Linear maps and matrices problem

In summary, the conversation discusses the use of the rank-nullity theorem to prove that any linear map from R2 to R2 is injective if and only if it is surjective. It also mentions a problem involving finding a non-zero polynomial that satisfies a specific equation involving a matrix and the identity matrix. A solution is provided for this problem using the characteristic polynomial of the matrix.
  • #1
sphlanx
11
0

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+a0I2x2 = I2x2 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!
 
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  • #2
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
Dick said:
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?
 
  • #4
Bump because deadline ends in 3 hours :) Thank you:P
 
  • #5
sphlanx said:

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+a0I2x2 = I2x2 Where I the identity matrix
Trivial answer: take [itex]a_4= a_3= a_2= a_1= 0[/itex] and [itex]a_0= 1[/itex]. 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 [itex]\lambda^2- 9\lamba+ 8[/itex] and, since every linear transformation makes its own characteristic polynomial equal to 0, [itex]A^2- 9A+ 8I= 0[/itex]. Take [itex]a_4= a_3= 0[/itex], [itex]a_2= 1[/itex]], [itex]a_1= -9[/itex], and [itex]a_0= 8[/itex].


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!
 

Related to Linear maps and matrices problem

1. What is the difference between a linear map and a matrix?

A linear map is a function that preserves the vector space structure, meaning it follows the rules of vector addition and scalar multiplication. A matrix, on the other hand, is a rectangular array of numbers used to represent a linear map in a specific coordinate system.

2. How do you determine if a given matrix is invertible?

A matrix is invertible if its determinant is non-zero. The determinant can be calculated by performing a series of operations on the matrix, such as row reduction. If the determinant is 0, the matrix is not invertible.

3. How do you find the eigenvalues and eigenvectors of a linear map?

The eigenvalues of a linear map are the values that, when multiplied by the corresponding eigenvectors, result in the same vector. To find the eigenvalues and eigenvectors, you can solve the characteristic equation det(A - λI) = 0, where A is the matrix representing the linear map and λ is the eigenvalue.

4. Can a linear map have more than one matrix representation?

Yes, a linear map can have multiple matrix representations depending on the choice of basis for the vector space. The matrix representation of a linear map is not unique, but the linear map itself is unique.

5. How are matrices used to solve systems of linear equations?

Matrices are used to represent systems of linear equations in a compact and efficient way. By performing row operations on the matrix, we can reduce it to its reduced row echelon form, which reveals the solution to the system of equations. This method is known as Gaussian elimination.

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