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In Exercises 21 and $22,$ all vectors and subspaces are in $\mathbb{R}^{n} .$ Mark each statement True or False. Justify each answer.a. If $\mathbf{z}$ is orthogonal to $\mathbf{u}_{1}$ and to $\mathbf{u}_{2}$ and if $W=$ $\operatorname{Span}\left\{\mathbf{u}_{1}, \mathbf{u}_{2}\right\},$ then $\mathbf{z}$ must be in $W^{\perp}$ .b. For each $\mathbf{y}$ and each subspace $W,$ the vector $\mathbf{y}-\operatorname{proj}_{W} c. The orthogonal projection $\hat{\mathbf{y}}$ of $\mathbf{y}$ onto a subspace $W$ can sometimes depend on the orthogonal basis for $W$ used to compute $\hat{\mathbf{y}}$ .d. If $\mathbf{y}$ is in a subspace $W,$ then the orthogonal projection of $\mathbf{y}$ onto $W$ is $\mathbf{y}$ itself.

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Calculus 3

Chapter 6

Orthogonality and Least Square

Section 3

Orthogonal Projections

Vectors

Johns Hopkins University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

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In mathematics, a vector (…

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In Exercises 19 and $20,$ …

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In Exercises 17 and $18,$ …

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In Exercises 21 and $22,$ …

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eso part, eh? You don't be minus b dot You equal zero. Well, we know that the dot product is commuted s. So that means that you dot b is equal to read about you. And these two things are part B for any scaler. See, uh, the norm of CVI is equal to see times the normal b Well, so we want to see if, uh, we can pull a scaler out of the norm. And if this holds true, So let's take c t equal Negative one on our vector Be to just be, say, 11 Let's compute uh, these two values So the norm of CV is equal to the norm of negative one negative one which is equal to the square root. Um, negative one squared postnegative one squared, which is a squared. But now, if we take, uh, c times the normal V, we get negative one times the norm of 11 which is negative one times the square root of one square post one squared, which is negative, squared or two. So these things aren't the same once he's negative. So this one iss who else? This person is true. Uh, next for part C. If X is or thought inal toe every vector in a subspace W then X is in, um, the Ortho Wagnalls base to W. So that's just the definition. Um, the space here, This just means that all the vectors that or thought minal to the space w closes just true. By definition, party says, if the normal view squared plus the norm of these squares is equal to the norm of you plus b squared then you movie or are orthogonal. So if we know that this is true, let's write out what this means. So the normal view squared is you, don't you? Normal v squared is we Dock B now on the right We have you close to me dot You cost me So we have you, don't you? Plus B, that me is equal to, um this very hand side we can essentially like foil it out So we have you got you Plus to you don't be plus me dot B um, so you don't use could cancel be Darby's can cancel. So we get zero equals two times you don't me. So this means that you don't be has to equal zero, which means that you and v our orthogonal. So this is true. And lastly, part e ah, for an m by n matrix A vectors in the null space of a our earth Ogle two factors in the roast base of it. Well, if we look in the section on here, um, three. It states that everything orthogonal to row space of a is equal to mall space of a. So this is just another way of saying exactly what the statement says. Um, so this is true because this is just stating they're on three, which is on page 3 55

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