# How to prove that $W_1\oplus W_2\oplus W_3=\mathbb{R^{5}}$

I've a question on this exercise:

Consider the following vector subspaces in $\mathbb{R^{5}}$

$W_1=\mathscr{L}((1,0,-1,5,0),(0,1,2,3,1))$

$W_2=\mathscr{L}((0,0,0,1,-1),(0,0,1,-1,0))$

$W_3=\mathscr{L}((0,0,0,0,1))$

Prove that $W_1\oplus W_2\oplus W_3=\mathbb{R^{5}}$.

In the solutions it says that it is proven by the fact that the matrix containing the vectors on the rows has rank 5.

I can't understand how can that be a proof since in general

$dim(W)=dim(W_1)+dim(W_2)+...+dim(W_k) \not \implies W=W_1\oplus W_2 \oplus ... \oplus W_k$

How can that be?

## 1 Answer

Since the matrix described has rank 5, the rows are linearly independent. That means the row vectors form a basis for $\mathbb{R}^5$. Also, that means that each respective subspace $W_i$ has a basis of linearly independent vectors, and since it so happens that the five row vectors are partitioned into bases of the subspaces $W_i$, the result follows. (i=1,2,3)

Indeed, this is a lucky special case.

• Thanks! It's a particular case, but, to be rigorous, does the fact that the rank is 5 prove that, taking sigularly every $W_i$, its intersection with the sum of the other remaining subspaces is empty? (this is indeed the definition of direct sum when more than two subspaces are concerned) – Gianolepo Dec 2 '15 at 23:31
• No, it is because 1. All vectors involved are linearly independent (which here follows from the rank being 5), and 2. All vectors involved are partitioned into the bases for $W_i$. In particular, none of the $W_i$ share basis vectors, which is why they only intersect in $\{0\}$. (Also, remember that any vector subspaces must contain the zero element per definition, so the smallest intersection possible is the set $\{0\}$) – MonadBoy Dec 3 '15 at 6:44