Linear Transformation and Determinant

[schaefera]

Homework Statement

Define L: R(mxm) to R(nxn). If L(A)=L(B), prove or disprove that det(A)=det(B).

Homework Equations

The Attempt at a Solution

I think I can prove that this is true.

L(A)=L(B) means that L(A)-L(B)=L(A-B)=0.

Now let C be the matrix representation of L. We have two possibilities:

1) C is nonsingular. If C is nonsingular, then C(A-B)=0, so A-B=0. Then det(A)-det(B)=0 and det(A)=det(B).

2) C is singular. Now I have an issue-- I don't know what to do!

Am I on the right path? Should I be disproving this?

 

[Dick]

Homework Statement

Define L: R(mxm) to R(nxn). If L(A)=L(B), prove or disprove that det(A)=det(B).

Homework Equations

The Attempt at a Solution

I think I can prove that this is true.

L(A)=L(B) means that L(A)-L(B)=L(A-B)=0.

Now let C be the matrix representation of L. We have two possibilities:

1) C is nonsingular. If C is nonsingular, then C(A-B)=0, so A-B=0. Then det(A)-det(B)=0 and det(A)=det(B).

2) C is singular. Now I have an issue-- I don't know what to do!

Am I on the right path? Should I be disproving this?

I'd say if you don't aren't told L is nonsingular then you should try and disprove your proposition. Case 2) is definitely a problem!

 

[HallsofIvy]

An obvious point is that if L is linear transformation the maps every matrix to the 0 matrix, then L(A)= L(B) for all matrices A and B.

 

[schaefera]

Does one counter-example suffice, then? Would the transformation sending everything to the 0 matrix be one?

If not, it should be simple enough to construct. Perhaps something like:

L(a,b,c,d)=(a+b,c,d,0) where that is a 2x2 matrix I wrote out in one line, then for (0,1,1,1) and (1,0,1,1) have the same image but different determinants.

Thanks!

 

[Dick]

Does one counter-example suffice, then? Would the transformation sending everything to the 0 matrix be one?

If not, it should be simple enough to construct. Perhaps something like:

L(a,b,c,d)=(a+b,c,d,0) where that is a 2x2 matrix I wrote out in one line, then for (0,1,1,1) and (1,0,1,1) have the same image but different determinants.

Thanks!

L(X)=0 certainly works and that's enough. If you'd rather have a less trivial example, yours works too. But one example suffices.