How to show that $\mathbb{R}^3$ is the direct sum of $W_1={\rm span}\{(1,1,1)\}$ and $W_2={\rm span}\{(1,0,0),(1,1,0)\}$?

Question

We write linear combination of two spans as $a(1,1,1)+b(1,0,0)+c(1,1,0)$. Any vector $v$ in $\mathbb{R}^3$ can be expressed as $(a+b+c, a+b, a)$. Also linear combination of $W_2$ cannot form vector in $W_1$ other than $(0,0,0)$, $W_1\cap W_2=(0,0,0)$, and therefore direct sum is $\mathbb{R}^3$. I am not sure if my method is correct, so could anyone help?

Answer 1

Note that you need to show that every vector in $\mathbb{R}^3$ can be expressed as

$$ (a + b + c, a + c, a) $$

and not the expression you wrote. Other then that, your method is correct (and you of course need to justify why $W_1 \cap W_2 = \{ (0,0,0) \}$). If you know the theorem that relates the dimension of the intersection $W_1 \cap W_2$ with the dimension of the sum $W_1 + W_2$, then it is enough to show that $W_1 \cap W_2 = \{ (0,0,0) \}$ as this shows that $W_1 + W_2$ is three-dimensional.

Answer 2

Since $\dim W_1=1$ and $\dim W_2=2$, if $\mathbb{R}^3=W_1+W_2$ the dimension formula gives $$ \dim(W_1\cap W_2)=\dim\mathbb{R}^3-\dim W_1-\dim W_2=0 $$ So you just need to show that $$ \mathbb{R}^3=\operatorname{Span}\{(1,1,1),(1,0,0),(1,1,0)\} $$ The matrix $$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{bmatrix} $$ clearly has rank $3$.

Answer 3

One simple way is the classical method to put vectors $$(1,1,1),(1,0,0),(1,1,0)$$ as column in a matrix $A$ and verify that $$\forall b \in \mathbb{R^3}$$

$$Ax=b$$

has solution.

That is $\iff$ $A$ is not a singular matrix.

Thus your consideration is correct as $W_1$ and $W_2$ are independent subspaces and $dim(W_1)+dim(W_2)=3$ thus they span $\mathbb{R^3}$.