3
$\begingroup$

My Question is twofold.

First in Linear Algebra Done Right Axler states the following

The set $U_1+U_2+U_3+\dots +U_m$ is the smallest subspace which contains $U_1, U_2, U_3, \dots , U_m$ but he seems to state the proof in words only so i tried to proof it myself

Is the following correct?

Proof. Given a vector space $V$ consider the following set. $$A=\{S\mid\forall j\in\{1,2,3,\dots ,m\}(U_j\subseteq S)\}\tag{1}$$ where $S$ in $(1)$ is a subspace of $V$.

We seek to prove the following proposition $$\sum_{j=0}^m U_j\in A\ , \ \forall Y\in A \left(\sum_{j=0}^m U_j\subseteq Y\right) \tag{2}$$

It is fairly easy to see that $\sum_{j=0}^m U_j\in A$.

Let $Y$ be arbitrary and assume that $Y\in A$. Consider an arbitrary $x\in\sum_{j=0}^m U_j$ so that $x=\sum_{j=0}^m u_j$ where $$\forall j\in \{1,2,3,\dots , m\}(u_j\in U_j)\tag{3}$$ as a consequence of $(3)$ we see that $$\forall j\in \{1,2,3,\dots , m\}(u_j\in Y)\tag{4}$$ and since $Y$ is a subspace and therefore closed under addition it follows that $x\in Y$ since $x$ was arbitrary we may now conclude that $\sum_{j=0}^{m}U_j\subseteq Y$.

$\blacksquare$

Second Axler defines the notion of Direct Sum of subspaces as follows The sum $U_1+U_2+\dots +U_m$ is the direct sum if each element of $U_1+U_2+\dots +U_m$ can be written in only one way as $u_1+u_2+\dots +u_m$ where each $u_j$ is in $U_j$

What does he mean by only one way?

$\endgroup$
2
$\begingroup$

Only one way’ means that if $$u_1+u_2+\dots+u_m=v_1+v_2+\dots+v_m,\quad\text{where}\; u_i, v_i\in U_i$$ then for each $i=1,\dots,m$, one has $u_i=v_i$.

Your proof other the first assertion is correct, but why do you think a proof ‘only in words’ is not a real proof? I think that it's a better proof when possible, on the contrary.

$\endgroup$
1
  • 1
    $\begingroup$ Strongly agree that a proof in words is better, evidenced by the comment in the other answer posted here, "I can't quite follow your argument for the first part." Proofs are about communication among humans, for which natural human language is uniquely well-adapted. $\endgroup$ – G Tony Jacobs Jul 20 '17 at 15:37
2
$\begingroup$

I can't quite follow your argument for the first part. The idea for the proof is that the sum of $n$ subspaces $U_1+U_2+\cdots+U_n$ is the smallest subspace containing $U_1\cup U_2\cup\cdots\cup U_n$ in that it is created by taking the elements of $U_1\cup U_2\cup\cdots \cup U_n$ and adjoining all of their linear combinations.

As for your second question: A sum $U=U_1+U_2+\cdots+U_m$ is a direct sum if given any $u\in U$ written as $u=u_1+u_2+\cdots+u_n$ for $u_i\in U_i, 1\le i\le n,$ this representation is unique. I.e., if also $ u=v_1+v_2+\cdots+v_n$ for $v_i\in U_i, 1\le i\le n$, we have $u_i=v_i.$

Alternatively, we can say that the map $\varphi:(U_1,\ldots, U_n)\to U_1+\cdots+U_n$ given by $\varphi(u_1,\ldots, u_n)=u_1+\cdots+u_n$ is injective.

$\endgroup$
2
  • $\begingroup$ So if we had a vector space $R^{2}$ and if we let the $x$ and $y$ axis embody our sub spaces $U_1$ and $U_2$ then given any vector $v\in U=U_1+U_2$ will have unique coordinates is that what your telling me? $\endgroup$ – Atif Farooq Jul 20 '17 at 15:49
  • 1
    $\begingroup$ Well, in particular if we have that $\mathbf{R}^2$ is the direct sum of $\mathbf{R}\times \{0\}$ and $\{0\}\times \mathbf{R}$. But, yes. $\endgroup$ – Alekos Robotis Jul 20 '17 at 15:57

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.