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## #1 2019-05-21 05:37:56

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 37

### Vector problem, help ASAP

(a) Show that any two-dimensional vector can be expressed in the form

where
and
are real numbers.

(b) Let
and
be non-zero vectors. Show that any two-dimensional vector can be expressed in the form:

where
and
are real numbers, if and only if of the vectors
and
, one vector is not a scalar multiple of the other vector.

I have no idea how to start; can you work me through how to prove it?

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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## #2 2019-05-21 06:54:16

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,241
Website

### Re: Vector problem, help ASAP

Hi !nval!d_us3rnam3,

Thanks for your post -- I fixed your LaTeX.

For part (a), suppose you've got some vector
in
. Then, you've just got to solve this pair of simultaneous equations for
and
:

Does that make sense? (Let me know if anything sounds confusing -- happy to help.)

For part (b), suppose that instead of
and
you have
and
. This gives you the system:

What sorts of conditions do you need here for that to have a solution?

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## #3 2019-05-21 10:24:57

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 37

### Re: Vector problem, help ASAP

For part (a), how do I prove this equation is the one I should use?
Also, for part (b), I don't understand what conditions you mean.

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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## #4 2019-05-21 22:32:32

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,241
Website

### Re: Vector problem, help ASAP

No worries -- let's look at part (a) first.

!nval!d_us3rnam3 wrote:

(a) Show that any two-dimensional vector can be expressed in the form

where
and
are real numbers.

Okay, let's take any two-dimensional vector, say,
. We want to know: can we find real numbers
and
so that:

If we 'multiply out' the left-hand side, we get:

Now, we can add two vectors just by adding the matching components, so that:

In other words, we want to find real numbers
and
so that:

which is exactly the same as solving the pair of simultaneous equations:

Remember, we're solving for
and
here. (Just pretend that
and
are any old real numbers.)

Let me know if this makes sense -- happy to explain anything further if you need more help.

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## #5 2019-05-22 11:55:34

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 37

### Re: Vector problem, help ASAP

I think I have part (a), but can you explain part (b) in a little more detail?
Thanks in advance!

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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## #6 2019-05-23 07:29:23

!nval!d_us3rnam3
Member
Registered: 2017-03-18
Posts: 37

### Re: Vector problem, help ASAP

Bump... can anyone else walk me through part (b)? I don't understand the scalar multiple bit.

Last edited by !nval!d_us3rnam3 (2019-05-23 07:43:17)

"If we wanna be great, we can't just sit on our hands" - 2017 NFL Bears draft

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## #7 2019-05-23 22:05:29

zetafunc
Moderator
Registered: 2014-05-21
Posts: 2,241
Website

### Re: Vector problem, help ASAP

Part (b) is similar, but bear in mind that you have to prove both directions. First, let's see if we can do the reverse direction (not scalar multiples => any vector can be represented in the form
).

Suppose that instead of
and
you have
and
. In other words, instead of having

you have this instead:

Another way of writing this is:

Now, let's say you wanted to solve this system for
and
. You'd need to find the inverse of
, right? But for that inverse to exist, the determinant can't be equal to 0. In other words, for that thing to have a solution, you must have:

i.e.

which tells you that
isn't a multiple of
. Remember, saying that
and
are scalar multiples of each other just means that you can find some real number
such that
, or in other words:

i.e.

Now, for the forwards direction (any vector can be expressed as
=>
and
aren't multiples of each other), try proving the contrapositive, i.e. show that if
for some real number
, then not every vector can be expressed in the form
. Let me know if you have any more questions!

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