $A = U \Sigma V^T$ be the SVD decomposition of $A$. Set $w = \frac{1}{\sigma_i}u_i (||\Sigma V^Tv||)^{-1}$, with $u_i$ some column of $U$, and $\sigma_i$ its corresponding singular value. First, notice that

$$||Av|| = ||U \Sigma V^T v|| = ||\Sigma V^T v||$$

And $$||A^{-1}w|| = ||\Sigma^{-1} U^T w|| = ||\Sigma^{-1} U^T \frac{1}{\sigma_i}u_i (||\Sigma V^Tv||)^{-1} || = || \frac{1}{\sigma_i}\Sigma^{-1} e_i (||\Sigma V^Tv||)^{-1} || = ||e_i (||\Sigma V^Tv||)^{-1}|| = |(||\Sigma V^Tv||)^{-1}| ||e_i|| = (||\Sigma V^Tv||)^{-1}$$

Hence, $$||A^{-1}w||^{-1} = ||\Sigma V^Tv|| = ||Av||$$

  • $\begingroup$ thank you, but it's not clear to me why it should be $||w||=1$.. $\endgroup$ – Francis Clochard Jun 3 '15 at 15:44
  • $\begingroup$ ouch!, sorry, forgot about that!. I will explore whether this is does happen or can be achieved with a similar approach. $\endgroup$ – Misguided Jun 3 '15 at 15:53
  • $\begingroup$ Well, this definitely doesn't work. Adding the requirement that $||w|| = 1$ requires that $||Av|| = |\frac{1}{\sigma_i}|$ $\endgroup$ – Misguided Jun 3 '15 at 15:59

Yes, the answer is $w=\frac{Av}{\|Av\|}$


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