# Linear maps and matrices problem

1. Dec 4, 2009

### sphlanx

1. The problem statement, all variables and given/known data
We are given a linear map f:R2->R2 f(x,y)=(x,3x+8y).

1)prove that any linear map R2->R2, if it is 1-1(injective??) it is also...well i dont 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)=a4x4+a3x3+a2x2+a1x+a0

so that: a4A4+a3A3+a2A2+a1A+a0I2x2 = I2x2 Where I the identity matrix

2. Relevant equations

3. The attempt at a solution

1) I think i can use the rank-nullity theorem to prove this, but i dont 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=dimR2. Since imf is a subspace of R2 and they have equal dimensions then imf=R2. I think this preety much does it. (ofcourse i will write it in a better way) Am I in the right direction?? I dont 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

Last edited: Dec 4, 2009
2. Dec 4, 2009

### Dick

For 1) sure, use rank-nullity. Injective means dim(ker(f))=0. Surjective means dim(im(f))=2. It should be easy to prove one implies the other from there, isn't it? The second problem is a little baffling. Sure there is such a polynomial. Put a0=1 and a1=a2=a3=a4=0. Are you sure that's what you meant to ask?

3. Dec 5, 2009

### sphlanx

Thanks about 1). About 2) the actual question is: "Is there a non-zero polyonymus that..??". It asks if it exists it doesnt say to prove that it exists. My bad. But are we sure that a non-zero polyonymus like the one in my original post doesnt exist?

4. Dec 5, 2009

### sphlanx

Bump because deadline ends in 3 hours :) Thank you:P

5. Dec 5, 2009

### HallsofIvy

Staff Emeritus
Trivial answer: take $a_4= a_3= a_2= a_1= 0$ and $a_0= 1$. That works, doesn't it? And proves that there exist such a polynomial. Did mean to make that equal to 0 rather than the identity matrix? That's just a trifle harder. The "characteristic polynomial" of A is $\lambda^2- 9\lamba+ 8$ and, since every linear transformation makes its own characteristic polynomial equal to 0, $A^2- 9A+ 8I= 0$. Take $a_4= a_3= 0$, $a_2= 1$], $a_1= -9$, and $a_0= 8$.