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!]]>(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.
]]>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?
]]>I have no idea how to start; can you work me through how to prove it?
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