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The problem reads:

Let $\mathcal V$ be the subspace $\mathcal span \ \mathbf x$, where $\mathbf x$=[3 2 1]$^T$. Find the best approximation in $\mathcal V$ $\mathcal to \ \mathbf y$ =[1 2 3]$^T$ with respect to the following norms:

A. $\Vert\cdot\Vert_1$

B. $\Vert\cdot\Vert_2$

C. $\Vert\cdot\Vert_\infty$

I'm beginning with A. I'm assuming I will minimize $\alpha$ in the following expression: $\Vert\alpha \mathbf x - \mathbf y\Vert_1$, which leads us to |3$\alpha$−1|+|2$\alpha$-2|+|$\alpha$−3|. How exactly do I separate it and graph it in order to find the best $\alpha$? And how do we get the best approximation using $\alpha$?

1 Answers 1

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  • case A

When you graph the function $f : \alpha \mapsto \lVert\alpha x - y\rVert_1$ you have to take into account that this function is a piecewise linear function. So you have to look up the points where the function is not differentiable. These points are obviously $1/3, 1$ and $3$. So the function $f$ can be described as : $$ f(\alpha) \mapsto \begin{cases} -(3\alpha - 1) -(2\alpha-2) -(\alpha - 3) &= -6\alpha +6 & \text { for }&\alpha \leq 1/3 \\ (3\alpha - 1) -(2\alpha-2) -(\alpha - 3) &= 4 & \text { for }&1/3 \leq \alpha \leq 1 \\ (3\alpha - 1) +(2\alpha-2) -(\alpha - 3) &= 4\alpha & \text { for }& 1 \leq \alpha \leq 3 \\ (3\alpha - 1) +(2\alpha-2) +(\alpha - 3)) &= 6\alpha -6 & \text { for }& 3 \leq \alpha \\ \end{cases} $$ Since the function is linear between these points we see that an absolute minimum is reached for all the points $/3 \leq \alpha \leq 1 $

  • case B

Solve $$\frac{d}{d\alpha}\sqrt{(3\alpha - 1)^2 +(2\alpha-2)^2 +(\alpha - 3)^2} = 0 \\ \text{ or : } \frac{d}{d\alpha}\sqrt{14\alpha^2-20\alpha+14} = 0 $$

  • case C Like in case A we have again a piecewise linear function: $$f(\alpha) \mapsto \max(\vert 3\alpha - 1\vert, \vert 2\alpha-2 \vert,\vert\alpha - 3\vert)$$ But the approch is a little more complicated. Let $a = 3\alpha - 1, b = 2\alpha-2, c = \alpha - 3$. If we pose $\vert \cdot \vert = \sqrt{(\cdot)^2}$ then the points where one of these functions takes over from another are the roots of solutions of the equations $a^2 = b^2$, $a^2 = c^2$ and $b^2=c^2$, giving $-1,3/5,4,5/3$. To see which function takes over we look at the following table of values (I added and subtracted a small value $1/11$ (eleven because the denominator doesn't factor out); $$ \begin{array}{} \alpha && a & b & c & \max \\ -1 - 1/11 && 47/11& 46/11& 45/11& 4.27\\ -1 && 4 & 4 & 4 & 4.00 \\ -1+1/11 && 41/11& 42/11& 43/11 & 3.91 \\ 3/5 - 1/11 && 29/55& 54/55& 137/55& 2.49\\ 3/5 && 4/5& 4/5& 12/5& 2.40 \\ 3/5 + 1/11 && 59/55& 34/55& 127/55 & 2.31 \\ 1-1/11 && 19/11& 2/11& 23/11& 2.10 \\ 1 && 2 & 0 & 2 & 2.00\\ 1 + 1/11 && 25/11& 2/11& 21/11& 2.27 \\ 5/3 - 1/11 && 123/33& 38/33& 47/33 & 3.72\\ 5/3 && 4& 4/3& 4/3& 4.00\\ 5/3 + 1/11 && 141/33& 50/33& 41/33& 4.27 \\ \end{array} $$ Since the function is lineair between these points we see that an absolute minimum is reached when $\alpha = 1$.