1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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) $\endgroup$ – Gianolepo Dec 2 '15 at 23:31
  • $\begingroup$ 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 $\{0\}$) $\endgroup$ – MonadBoy Dec 3 '15 at 6:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.