# $\sup \left\| A x + B y + C z \right\|$ subject to $\left\|x\right\| = \left\|y\right\| = \left\|z\right\| = 1$

I'm interested in finding

$$\sup \left\| A x + B y + C z \right\|$$ subject to $$\left\|x\right\| = \left\|y\right\| = \left\|z\right\| = 1$$

where $$A$$, $$B$$, $$C$$ and $$x$$, $$y$$, $$z$$ are real matrices and vectors, respectively, of compatible sizes, and the norms are Euclidean.

What is this problem called in the literature? (If there is no established name, does this problem reduce to a well-known one?)

(If there was only one constraint, this would be the induced matrix norm)

$$[x, y, z]^{T}= \alpha_{1} v^{T}_{1} + \alpha_{2} v^{T}_{2}+\alpha_{3} v^{T}_{3}$$, where $$v_{i}$$ is the $$i$$-the right singular vector of $$[A,B,C]$$ (i.e. singular vector corresponding to $$i$$-th largest singular value).

This assumes that the optimal $$[x, y, z]^{T}$$ is a linear combination of the first 3 right singular vectors of $$[A, B, C]$$. However, this assumption is incorrect. As a counterexample, consider

$$A = \begin{bmatrix}2&0&0\\0&2&0\\0&0&2\\0&0&0\\0&0&0\end{bmatrix}$$

$$B = \begin{bmatrix}0\\0\\0\\1\\0\end{bmatrix}$$

$$C = \begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$$

Hint:

$$\left \| Ax+By+Cz \right \| = \left \| \left [ A, B, C \right ] \left [ \begin{matrix} x\\ y\\ z \end{matrix} \right ]\right \|$$

$$[x, y, z]^{T}= \alpha_{1} v^{T}_{1} + \alpha_{2} v^{T}_{2}+\alpha_{3} v^{T}_{3}$$, where $$v_{i}$$ is the $$i$$-the right singular vector of $$[A,B,C]$$ (i.e. singular vector corresponding to $$i$$-th largest singular value). Denoting $$[v_{ix},v_{iy},v_{iz}]^{T}=v_{i}$$ to mark the partition of $$v_{i}$$ corresponding to $$x$$, $$y$$ and $$z$$.

Now solve: $$||x||^{2}=1 \rightarrow ||(\alpha_{1} v_{1x} + \alpha_{2} v_{2x}+\alpha_{3} v_{3x})||=1$$

Similarly , $$||(\alpha_{1} v_{1y} + \alpha_{2} v_{2y}+\alpha_{3} v_{3y})||=1$$

$$||(\alpha_{1} v_{1z} + \alpha_{2} v_{2z}+\alpha_{3} v_{3z})||=1$$

Three equations to solve for three unknowns $$\alpha_{1}, \alpha_{2}$$, and $$\alpha_{3}$$. Then, your answer is $$\sqrt{\alpha^{2}_{1} \sigma^{2}_{1} + \alpha^{2}_{2} \sigma^{2}_{2} + \alpha^{2}_{3} \sigma^{2}_{3}}$$, where $$\sigma_{i}$$ is the singular value corresponding to $$i$$-the largest singular vectors.

• hint: $[x, y, z]^{T}$ is a linear combination of three orthogonal vectors. – piyush_sao Oct 9 '18 at 6:59
• This answer is still incorrect. See the refutation above. – MaxB Oct 10 '18 at 2:21
• Good counterexample. In that case, you can not solve for $\alpha_{i}$. So in such cases, choose $v_{2}$ and $v_{3}$, such that this system is solvable. – piyush_sao Oct 10 '18 at 16:12
• Your whole assumption is incorrect. This could be a $10^9$-dimensional matrix, and you are assuming that the optimal solution belongs to your 3-dimensional subspace. This assumption is baseless. – MaxB Oct 10 '18 at 17:51
• Do you not see that your answer is incorrect? If you do, please delete it: math.meta.stackexchange.com/questions/4731/… – MaxB Oct 12 '18 at 0:23