Dimension of a direct sum

Proof. We assert that if $\{x_1,\dots ,x_n\}$ is a basis in $𝒰$ , and if $\{y_1,\dots ,y_m\}$ is a basis in $𝒱$ , then the set $\{x_1,\dots ,x_n,y_1,\dots ,y_m\}$ (or, more precisely, the set $\{⟨x_1,0⟩,\dots ,⟨x_n,0⟩,⟨0,y_1⟩,\dots ,⟨0,y_m⟩\}$ ) is a basis in $𝒲$ . The easiest proof of this assertion is to use the implication (1) $⟹$ (3) from the theorem of the preceding section. Since every $z$ in $𝒲$ may be written in the form $z=x+y$ , where $x$ is a linear combination of $x_1,\dots ,x_n$ and $y$ is a linear combination of $y_1,\dots ,y_m$ , it follows that our set does indeed span $𝒲$ . To show that the set is also linearly independent, suppose that $${\alpha }_1{x}_1+\cdots +{\alpha }_n{x}_n+{\beta }_1{y}_1+\cdots +{\beta }_m{y}_m=0.$$ The uniqueness of the representation of $0$ in the form $x+y$ implies that $${\alpha }_1{x}_1+\cdots +{\alpha }_n{x}_n={\beta }_1{y}_1+\cdots +{\beta }_m{y}_m=0,$$ and hence the linear independence of the $x$ ’s and of the $y$ ’s implies that $${\alpha }_1=\cdots ={\alpha }_n={\beta }_1=\cdots ={\beta }_m=0.$$ ◻

Proof. Let $\{x_1,\dots ,x_n\}$ be any basis in $𝒰$ ; by the theorem of Section: Bases we may find a set $\{y_1,\dots ,y_m\}$ of vectors in $𝒲$ with the property that $\{x_1,\dots ,x_n,y_1,\dots ,y_m\}$ is a basis in $𝒲$ . Let $𝒱$ be the subspace spanned by $y_1,\dots ,y_m$ ; we omit the verification that $𝒲=𝒰\oplus 𝒱$ . ◻