A Second Course in Linear Algebra

Choose any vector \(\vect{v}\in V\text{.}\) Then by Theorem VRRB there are unique scalars, \(\scalarlist{a}{n}\) such that

where we have implicitly defined \(\vect{u}\) and \(\vect{w}\) in the last line. It should be clear that \(\vect{u}\in{U}\text{,}\) and similarly, \(\vect{w}\in{W}\) (and not simply by the choice of their names).

Suppose we had another decomposition of \(\vect{v}\text{,}\) say \(\vect{v}=\vect{u}^\ast+\vect{w}^\ast\text{.}\) Then we could write \(\vect{u}^\ast\) as a linear combination of \(\vect{v}_1\) through \(\vect{v}_m\text{,}\) say using scalars \(\scalarlist{b}{m}\text{.}\) And we could write \(\vect{w}^\ast\) as a linear combination of \(\vect{v}_{m+1}\) through \(\vect{v}_n\text{,}\) say using scalars \(\scalarlist{c}{n-m}\text{.}\) These two collections of scalars would then together give a linear combination of \(\vect{v}_1\) through \(\vect{v}_n\) that equals \(\vect{v}\text{.}\) By the uniqueness of \(\scalarlist{a}{n}\text{,}\) \(a_i=b_i\) for \(1\leq i\leq m\) and \(a_{m+i}=c_{i}\) for \(1\leq i\leq n-m\text{.}\) From the equality of these scalars we conclude that \(\vect{u}=\vect{u}^\ast\) and \(\vect{w}=\vect{w}^\ast\text{.}\) So with both conditions of Definition Definition 1.2.1 fulfilled we see that \(V=U\ds W\text{.}\)

If \(U=V\text{,}\) then choose \(W=\set{\zerovector}\text{.}\) Otherwise, choose a basis \(B=\set{\vectorlist{v}{m}}\) for \(U\text{.}\) Then since \(B\) is a linearly independent set, Theorem ELIS tells us there is a vector \(\vect{v}_{m+1}\) in \(V\text{,}\) but not in \(U\text{,}\) such that \(B\cup\set{\vect{v}_{m+1}}\) is linearly independent. Define the subspace \(U_1=\spn{B\cup\set{\vect{v}_{m+1}}}\text{.}\)

We can repeat this procedure, in the case were \(U_1\neq V\text{,}\) creating a new vector \(\vect{v}_{m+2}\) in \(V\text{,}\) but not in \(U_1\text{,}\) and a new subspace \(U_2=\spn{B\cup\set{\vect{v}_{m+1},\,\vect{v}_{m+2}}}\text{.}\) If we continue repeating this procedure, eventually, \(U_k=V\) for some \(k\text{,}\) and we can no longer apply Theorem ELIS. No matter, in this case \(B\cup\set{\vect{v}_{m+1},\,\vect{v}_{m+2},\,\dots,\,\vect{v}_{m+k}}\) is a linearly independent set that spans \(V\text{,}\) i.e. a basis for \(V\text{.}\)

Define \(W=\spn{\set{\vect{v}_{m+1},\,\vect{v}_{m+2},\,\dots,\,\vect{v}_{m+k}}}\text{.}\) We now are exactly in position to apply Theorem Theorem 1.2.3 and see that \(V=U\ds W\text{.}\)

The first condition is identical in the definition and the theorem, so we only need to establish the equivalence of the second conditions.

Assume that \(V=U\ds W\text{,}\) according to Definition Definition 1.2.1. By Property Z, \(\zerovector\in V\) and \(\zerovector=\zerovector+\zerovector\text{.}\) If we also assume that \(\zerovector=\vect{u}+\vect{w}\text{,}\) then the uniqueness of the decomposition gives \(\vect{u}=\zerovector\) and \(\vect{w}=\zerovector\text{.}\)

⇐ Suppose that \(\vect{v}\in V\text{,}\) \(\vect{v}=\vect{u}_1+\vect{w}_1\) and \(\vect{v}=\vect{u}_2+\vect{w}_2\) where \(\vect{u}_1,\,\vect{u}_2\in U\text{,}\) \(\vect{w}_1,\,\vect{w}_2\in W\text{.}\) Then

By Property AC, \(\vect{u}_1-\vect{u}_2\in U\) and \(\vect{w}_1-\vect{w}_2\in W\text{.}\) We can now apply our hypothesis, the second statement of the theorem, to conclude that

which establishes the uniqueness needed for the second condition of the definition.

The first condition is identical in the definition and the theorem, so we only need to establish the equivalence of the second conditions.

