Prove "dim U \geq dim V - dim W" Linear Map Question

[jimmypoopins]

Homework Statement

Suppose that V and W are finite dimensional and that U is a subspace of V. Prove that there exists $$T \in L(V,W)$$ such that null T = U if and only if $$dim U \geq dim V - dim W.$$

Homework Equations

thm: If $$T \in L(V,W)$$, then range T is a subspace of W.

thm: If V is a finite dimensional vector space and $$T \in L(V,W)$$ then range T is a finite-dimensional subspace of W and dim V = dim null T + dim range T.

The Attempt at a Solution

forward direction: by thm, range T is a subspace of W implies that
$$dim range T \leq dim range W$$.

by thm, dim V = dim null T + dim range T
dim V = dim U + dim range T (since U = null T)
dim V - dim range T = dim U

$$dim V - dim W \leq dim U$$ since $$dim range T \leq dim range W$$.

i think the forward direction is good. comments?

backward direction:
we have $$dim V - dim W \leq dim U$$. Let $$(u_{1},...,u_{n})$$ be a basis for U. extend this to a basis for V: $$(u_{1},...,u_{n},u_{n+1},...u_{m})$$. then dim U = n, and dim V = m. Then any $$v \in V$$ can be written as $$a_{1}u_{1}+...+a_{m}u_{m}$$.

I think I'm in the right direction but I'm confused as to what to do. since we have dim V - dim W is less than dim U, i want to say that dim W is greater than or equal to m, but i don't know how to define T so that null T = U. If i make all of the T(u_i} in the basis 0, then null T = U, but how does that relate to the relation of $$dim V - dim W \leq dim U$$?

thanks.

 

[HallsofIvy]

Homework Statement

Suppose that V and W are finite dimensional and that U is a subspace of V. Prove that there exists $$T \in L(V,W)$$ such that null T = U if and only if $$dim U \geq dim V - dim W.$$

Homework Equations

thm: If $$T \in L(V,W)$$, then range T is a subspace of W.

thm: If V is a finite dimensional vector space and $$T \in L(V,W)$$ then range T is a finite-dimensional subspace of W and dim V = dim null T + dim range T.

The Attempt at a Solution

forward direction: by thm, range T is a subspace of W implies that
$$dim range T \leq dim range W$$.

by thm, dim V = dim null T + dim range T
dim V = dim U + dim range T (since U = null T)
dim V - dim range T = dim U

$$dim V - dim W \leq dim U$$ since $$dim range T \leq dim range W$$.

i think the forward direction is good. comments?

Yes, thatis correct. I had to stop and think about it for a moment. Since range T is a subset of W, dim Range T$\le$ dim W. Therefore, -dim Range T$\ge$ -dim W, therefore dim V- dim W$\le$dimV- dim Range T= dim U.

backward direction:
we have $$dim V - dim W \leq dim U$$. Let $$(u_{1},...,u_{n})$$ be a basis for U. extend this to a basis for V: $$(u_{1},...,u_{n},u_{n+1},...u_{m})$$. then dim U = n, and dim V = m. Then any $$v \in V$$ can be written as $$a_{1}u_{1}+...+a_{m}u_{m}$$.

I think I'm in the right direction but I'm confused as to what to do. since we have dim V - dim W is less than dim U, i want to say that dim W is greater than or equal to m, but i don't know how to define T so that null T = U. If i make all of the T(u_i} in the basis 0, then null T = U, but how does that relate to the relation of $$dim V - dim W \leq dim U$$?

thanks.

Think about what you want to prove: if dim V- dim W$\le$ dim U, then there exists a linear transformation T such that null T= U. dim V- dim W$\le$ dim U means dim V- dim U$\le$ dim W. Choose a basis for U, extend it to a basis for V. Now, define T so that Tu= 0 for any basis vector of U. Tv, for v a basis vector for v not in the basis for U, can be anything in W.