Linear maps and matrices problem

[sphlanx]

Homework Statement

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 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)=a₄x⁴+a₃x³+a₂x²+a₁x+a₀

so that: a₄A⁴+a₃A³+a₂A²+a₁A+a₀I_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=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 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!

 

[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?

 

[sphlanx]

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?

Thanks about 1). About 2) the actual question is: "Is there a non-zero polyonymus that..??". It asks if it exists it doesn't say to prove that it exists. My bad. But are we sure that a non-zero polyonymus like the one in my original post doesn't exist?

 

[sphlanx]

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

 

[HallsofIvy]

Homework Statement

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 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)=a₄x⁴+a₃x³+a₂x²+a₁x+a₀

so that: a₄A⁴+a₃A³+a₂A²+a₁A+a₀I_2x2 = I_2x2 Where I the identity matrix

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 $a_0= 8$.

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=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 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!