Section 1.8

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1. Let be linear maps between finite-dimensional vector spaces that satisfy . Is it true that ?

Solution:
No. in general composition of functions is not commutative. By the theorem that any linear map can be expressed as a matrix, finding a counterexample comes down to finding two matrices such that but . Here is one example of functions: . . Then we have but so but .


4. Show that a linear map is one-to-one if and only if implies .

Proof:
First note that for any linear map because .


Proof: Suppose that is one-to-one. Then if we have by the note above so that we must have . Therefore implies . Now suppose that implies . If then by linearity of we have . But then by hypothesis that means which implies . Therefore is one-to-one.


6. Let be finite-dimensional and assume that

are linear operators. Show that if then at least one of the are not one-to-one.

Proof:
I will use proof by contrapositive. The equivalent statement would then be "`If all of the are one-to-one, then . Then this becomes very easy if you know the fact from set theory that the composition of one-to-one functions is a one-to-one function. This gives the following. Suppose that are all one-to-one. Then is also a one-to-one function and so the only input that will give an output of 0 is the input from problem 4. Therefore and we are done.


If you don’t know the fact from set theory you can prove it as follows. Suppose are one-to-one functions. Consider the function . Then to show this new function is one-to-one assume that . Then . But since is one-to-one that means the inputs to must be the same or in other words . But then is one-to-one so that means and therefore is one-to-one.

13. Consider the map defined by ( a) Show that this is -linear and one-to-one, but not onto. Find an example of a matrix in that does not come from .

Proof:
To show this is linear let . Then:


Similarly if and then:

Therefore is -linear.
Now to show is not onto we notice that any matrix in the image of has top left and bottom right coordinate the same. So the simple matrix cannot possibly be in the image of . Therefore is not onto.