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OpenStudy (anonymous):

) If x is orthogonal to both u and v, then x is orthogonal to every vector in span(u; v). True | False

OpenStudy (anonymous):

What does it mean to be orthogonal?

OpenStudy (anonymous):

Nice tie btw.

OpenStudy (anonymous):

perpendicular, basically

OpenStudy (anonymous):

the inner product <x, v> and <x, u> = 0 when x does not equal v or u respectively

OpenStudy (anonymous):

True, but I was looking for more of a conceptual definition. It means essentially that x has no component that is in the same direction as u or v.

OpenStudy (anonymous):

Now what do we know about the vectors in the span of {u,v}

OpenStudy (anonymous):

when something spans it is all possible linear combinations of that set? is the answer true?

OpenStudy (anonymous):

oh it was a leading question, wow I'm tired to not have recognized that

OpenStudy (anonymous):

The answer is true. You can also prove it using the definition of the dot product for vectors.

OpenStudy (anonymous):

because if X is orthogonal to U, then X is orthogonal to kU? let k be any constant that scales u

OpenStudy (bobbyleary):

True.

OpenStudy (anonymous):

Exactly.

OpenStudy (anonymous):

would i be correct when i say that as a reason

OpenStudy (anonymous):

Yup :)

OpenStudy (anonymous):

where the flutter have you guys been while ive been failing this linear algebra class

OpenStudy (anonymous):

*webpage bookmarked*

OpenStudy (anonymous):

You just need to show that it is orthogonal to any vector kU + hV

OpenStudy (anonymous):

As all those vectors are the span of {u,v}

OpenStudy (anonymous):

he wouldnt ask me to show it that far as long as i could explain it to that point

OpenStudy (anonymous):

Fair enough, though it's easy to prove.. \[x \cdot v = 0\] \[x \cdot u = 0\] \[x \cdot (ku + hv) = x \cdot ku + x \cdot hv = k(x\cdot u) + h(x\cdot v) = 0 \]

OpenStudy (anonymous):

see he would either expect me to do that, or to explain it as we have already done. he considers both answers to be the same

OpenStudy (anonymous):

And they are. Nice work

OpenStudy (anonymous):

while we are here may i ask another question?

OpenStudy (anonymous):

Certainly. Though when threads get too long they tend to lag things up a bit in my browser.

OpenStudy (anonymous):

The orthogonal projection of y onto u is a scalar multiple of y. True | False? REASON:

OpenStudy (anonymous):

I'm not sure I understand what it means by orthogonal projection of y

OpenStudy (anonymous):

False. The if the projection is orthogonal then it is a rotation (in other words multiples only make vectors change direction or shorten but not rotate).

OpenStudy (anonymous):

Typically the projection of a vector onto another vector will not be a scalar multiple of the original vector. That would imply that the vector being projected onto is the same as the vector your projecting.

OpenStudy (anonymous):

I don't remember orthogonal projections though.. I certainly don't recall them with regard to rotations.

OpenStudy (anonymous):

Well, by rotation I simply mean you take it from one plane, say the x-y plane, and puts it in the z plane. In my words "rotated" out of the plane. A scalar multiple won't do that. Even on a projection.

OpenStudy (anonymous):

My equation here is <(y-cu), u)> = 0 which is equivalent to c= <v,u> / <u,u>

OpenStudy (anonymous):

Err v and y

OpenStudy (anonymous):

i mean v = y in that equation

OpenStudy (anonymous):

Ok then I'm gonna say false. the projection of y onto u is not a (non-zero) scalar multiple of y unless \(y = ku\)

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