Question

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

Answer 1

What does it mean to be orthogonal?

Answer 2

Nice tie btw.

Answer 3

perpendicular, basically

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

True.

Answer 12

Exactly.

Answer 13

would i be correct when i say that as a reason

Answer 14

Yup :)

Answer 15

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

Answer 16

*webpage bookmarked*

Answer 17

You just need to show that it is orthogonal to any vector

kU + hV

Answer 18

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

Answer 19

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

Answer 20

Fair enough, though it's easy to prove..

\[x \cdot v = 0\]
\[x \cdot u = 0\]
\[x \cdot (ku + hv) = x \cdot ku + x \cdot hv = k(x\cdot u) + h(x\cdot v) = 0 \]

Answer 21

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

Answer 22

And they are. Nice work

Answer 23

while we are here may i ask another question?

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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).

Answer 28

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.

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

Err v and y

Answer 33

i mean v = y in that equation

Answer 34

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