# Linear Algebra - isomorphism questions

1. Feb 13, 2010

### zeion

1. The problem statement, all variables and given/known data

Determine whether the following mappings f is onto or one-to-one. Is f an isomorphism?

a) f maps R2 into R2 and is defined by f(x,y) = (x-2y, x+y)

b) f maps R2 into R3 and is defined by f(x,y) = (x, y, x+y)

i) f maps R3 into P2(R), defined by f(a1, a2, a3) = a2 - a3x + (1-a1)x^2

j) f maps Rmxn into Rnxm and is defined by f(A) = A^T (transpose of A) for all A in Rmxn

2. Relevant equations

3. The attempt at a solution

Can I solve this by forming the matrix of the output?
ie I have x-2y = 0 and x+y = 0 as a matrix and row reduce, I get a 2x2 identity matrix and therefore
1) There are always solution for any augmented matrix, so f is onto
2) There are no parameters in the matrix and both variables always have a unique solution for every augmented matrix, so f is one-to-one

Is that right?

For b)

I get form the 3x2 matrix and row reduce, then a have third row of zero and no parameter, therefore f is not onto since there will always be some component of R3 that has no solution (hence the zero row). But f is 1-1 since both variables always have a unique solution for any augmented matrix (hence it has 2 leading ones).

Does that explanation seem right?

For i)

I form the matrix and row reduce to a I3x3, therefore f is onto and 1-1.

For j)

f is one-to-one since it forms the transpose of A and every row of A is mapped to every column of the image once.
It is onto since this applies to every matrix of Rmxn, therefore transposed into every matrix of Rnxm.

Last edited: Feb 13, 2010
2. Feb 14, 2010

### HallsofIvy

Staff Emeritus
Yes, although you don't need to do it with matrices. If a and b are any two numbers then we can solve f(x,y)= (a,b) as x- 2y= a, x+ y= b for x= (a+2b)/2, y= (a+b)/3, so this is onto. Also if $f(x_1,y_1)= f(x_2,y_2)$, $x_1- 2y_1= x_2- 2y_2$, $x_1+ y_1= x_2+ y_2$, then $x_1- x_2= y_2- y_1$ from the first equation and $x_2- x_1= -2(y_2- y_1)$. Then $y_2- y_1= -2(y_2- y_1)$ which leads to $y_1= y_2$ and then $x_1= x_2$ showing that this function is one-to-one.

You still need to show that f((a,b)+ (c,d))= f((a,b))+ f((c,d)) and f(r(a,b))= rf(a,b) for all real numbers a, b, c, d, and r to show that this is an isomorphism.

You need to say a bit more. You can use the fact that the transpose is "dual". That is, that
(AT)T= A.

This function is onto because, given any matrix B, it is the transpose of BT. It is one-to-one because, if
AT= BT
then, taking the transpose of each side, A= B.

And, again, just showing that a function is one-to-one and onto is NOT enough to show that it is an isomorphism. You must also show that the function is a homormorphism- that it "preserves" the operations.

Last edited: Feb 14, 2010