Question

4. There are hints to this problem on the second page. Let $b_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $b_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Let $W$ be a 1d subspace of $\mathbb{R}^2$. We want to find $W$ so that both $b_1$ and $b_2$ are as close to $W$ as possible, in the sense that \\ $\lambda = ||z_1||^2 + ||z_2||^2$ \\ is minimized where $z_i = b_i - \text{proj}_W b_i$. Note that we can say $W = \text{Span}\left\{ \begin{bmatrix} k \\ l \end{bmatrix} \right\}$ where $k$ and $l$ are constants to be determined. (a) Express $\lambda$ as a function of $l$ and $k$. Simplify your answer as much as possible (there is a fairly answer). (b) Find values of $k$ and $l$ such that $\lambda$ is minimized. This can be done exactly by hand, but you may use a computer if you wish.

          4. There are hints to this problem on the second page. Let $b_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $b_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Let $W$ be a 1d subspace of $\mathbb{R}^2$. We want to find $W$ so that both $b_1$ and $b_2$ are as close to $W$ as possible, in the sense that \\
$\lambda = ||z_1||^2 + ||z_2||^2$ \\
is minimized where $z_i = b_i - \text{proj}_W b_i$. Note that we can say $W = \text{Span}\left\{ \begin{bmatrix} k \\ l \end{bmatrix} \right\}$ where $k$ and $l$ are constants to be determined.

(a) Express $\lambda$ as a function of $l$ and $k$. Simplify your answer as much as possible (there is a fairly answer).
(b) Find values of $k$ and $l$ such that $\lambda$ is minimized. This can be done exactly by hand, but you may use a computer if you wish.

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$4. There are hints to this problem on the second page. Let b1 = < b m a t r i x > and b2 = < b m a t r i x >. Let W be a 1d subspace of ℝ^2. We want to find W so that both b1 and b2 are as close to W as possible, in the sense that λ = ||z1||^2 + ||z2||^2 is minimized where zi = bi - projW bi. Note that we can say W = Span{ < b m a t r i x > } where k and l are constants to be determined. (a) Express λ as a function of l and k. Simplify your answer as much as possible (there is a fairly answer). (b) Find values of k and l such that λ is minimized. This can be done exactly by hand, but you may use a computer if you wish.$

Added by James U.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

06/12/2023

Step 1

Step 1: Given that W is the span of a unit vector, we can write W as W = Span{u} where u is a unit vector. Show more…

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Hint: You may assume that W is the span of a unit vector. Write the meaning of this constraint and use it repeatedly to simplify your answer in a. Your final answer will then be a polynomial of three terms. For b, how about representing your two variables in terms of one? Then you can use single variable optimization. Double angle formulas may be useful. 4. There are hints to this problem on the second page. Let bi: and b Let W be a 1d subspace of R2. We want to find W so that both bi and b2 are as close to W as possible, in the sense that x=z1|2+|z2|2 is minimized where zi = bi - projwb;. Note that we can say W = Spar where k and are constants to be determined. (a) Express X as a function of and k. Simplify your answer as much as possible (there is a fairly answer). (b) Find values of k and such that is minimized. This can be done exactly by hand, but you may use a computer if you wish.

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Transcript

00:01 In this question here, let's see a part.

00:03 So, here in this one vector is perpendicular to other perpendicular to other to other.

00:22 So, here it is i plus j plus k i minus j plus k.

00:31 So, these two vectors, so here i, j, k is 1, 1, 1 and here another is 1, minus 1 and 1.

00:48 So, here when we calculate it, it come i cap multiply by 2 plus j cap multiply by 0 plus k cap multiply by minus 2.

01:04 So, here when we calculate it, it will come 2, 0, minus 2 perpendicular to 1, 1, 1 and 1, minus 1, 1.

01:23 So, here this is our answer for a part.

01:27 So, here this is our answer.

01:30 Now, let's see b part.

01:32 So, here in this question, theta is equal to 65 degree and cos theta is equal to 0 .42.

01:43 So, here 1, 1, 1 dot 1, minus 1, a.

01:55 So, cos theta is equal to here it is a divided by under root 3 multiply by under root 2 plus a square.

02:06 So, here it is equal to 0 .42.

02:10 So, we can write as a square divided by 2 plus a square is equal to 0 .5292.

02:20 So, here when we calculate it, we get a square is equal to 1 .0584 plus 0 .529 a square.

02:31 So, here a square is equal to 2 .248.

02:37 Therefore, the value of a will come 1 .499.

02:42 So, here it is 1 vector, 1 vector, 1, 1, minus 1, 1 .499...

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