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All vectors are in $\mathbb{R}^{n}$ . Mark each statement True or False. Justify each answer.a. Not every linearly independent set in $\mathbb{R}^{n}$ is an orthogonal set.b. If y is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.c. If the vectors in an orthogonal set of nonzero vectors are normalized, then some of the new vectors may not be orthogonal.d. A matrix with orthonormal columns is an orthogonal matrix.e. If $L$ is a line through $\mathbf{0}$ and if $\hat{\mathbf{y}}$ is the orthogonal projection of $\mathbf{y}$ onto $L,$ then $\|\hat{\mathbf{y}}\|$ gives the distance from $\mathbf{y}$ to $L .$

A. TrueB. TrueC. FalseD. FalseE. False

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 2

Orthogonal Sets

Vectors

Missouri State University

Campbell University

Baylor University

University of Michigan - Ann Arbor

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Okay, This question asks us various true false questions relating to orthogonal matrices. So are a is to false saying that not all linearly independent sets Arthur Ivan all So this is saying, Yeah, a linearly independent set isn't necessarily orthogonal. And this is true because if we just think of the vectors 11 and 13 they're definitely linearly independent, but 11 dotted with 13 is equal to four. So they're not orthogonal. So that's a nice little counter example. Then Part B says that if why is a linear combination of vectors from the north agonal sent the weights can be computed without row operations, and this is also true. So if we say that a vector V is equal to a one x one plus a two x two plus dot, dot, dot dot We can calculate these coefficients just by taking the projection of the vector onto each of the basis. So, for example, a one would be equal to the dotted with X one over the magnitude of X one squared, and this could be done for any of the vectors because all we're doing since their orthogonal, none of them overlap with their projections. So we just need to calculate the component of the that's projected under each of the basis vectors. And we don't have to do anything with the Matrix to find that. Then Part C says if vectors in an orthogonal set of non zero vectors air normalized than some of the new vectors may not be orthogonal. Okay, So, for example, let's say we have a vector, the one and another vector V two such that their orthogonal so v one dotted with the to is equal to zero. Then it says we normalize each of these factors. So v one over the magnitude of the one started with V two over the magnitude of the two, and we want to see if this dot product is still zero, and we can see that the DOT product allows us to pull out scaler multiples. So in actuality, we can write this as one over the magnitude of the one times one over the magnitude of V two, and then that is just a coefficient outside V one dotted with the to. And we see this is still equal to zero because this thing in brackets zero so the claim is false for C because we can just pull the magnitudes out front. And the only thing affecting or Thorgan ality are the vectors themselves, not their magnitudes, because they they just come along for the ride. Then for D were asked a matrix with Ortho Normal columns is an orthogonal matrix. So this is really tricky because you might want to say true. But actually this is false because the matrix must be. And by n so on, Lee Square matrices could be classified as orthogonal. So even so, the Ortho Normal column claim would be enough if it included the fact that the Matrix had to be square. So if it said there was a square matrix, then do you would be true? And then e says if we have a line through the origin and why hat is the orthogonal projection than the magnitude of why gives the distance from L. Yeah. So we say this is a line and this is a point, and we want the distance between these two points. So what we'd actually do is we would split this up this blue vector into a component parallel and a component perpendicular which I'll actually write this better. So this is the parallel component and this is the perpendicular component. And it is saying that the magnitude of this component right here is the distance to the line. And that's false because we want. We want the magnitude of the perpendicular component. So this is the distance we want the perpendicular, not the parallel, so it's false.

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