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Linear Algebra Done Right -Sheldon Axler, Third Edition, Section 1.C, Page 23.

Condition for a direct sum:

Suppose $U_1,U_2,\ldots,U_m$ are subspaces of $V$. Then $U_1+\cdots+U_m$ is a direct sum if and only if the only way to write $0$ as a sum $u_1+\cdots+u_m$, where each $u_j$ is in $U_j$, is by taking each $u_j=0$.

I don't seem to understand their proof.

They first write $v=u_1+\cdots+u_m$ (where $u_1\in U_1,u_2\in U_2,\ldots,u_m\in U_m$) and also suppose that $v=v_1+\cdots+v_m$ (where $v_1\in U_1,\ldots,v_m\in U_m$).

Next, they subtract the two equations and write:

$$0=(u_1-v_1)+\cdots+(u_m-v_m)$$

"Because $u_1-v_1\in U_1,\ldots,u_m-v_m\in U_m$, the equation above implies that each $u_j-v_j=0$. Thus $u_1=v_1,\ldots,u_m=v_m$." I didn't understand how they deduced this. Why should $u_1-v_1\in U_1,\ldots,u_m-v_m\in U_m$ imply that each $u_j-v_j=0$ ?

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  • $\begingroup$ I feel that it may be necessary that $\dim U_i=1$ $\endgroup$ – e.turatti Mar 16 '18 at 20:14
  • $\begingroup$ I think they are using implictly $U_i\cap U_j= \{0\}$. $\endgroup$ – Dog_69 Mar 16 '18 at 20:14
  • $\begingroup$ @e.turatti Exactly. They haven't mentioned that. $\endgroup$ – user400242 Mar 16 '18 at 20:14
  • $\begingroup$ Ok, I will think if it is necessary, but this is my first impression, and also that $U_i\cap U_j=0$ $\endgroup$ – e.turatti Mar 16 '18 at 20:16
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They are proving that $U_1 + \ldots + U_m$ is a direct sum if the only way to write $0 = z_1 + \ldots + z_m$ with $z_i \in U_i$ is when all $z_i = 0$. To prove that, they assume that the only way to write $0 = z_1 + \ldots + z_m$ is when all $z_i = 0$, and then they prove that $U_1 + \ldots + U_m$ is indeed a direct sum.

To prove that something is a direct sum, one needs to prove that each vector $v \in U_1 + \ldots + U_m$ has a unique representation of the form $ v = u_1 + \ldots + u_m$ with $u_i \in U_i$. So they assume that they have some vector $v$ has two representations $v = u_1 + \ldots + u_m = v_1 + \ldots + v_m$, and they show that these representations are actually equal, that is, there's no way to have two different representations, so that the representation must be unique.

The way they prove is to note that if $v = u_1 + \ldots + u_m = v_1 + \ldots + v_m$, then $0 = v - v = (u_1 + \ldots + u_m) - (v_1 + \ldots + v_m) = (u_1 - v_1) + ... + (u_m - v_m)$. But since $u_i, v_i \in U_i$, we also have $u_i - v_i \in U_i$. But now, the assumption is that the only way to write $0 = z_1 + \ldots + z_m$ is when all $z_i = 0$, thus $u_i - v_i = 0$ for all $i$.

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The proof starts off by showing that if we asssume $U_{1}+...+U_{m}$ is a direct sum, then the only way to write $0$ in this subspace is in the form $0 = u_{1} +...+u_{m}$, where each $u_{j}$ is equal to $0$. (This must be true since $0 \in U_{1},..., 0 \in U_{m}$, therefore one way to write $0$ would be in the form $0 = 0+...+0$, since we assume that $U_{1}+...+U_{m}$ is a direct sum, this means that this must be the only representation). This means that every direct sum must have 0=0+...+0, as the only representation of 0. (This will be important for the rest of the proof). We have shown that this is a requirement for a direct sum.

We have shown that the condition $0 = 0+...+0$, being the only possible representation of $0$ is required for $U_{1}+...+U_{m}$, to be a direct sum, now we must show that this condition implies that every vector has a unique representation. This is where we introduce two representations of some vector $v \in U_{1}+...+U{m}$, $v = u_{1}+...+u_{m}$ and $v = v_{1}+...+v_{m}$. Notice that subtracting the two representations results in $0 = (u_{1}-v_{1})+...+(u_{m}-v_{m})$. We proved earlier that for $U_{1}+...+U_{m}$ can only be a direct sum if the only representation of $0$ is in the form $0+...+0$, therefore $u_{j}-v_{j}$ must equal $0$ for all $j$ inbetween 1 and $m$. This shows that $u_{j} = v_{j}$ for all $j$ between 1 and $m$, this shows that every vector has one and only one representation based off of the zero as a sum of zero's result.

This shows that being a direct sum implies that $0$'s only representation is through a sum of $0$'s, and we have also shown that having this representation of $0$ be the only one implies that every vector has a unique representation, hence $U_{1}+...+U_{m}$ is a direct sum. Therefore, the two imply each other, proving the proposition brought about in the text.

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The sum $U_1 + \cdots + U_m$ is direct if $$U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m) = \{0\}$$ for all $j \in \{1, \ldots, m\}$.

Assume that the intersections are trivial. If $$0 = u_1 + \cdots + u_m$$

is a representation of $0$ with $u_i \in U_i$ then $$-u_j = u_1 + \cdots + u_{j-1} + u_{j+1} + u_n$$

The left side is in $U_j$ and the right side is in $U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m$. Thus, $$-u_j \in U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m) = \{0\}$$

so $u_j = 0$. Since this is valid for all $j \in \{1, \ldots, m\}$, we conclude that $0$ has only the trivial representation $$0 = 0 + \cdots + 0$$

Conversely, assume that $0$ has only the trivial representation.

Take $x \in U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m)$ for some $j \in \{1, \ldots, m\}$.

Since $x \in U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m$ there exists a decomposition

$$x = \underbrace{v_1}_{\in U_1} + \cdots + \underbrace{v_{j-1}}_{\in U_{j-1}} + \underbrace{v_{j+1}}_{\in U_{j+1}} + \cdots + \underbrace{v_m}_{\in U_m}$$

We have

$$0 = x - x = \underbrace{v_1}_{\in U_1} + \cdots + \underbrace{v_{j-1}}_{\in U_{j-1}} + \underbrace{-x}_{\in U_j} + \underbrace{v_{j+1}}_{\in U_{j+1}} + \cdots + \underbrace{v_m}_{\in U_m}$$

which is a representation of zero. Hence, it must be trivial so $x = 0$. We conclude that all intersections

$$U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m)$$

for $j \in \{1, \ldots, m\}$ are trivial.

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    $\begingroup$ This is completely wrong. $U_i \cap U_j = \{0\}$ is not sufficient condition for a sum to be direct. For example, consider $U_1 = span([1, 0]), U_2 = span([0, 1]), U_3 = span([1, 1])$. Then intersection of any par is zero, but the sum $U_1 + U_2 + U_3$ is not direct. $\endgroup$ – xyzzyz Mar 17 '18 at 0:07
  • $\begingroup$ @xyzzyz Thanks, you are right. I wrongly assumed that $$U_i \cap U_j = \{0\}, \forall i \ne j \implies U_j \cap (U_1 + \cdots + U_{j-1} + U_{j+1} + \cdots + U_m) = \{0\}, \forall j$$ I fixed it now. $\endgroup$ – mechanodroid Mar 17 '18 at 14:02

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