Given surjective ##T:V\to W##, find isomorphism ##T|_U## of U onto W.

[zenterix]

Homework Statement

Axler, "Linear Algebra Done Right", Problem 8 Section 3D

Suppose ##V## is finite-dimensional and ##T:V\to W## is a surjective linear map of ##V## onto ##W##. Prove that there is a subspace ##U## of ##V## such that ##T|_U## is an isomorphism of ##U## onto ##W##. (Here ##T|_U## means the function ##T## restricted to ##U##. In other words, ##T|_U## is the function whose domain is ##U## with ##T|_U## defined by ##T|_U(u)=Tu## for every ##u\in U##.)

Relevant Equations

Since ##T## is surjective then ##\text{range}(T)=W##.

I will use a proof by cases.

Case 1: dim V = dim W

Then ##T=T|_V## is an isomorphism of ##V## onto ##W##. The reason for this is that it is possible to prove that if ##T## is surjective, which it is, then it is also injective and so it is invertible (hence an isomorphism).

Case 2: dim V < dim W

This case is impossible because a linear map to a larger dimensional space is not surjective.

Case 3: dim V > dim W

$$\text{dim} V = \text{dim null} T + \text{dim range} T = \text{dim null} T+\text{dim} W$$

Let ##U## be such that ##V=U \bigoplus \text{null} T##, that is, the complement of ##\text{null} T##.

Then

$$\text{dim} U + \text{dim null} T = \text{dim} V$$

$$\text{dim} U = \text{dim} W$$

##U## and ##W## are isomorphic and ##T|_U\in L(U,W)## is an isomorphism because it is surjective

Pictorially, we have something like

In each case, we can infer the desired result.

 

[nuuskur]

Yes, it suffices to consider any complement ##U## of ##\mathrm{Ker}\,T##, i.e, ##U\oplus \mathrm{Ker}\,T \cong V##. Then ##T\vert _U :U\to W## is automatically an isomorphism. To find ##U##, it suffices to take any basis of ##\mathrm{Ker}\,T##, extend it to a basis in ##V## and throw out the null part. In the special case of ##\dim V=\dim W##, any basis of ##\mathrm{Ker}\,T## would have to be empty.