Could anyone walk me through these two? I need some help getting going.
1) Prove that the map defined by is a linear transformation and determine if is one-to-one.
2) Let . Prove that the range of is a subspace of .
Could anyone walk me through these two? I need some help getting going.
1) Prove that the map defined by is a linear transformation and determine if is one-to-one.
2) Let . Prove that the range of is a subspace of .
Take two polynomials . Therefore,
Can you finish this?
Note that is not one to one. Can you think of an easy counterexample?
Let . Thus, and for some .2) Let . Prove that the range of is a subspace of .
Now note that
Since , we can now say that is a subspace of .
Does this make sense?
I'm a little confused as to where and came from.
If I'm interpreting the problem right, a counter example would be because then the transformation would lead to
EDIT: Technically, I guess it would be as a counter example. Gotta keep the variables straight.
Again, I'm confused about and . Are they just arbitrary constants? If so, the math makes sense to me, but I don't see how it proves that it's a subspace.Let . Thus, and for some .
Now note that
Since , we can now say that is a subspace of .
Does this make sense?
To clarify, a and b are arbitrary constants. Now, to show that injectivity fails, you want to pick polynomials . Don't make it too hard.
Yes, a and b are arbitrary constants. Recall that if and when we have , then is a subspace of .Again, I'm confused about and . Are they just arbitrary constants? If so, the math makes sense to me, but I don't see how it proves that it's a subspace.
In this case we let and .
Does this clarify things?
woops...forgot we were working in dimensions. So something simple like, in your example in the quote box above, setting
To finish the proof, I just add:
Then I do the same thing with an arbitrary scalar multiple, right?
This part makes sense now. Thank you.Yes, a and b are arbitrary constants. Recall that if and when we have , then is a subspace of .
In this case we let and .
Does this clarify things?
The second to last line isn't correct. We can't assume its linear! However, note that .
Taking takes care of the constant muliple part as well. So you only need to check this linear combination, instead of the sum and constant multiple parts separately (in other words, its easier and quicker to do both steps in one step).
Does this make sense?