Assume that \(V=U\ds W\text{,}\) according to Definition Definition 1.2.1. By Property Z and Definition SI, \(\set{\zerovector}\subseteq U\cap W\text{.}\) To establish the opposite inclusion, suppose that \(\vect{x}\in U\cap W\text{.}\) Then, since \(\vect{x}\) is an element of both \(U\) and \(W\text{,}\) we can write two decompositions of \(\vect{x}\) as a vector from \(U\) plus a vector from \(W\text{,}\)

By the uniqueness of the decomposition, we see (twice) that \(\vect{x}=\zerovector\) and \(U\cap W\subseteq\set{\zerovector}\text{.}\) Applying Definition SE, we have \(U\cap W=\set{\zerovector}\text{.}\)

⇐ Assume that \(U\cap W=\set{\zerovector}\text{.}\) And assume further that \(\vect{v}\in V\) is such that \(\vect{v}=\vect{u}_1+\vect{w}_1\) and \(\vect{v}=\vect{u}_2+\vect{w}_2\) where \(\vect{u}_1,\,\vect{u}_2\in U\text{,}\) \(\vect{w}_1,\,\vect{w}_2\in W\text{.}\) Define \(\vect{x}=\vect{u}_1-\vect{u}_2\text{.}\) then by Property AC, \(\vect{x}\in U\text{.}\) Also

So \(\vect{x}\in W\) by Property AC. Thus, \(\vect{x}\in U\cap W =\set{\zerovector}\) (Definition SI). So \(\vect{x}=\zerovector\) and

yielding the desired uniqueness of the second condition of the definition.

Let \(R=\set{\vectorlist{u}{k}}\) and \(S=\set{\vectorlist{w}{\ell}}\text{.}\) Begin with a relation of linear dependence (Definition RLD) on the set \(R\cup S\) using scalars \(\scalarlist{a}{k}\) and \(\scalarlist{b}{\ell}\text{.}\) Then,

where we have made an implicit definition of the vectors \(\vect{u}\in U\text{,}\) \(\vect{w}\in W\text{.}\)

Applying Theorem Theorem 1.2.5 we conclude that

Now the linear independence of \(R\) and \(S\) (individually) yields

Forced to acknowledge that only a trivial linear combination yields the zero vector, Definition LI says the set \(R\cup S\) is linearly independent in \(V\text{.}\)

We will establish this equality of positive integers with two inequalities. We will need a basis of \(U\) (call it \(B\)) and a basis of \(W\) (call it \(C\)).

First, note that \(B\) and \(C\) have sizes equal to the dimensions of the respective subspaces. The union of these two linearly independent sets, \(B\cup C\) will be linearly independent in \(V\) by Theorem Theorem 1.2.7. Further, the two bases have no vectors in common by Theorem Theorem 1.2.6, since \(B\cap C\subseteq\set{\zerovector}\) and the zero vector is never an element of a linearly independent set (Exercise LI.T10). So the size of the union is exactly the sum of the dimensions of \(U\) and \(W\text{.}\) By Theorem G the size of \(B\cup C\) cannot exceed the dimension of \(V\) without being linearly dependent. These observations give us \(\dimension{U}+\dimension{W}\leq\dimension{V}\text{.}\)

Grab any vector \(\vect{v}\in V\text{.}\) Then by Theorem Theorem 1.2.6 we can write \(\vect{v}=\vect{u}+\vect{w}\) with \(\vect{u}\in U\) and \(\vect{w}\in W\text{.}\) Individually, we can write \(\vect{u}\) as a linear combination of the basis elements in \(B\text{,}\) and similarly, we can write \(\vect{w}\) as a linear combination of the basis elements in \(C\text{,}\) since the bases are spanning sets for their respective subspaces. These two sets of scalars will provide a linear combination of all of the vectors in \(B\cup C\) which will equal \(\vect{v}\text{.}\) The upshot of this is that \(B\cup C\) is a spanning set for \(V\text{.}\) By Theorem G, the size of \(B\cup C\) cannot be smaller than the dimension of \(V\) without failing to span \(V\text{.}\) These observations give us \(\dimension{U}+\dimension{W}\geq\dimension{V}\text{.}\)

Suppose that \(\vect{v}\in V\text{.}\) Then due to \(V=U\ds W\text{,}\) there exist vectors \(\vect{u}\in U\) and \(\vect{w}\in W\) such that \(\vect{v}=\vect{u}+\vect{w}\text{.}\) Due to \(W=X\ds Y\text{,}\) there exist vectors \(\vect{x}\in X\) and \(\vect{y}\in Y\) such that \(\vect{w}=\vect{x}+\vect{y}\text{.}\) All together,

which would be the first condition of a definition of a 3-way direct product.

Now consider the uniqueness. Suppose that

Because \(\vect{x}_1+\vect{y}_1\in W\text{,}\) \(\vect{x}_2+\vect{y}_2\in W\text{,}\) and \(V=U\ds W\text{,}\) we conclude that

From the second equality, an application of \(W=X\ds Y\) yields the conclusions \(\vect{x}_1=\vect{x}_2\) and \(\vect{y}_1=\vect{y}_2\text{.}\) This establishes the uniqueness of the decomposition of \(\vect{v}\) into a sum of vectors from \(U\text{,}\) \(X\) and \(Y\text{.}\)