Question

Two subspacesV andW of${\mathbf{ℝ}}^{\mathbf{n}}$ are called complements if any vector$\overrightarrow{\mathbf{x}}$ in${\mathbf{ℝ}}^{\mathbf{n}}$ can be expressed uniquely as $\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{v}}\mathbf{+}\overrightarrow{\mathbf{w}}$, where$\overrightarrow{\mathbf{v}}$ in V and$\overrightarrow{\mathbf{w}}$ is in W. Show thatV andW are complements if (only if) can$\mathit{V}\mathbf{\cap }\mathit{W}\mathbf{=}\left\{\overrightarrow{0}\right\}$ and $\mathbf{dim}\mathbf{\left(}\mathit{V}\mathbf{\right)}\mathbf{+}\mathbf{dim}\mathbf{\left(}\mathit{W}\mathbf{\right)}\mathbf{=}\mathit{n}$.

Short answer

Thus, it is proved thatV areW complements if and only if$V\cap W=\left\{0\right\}$ and role="math" localid="1660365353877" $dimV+dimW=n$.

Step-by-step solution

Definition of Complement subspaces.

By definition of uniqueness, letV andW are complements of X. The map$T:X\to V\times W$ given by $T\left(x\right)=\left(v,w\right)$, wherev,w are the unique vectors inV andW respectively so that $v+w=x$.

Showing that V∩W={0}.

Use the rank nullify theorem on T, observe that $dimV+dimW=dim(V\times W)$.

Let $a\in V\times W$. Then $-a\in V\times W$, but then

$\begin{array}{l}0=0+0\\ =a+\left(-a\right)\end{array}$

Thus, 0 is not uniquely written as a sum of terms fromV and W.

Unless, it gives:

$\begin{array}{l}a=-a\\ \Rightarrow a=0\end{array}$

Thus, $V\cap W=\left\{0\right\}$.

Showing that dim(V)+dim(W)=n.

Now, consider$S:V\times W\to X$ given by $S\left(\right(v,w\left)\right)=v+w$.

Let$n=dim(V\times W)$

Thus,S is a bijection, so every elementx inX is uniquely the image of some $(v,w)\in V\times W$, so that $v+w=x$.

Thus,V areW complements.

Hence, it is proved that V and Ware complements if and only if$V\cap W=\left\{0\right\}$ and $dimV+dimW=n$.