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

What does it mean to be orthogonal?

Nice tie btw.

perpendicular, basically

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

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.

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

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

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

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

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

True.

Exactly.

would i be correct when i say that as a reason

Yup :)

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

*webpage bookmarked*

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

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

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

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

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

And they are. Nice work

while we are here may i ask another question?

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

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

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

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).

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.

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

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.

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

Err v and y

i mean v = y in that equation

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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