Thread: Proof for Isomorphic Linear Operator

1. Proof for Isomorphic Linear Operator

The following linear algebra problem on linear transformations/isomorphisms was confusing me so any help would be appreciated!

Let R^n -> R^n be a linear operator with matrix A (using the standard basis for R^n). Prove that L is an isomorphism if and only if the columns of A are linearly independent.

2. Originally Posted by buckaroobill
The following linear algebra problem on linear transformations/isomorphisms was confusing me so any help would be appreciated!

Let R^n -> R^n be a linear operator with matrix A (using the standard basis for R^n). Prove that L is an isomorphism if and only if the columns of A are linearly independent.
Let L be the linear operator.

"Isomorphism --> Linear independence".
Isomorphism (for vector spaces) means a bijective linear operator. Bijective means both injective and surjective. Let b in R^n then there exists a unique x in R^n such that L(x)=b. That is, Ax = b. (Why?).
Thus, the linear system of equations,
Ax = b has a unique solution for every column vector b.
Thus, det(A) != 0.
Thus, the column and row vectors in A are linearly independent.
Q.E.D.

See if you can do the converse.