Theorem 26:

26

(i) \iff (ii)

Suppose V = U \oplus W, then U \cap W = \{0\} and all v \in V can be written in the form v = u + w for u \in U, v \in W. Suppose that there existed two such decompositions of a particular v, eg. v = u_1 + w_1 and v = u_2 + w_2 for u_1 \ne u_2 and w_1 \ne w_2. Then u_1 - u_2 = w_2 - w_1. But the LHS is in U and the RHS is in W. Since U \cap W = \{0\} we necessarily have u_1 - u_2 = 0 and w_2 - w_1 = 0, ie. u_1 = u_2 and w_2 = w_1, and so such an expression is unique.

Now suppose (ii). We have that V = U + W, it just remains to establish that U \cap W = \{0\}. Clearly \{0\} \subseteq U \cap W since U, W \le V. Suppose (U \cap W) \setminus \{0\} \ne \emptyset and pick v \in (U \cap W) \setminus \{0\}. Then we could write, say 3v = v + 2v with v \in U and 2v \in W and 3v = 2v + v with 2v \in U and v \in W. Since v \ne 0 we have v \ne 2v so we’ve apparently found two distinct decompositions, impossible by premise. So U \cap W = \{0\}.

(i) \iff (iii)

Suppose that V = U \oplus W. Then V = U + W and U \cap W = \{0\}. So \mathrm{dim}(U + W) + \mathrm{dim}(U \cap W) = \mathrm{dim}(V) + 0 = \mathrm{dim}(V). Using the dimension formula we have \mathrm{dim}(U + W) + \mathrm{dim}(U \cap W) = \mathrm{dim}(U) + \mathrm{dim}(W), so \mathrm{dim}(V) = \mathrm{dim}(U) + \mathrm{dim}(W).

Suppose (iii) Then \mathrm{dim}(U + W) = \mathrm{dim}(U) + \mathrm{dim}(W) = \mathrm{dim}(U + W) + \mathrm{dim}(U \cap W) using the dimension formula. So \mathrm{dim}(U \cap W) = 0, so U \cap W = \{0\}. Since V = U + W also, we have V = U \oplus W.

(i) \iff (iv)

Suppose V = U \oplus W. Then \mathrm{dim}(U + W) + \mathrm{dim}(U \cap W) = \mathrm{dim}(V), since V = U + W and U \cap W = \{0\}. Then using the dimension formula we have \mathrm{dim}(V) = \mathrm{dim}(U) + \mathrm{dim}(W).

Suppose (iv) We have \mathrm{dim}(V) = \mathrm{dim}(U) + \mathrm{dim}(W) = \mathrm{dim}(U \cap W) + \mathrm{dim}(U + W) by the dimension formula. Since \mathrm{dim}(U \cap W) = 0, we have \mathrm{dim}(V) = \mathrm{dim}(U + W). Since U + W \le V we have V = U + W. Since U \cap W also we have V = U \oplus W.

(i) \iff (v)

Suppose V = U \oplus W. Then we can write any v \in V in the form v = u + w. Let \{u_1, \ldots, u_m\} be a basis for U and \{w_1, \ldots, w_n\} be a basis for W. Then we can write:

\displaystyle u = \sum_{i = 1}^m \alpha_i u_i

and:

\displaystyle w = \sum_{i = 1}^n \beta_i w_i

So that:

\displaystyle v = \sum_{i = 1}^m \alpha_i u_i + \sum_{i = 1}^n \beta_i w_i

So certainly u_1, \ldots, u_m, w_1, \ldots, w_n spans V. These are necessarily linearly independent. Suppose not, then there exists non-zero coefficients \langle c_i\rangle, \langle d_i\rangle such that:

\displaystyle 0 = \sum_{i = 1}^m c_i u_i + \sum_{i = 1}^n d_i w_i

but then:

\displaystyle \sum_{i = 1}^m c_i u_i = -\sum_{i = 1}^n d_i w_i

So \sum c_i u_i \in W and \sum d_i w_i \in U. But U \cap W = \{0\}, so these sums must both be zero. Linear independence of \langle u_i\rangle and \langle w_i\rangle then requires c_i = d_j = 0. So u_1, \ldots, u_m, w_1, \ldots, w_n is linearly independent and so a basis for V.

Now suppose (v). Since u_1, \ldots, u_m, w_1, \ldots, w_n is a basis for V we can write each v \in V in the form:

\displaystyle v = \sum_{i = 1}^m \alpha_i u_i + \sum_{i = 1}^n \beta_i w_i

with the u_i and w_i s linearly independent.

Clearly \sum \alpha_i u_i \in U and \sum \beta_i w_i \in W, so V = U + W. We want to now show that U \cap W = \{0\}. The fact that \{0\} \subseteq U \cap W follows from the fact that U, W \le V. Suppose that (U \cap W) \setminus \{0\} \ne \emptyset, and let x \in (U \cap W) \setminus \{0\} \ne \emptyset. Then we would be able to write:

\displaystyle \sum_{i = 1}^m c_i u_i = \sum_{i = 1}^n d_i w_i

for c_i, d_i not all zero. But then:

\displaystyle \sum_{i = 1}^m c_i u_i - \sum_{i = 1}^n d_i w_i = 0

contradicting linear independence of u_1, \ldots, u_m, w_1, \ldots, w_n. So U \cap W = \{0\} and V = U \oplus W.

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Really strong!!!