Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question copied from "Linear Algebra - Friedberg". Can anyone explain the procedure, i.e. the strategy, of how to prove the statement you're asked to prove.

The question is:

Let W1 and W2 be subspaces of a vector space V . Prove that V is the direct sum of W1 and W2 if and only if each vector in V can be uniquely written as x1 + x2 where x1 ∈ W1 and x2 ∈ W2.

My swing at it: $$V = W_1 \oplus W_2 \ \ \ \ \ <=> \ \ \ \ \ V = \{x_1 + x_2: x_1 \in W_1, x_2 \in W_2\}$$

I don't know how to proceed. In reality, the answer seems so obvious to me, I just don't know how to put it down on paper.

share|cite|improve this question
What's the definition of "direct sum" that you have? – DonAntonio Dec 9 '12 at 4:14
@DonAntonio Two vectors being added together: $(1,2) \oplus (2,3) = (3, 5)$ – CodyBugstein Dec 9 '12 at 4:17
That's not direct sum of vector spaces, that's vector addition. The definition I've usually seen is that we say that $V = W_1 \oplus W_2$ if $V = W_1 + W_2$ and $W_1 \cap W_2 = \{0\}$. – Javier Dec 9 '12 at 4:23
That's not any definition of direct sum but only the definition of sum. Check your notes. – DonAntonio Dec 9 '12 at 4:23
@DonAntonio Yes, I had mistaken the definition. I stand corrected. – CodyBugstein Dec 9 '12 at 4:34
up vote 4 down vote accepted

$$(1)\;\;\;\;\;V=W_1\oplus W_2\Longrightarrow W_1\cap W_2=\{0\}\,\,,\,\,V=W_1+W_2$$

Supose that for some vector $\,v\in V\,$ we have two expressions -- [directly, no reduction ad absurdum as we don't assume the expressions are different] --

$$v=w_1+w_2=u_1+u_2\,\,,\,\,w_i,u_i\in W_i\,\,,\,i=1,2\Longrightarrow\,\,\text{-- [gather similar terms] --} $$

$$w_i-u_1=u_2-w_2\in W_1\cap W_2=\{0\}\Longrightarrow w_1=u_1\,\,,\,w_2=u_2$$

and the expression is unique

$$(2)\;\;\;\;\;\;\;\;V=W_1+W_2\,\,\,\text{and every vector in}\,\,V\,\,\text{ has a unique expression} \,\,v=w_1+w_2$$

$$w_i\in W_i\,\,,\,i=1,2$$

$$x\in W_1\cap W_2\Longrightarrow x=0+x=x+0\,\,\text{are two expressions for this vector}\Longrightarrow x=0$$

[Whatever is in the intersection gives us a straightforward contradiction to the assumption of trivial intersection ...unless... it is the zero vector]

share|cite|improve this answer
On the fourth line, did you mean $u_2-w_2$ rather than $u_2-W_2$? – CodyBugstein Dec 9 '12 at 4:36
Yup, typo. I fix it now – DonAntonio Dec 9 '12 at 4:43
Thanks! Can you explain how you came up with the proof? What is your approach to this sort of question? – CodyBugstein Dec 9 '12 at 4:51
Well, I'm extremely intelligent and resourceful and, just kidding: this is a common exercise in basic linear algebra, and I think about 115% of all first year students of mathematics deal with it in this or that fashion. I happen to have some experience since I graduated long ago, and I've even taught this stuff several times, so... – DonAntonio Dec 9 '12 at 4:55
Ok, I'll try to add something between square parentheses []. – DonAntonio Dec 9 '12 at 5:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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