Proving that $\dim(U ⊕ W ) = \dim U + \dim W$

Question

I think I can use the theorem that states that $$\dim U + W = \dim U + \dim W - \dim(U \cap W),$$ but I'm not sure how to use it.

Answer 1

Hint: I think you meant: $$\dim(U+W) = \dim U + \dim W - \dim(U \cap W).$$If we're dealing with $U \oplus W$, then $U \cap W = \{0\}$. Certainly you know $\dim \{0\}$ ?

Answer 2

Show that

If $B$ is a basis for $V$ and $C$ is a basis for $W$, then $B\cup C$is a basis for $V\oplus W$.

It will follow at once from the definition of dimension that $\dim V\oplus W=\dim V+\dim W$.

Answer 3

You don't need that. Furthermore what you're asked to prove is used to prove the formula you mention.

Actually, you just have to observe any two non-zero vectors $u\in U$, $\;w\in W$ are linearly independent: indeed, if $\lambda u+\mu w=0$, then $\lambda u=-\mu w\in U\cap W=0$, hence $\lambda=\mu=0$.

Now if $\mathcal B=\{e_1,\dots,e_n\}$ is a basis for $U$, $\mathcal C=\{e'_1,\dots,e'_p\}$ is a basis for $W$, then $\mathcal B\cup\mathcal C$ is a system of generators of $U\oplus W$. If there is a linear relation: $$\lambda_1 e_1+\dots+\lambda_n e_n+\mu_1e'_1+\dots+\mu_pe'_p=0$$ then

• $\lambda_1 e_1+\dots+\lambda_n e_n=0$, whence $\lambda_1=\dots=\lambda_n=0$,
• $\mu_1e'_1+\dots+\mu_pe'_p=0$, whence $\mu_1=\dots=\mu_p=0$.

Answer 4

$U ⊕ W$ is the direct sum of $U$ and $V$ which could be two subspaces of a vector space (the "internal direct sum") or could be two independent vector spaces over a common field ("the external direct sum").

Let's work with the second case, in which event one way to represent $U ⊕ W$ is as the set of ordered pairs $(u, v) $ with $u \in U $ and $v \in V$. Vector addition of ordered pairs uses the additions in $U$ and $V$ respectively for the first and second elements of the pairs.

If $U$ and $V$ have bases $\{u_i\}$ and $\{v_j\}$ respectively then an ordered pair $(u, v) $ can be represented as $\Sigma \mu_i.u_i + \Sigma \nu_j.v_j$ and so $\{u_i\} \cup \{v_j\}$ is a basis for $U ⊕ W$ and therefore $\dim(U ⊕ W ) = \dim U + \dim W$

(Note: this is true whether the spaces are finite or infinite or a combination: the subtraction $\dim(U ⊕ W ) - \dim U = \dim W$ is not guaranteed unless $U$ is finite.