Proving that two subspaces are complementary?

Question

Pardon me if I have the terminology wrong here, I'm not totally sure if this is what it's called in English.

I want to show that two sets are complementary in $\mathbb R^4$. From what I understand, this is equivalent to show that two sets are direct sum of $\mathbb{R}^4$. Is that true?

Here are my subspaces:

$H = \{(a,b,c,d)\in \mathbb{R}^4 \mid a + b + c + d = 0\}$

$\operatorname{span}((1,1,1,1))$

For starters, I proved that $H \cap \operatorname{span}((1,1,1,1))= \{(0,0,0,0)\}$

by saying that $\operatorname{span}((1,1,1,1))$ can be described as $\{(x,y,z,t)\in \mathbb{R}^4 \mid x = y = z = t\}$. Thus, $H \cap \operatorname{span}((1,1,1,1))= \{(x,y,z,t) \mid x + y + z + t = 0, x = y = z = t\}=\{(x,x,x,x)\mid 4x = 0\} = \{(0,0,0,0)\}$

Is this the right way to show this?

Now, to show that $H + \operatorname{span}((1,1,1,1)) = \mathbb{R}^4$:

I can say that $$ H + \operatorname{span}((1,1,1,1)) = \{(a,b,c,d)\in \mathbb{R}^4 \mid a + b + c + d = 0\} + \{(x,y,z,t)\in \mathbb{R}^4 \mid x = y = z = t\}. $$ I sort of have an idea of what I should be doing at this point, but I'm not too sure I'm even doing this properly.

How do I go about finishing my proof?

Furthermore, is showing that two subspaces are complementary equivalent to showing that their sum creates a basis?

Sorry for the poor notation. Any help is appreciated. Thank you.

Answer 1

Yes, you have to prove that

1. $H\cap\operatorname{span}((1,1,1,1))=\{(0,0,0,0)\}$
2. $H+\operatorname{span}((1,1,1,1))=\mathbb{R}^4$

You can simplify the proof of the first fact by taking $a(1,1,1,1)=(a,a,a,a)\in\operatorname{span}((1,1,1,1))$ and imposing the condition that $(a,a,a,a)\in H$, that is, $$ a+a+a+a=0 $$ which of course implies $a=0$.

The second part can be done in several ways. One is to take $(x,y,z,w)\in\mathbb{R}^4$ and trying to express it in the form $$ (x,y,z,w)=(a,b,c,d)+r(1,1,1,1) $$ where $a+b+c+d=0$. This becomes a linear system $$ \begin{cases} a+r=x \\ b+r=y \\ c+r=z \\ d+r=w \\ a+b+c+d=0 \end{cases} $$ The matrix is $$ \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 1 & x \\ 0 & 1 & 0 & 0 & 1 & y \\ 0 & 0 & 1 & 0 & 1 & z \\ 0 & 0 & 0 & 1 & 1 & w \\ 1 & 1 & 1 & 1 & 0 & 0 \end{array}\right] $$ and a simple elimination brings it in the form $$ \left[\begin{array}{ccccc|c} 1 & 0 & 0 & 0 & 1 & x \\ 0 & 1 & 0 & 0 & 1 & y \\ 0 & 0 & 1 & 0 & 1 & z \\ 0 & 0 & 0 & 1 & 1 & w \\ 0 & 0 & 0 & 0 & -4 & -x-y-z-w \end{array}\right] $$ which shows that the system has a (unique) solution.

Another way is to notice that $\dim H=3$ and $\dim\operatorname{span}((1,1,1,1))=1$. Since $H\cap\operatorname{span}((1,1,1,1))=\{(0,0,0,0)\}$, we know that $$ \dim\bigl(H+\operatorname{span}((1,1,1,1))\bigr)=3+1=4 $$ so the sum is $\mathbb{R}^4$. Why is $\dim H=3$? Because a basis is $$ \{(1,0,0,-1),(0,1,0,-1),(0,0,1,-1)\} $$