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

(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?

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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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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
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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.

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

!nval!d_us3rnam3 wrote:

Okay, let's take any two-dimensional vector, say, . We want to know: can we find real numbers and so that:(a) Show that any two-dimensional vector can be expressed in the form

where and are real numbers.

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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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
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I think I have part (a), but can you explain part (b) in a little more detail?

Thanks in advance!

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**!nval!d_us3rnam3****Member**- Registered: 2017-03-18
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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)*

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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!**LearnMathsFree: Videos on various topics.New: Integration Problem | Adding FractionsPopular: Continued Fractions | Metric Spaces | Duality**

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