- #1

sphlanx

- 11

- 0

## Homework Statement

We are given a linear map f:R

^{2}->R

^{2}f(x,y)=(x,3x+8y).

1)prove that any linear map R

^{2}->R

^{2}, if it is 1-1(injective??) it is also...well i don't know the word! It is the property that every element of the destination set is imaged by the source set. (candidate is "surjective" but wikipedia confused me a bit). We also have to prove it vice versa.

2)If A is the matrix of f for the standar basis of R2, prove that there is a non-zero polyonymus h(x)=a

_{4}x

^{4}+a

_{3}x

^{3}+a

_{2}x

^{2}+a

_{1}x+a

_{0}

so that: a

_{4}A

^{4}+a

_{3}A

^{3}+a

_{2}A

^{2}+a

_{1}A+a

_{0}I

_{2x2}= I

_{2x2}Where I the identity matrix

## Homework Equations

## The Attempt at a Solution

1) I think i can use the rank-nullity theorem to prove this, but i don't know what can I take as a given. If f is 1-1 then dimkerf=0(because if it wasnt more than 1 elements of source would image at 0, but is this enough?). If kerf=0 then dimimf=dimR

^{2}. Since imf is a subspace of R

^{2}and they have equal dimensions then imf=R

^{2}. I think this preety much does it. (ofcourse i will write it in a better way) Am I in the right direction?? I don't know if i can use the same logic to prove the opossite though (that when it is surjective it is also injective)

2)No clue. This makes me thinks of eigenvalues but we have not be taught about this stuff yet. I have also heard that when you see a matrix raised at a power determinants may come in handy. But still no clue. Thanks in advance.

P.S. My post is messed up but its 5:00 morning hours in Greece and i am quite tired

P.S. Deadline is coming so i would appreciate a fast reply. Thanks in advance!

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