What is the formal definition of direct sum of operators and functions?

Is it always some thing like the following?

$A: U \to U, B: V \to V$ be continuous and let $T:U\oplus V\to U\oplus V$ $$T(x) = (A \oplus B)(u \oplus v)= A(u) \oplus B(v)$$.

$X$ is a topological vector space and $U,V$ are subspaces.

Can we use direct sum to describe the following relations?

1) $A: U \to U, B: V \to U$ be continuous and define: $$A\oplus B =T:U\oplus V\to U, u\oplus v\mapsto T(u\oplus v)$$

For example, $T(u\oplus v)=A(u) + B(v)$

2) $A: U \to U, B: U \to V$ be continuous and define $A\oplus B =T:U\to U\oplus V, u\mapsto T(u\oplus B(u))$



[1] http://cc.bingj.com/cache.aspx?q=direct+sum+of+hilbert+spaces+operators&d=4885758912236128&mkt=en-US&setlang=en-US&w=kmBEdXxmF9OtyUjV8AanzoUQsyc0N2mZ

[2]Bound of direct sum of operators

[3]Considering operators on the direct sum of Hilbert spaces as operator valued matrices


1 Answer 1


When one wishes to generalize direct sum, one should read about coproducts. This gives your first display $(A \oplus B)(x \oplus y) = A(x) \oplus B(y)$. We can widen your hypotheses a little to $A: M \rightarrow N$, $B: R \rightarrow S$, $x \oplus y \in M \oplus R$, giving $(A \oplus B): (M \oplus R) \rightarrow (N \oplus S)$.

I see no defect in your expression 1, but your expression 2 has a small omission. You write $u \mapsto T(u)$, but $u \in U$ and $T:U \oplus V \rightarrow U \oplus V$, so the argument passed to $T$ isn't quite of the right type. Instead, $u \mapsto T(u \oplus B(u))$ has matching types (and uses $B$).

Edit 20170719T2131Z : Let's avoid repetition in your first display. Let \begin{align} A &: U \rightarrow V \text{,} \tag{1} \\ B &: W \rightarrow X \text{.} \end{align} This forces \begin{align*} A \oplus B &: U \oplus W \rightarrow V \oplus X \text{.} \tag{2} \end{align*} That is, the direct sum of operators is always a map from the direct sum of the domains to the direct sum of the codomains.

Once you write (1), $A \oplus B$ is completely determined, so if you write $T = A \oplus B$, you do not need to specify its domain and codomain; you already have.

In your two follow-on expressions, you are trying to break the consequence (2). In your expression 1, you have $A: U \rightarrow U$ and $B:V \rightarrow U$. This forces $A \oplus B$ is a map from $U \oplus V$ to $U \oplus U$. Since $U \oplus U$ is unlikely to be the same as $U$ (If you are working with vector spaces and $U = \mathbb{R}$, then $U \oplus U$ looks remarkably like $\mathbb{R}^2$.), we have to do more work. We could make a new map $c:U \oplus U \rightarrow U : u_1 \oplus u_2 \mapsto u_1 + u_2$ (where I have assumed that in your setting, we can add elements of $U$ together) and have $(c \,\circ (A \oplus B))$ which is a sequence of maps : \begin{align*} U \oplus V \overset{A \oplus B}\longrightarrow U \oplus U \overset{c}\rightarrow U \text{.} \end{align*} Maybe this isn't the right choice for your application, but for whatever is the right choice, you have to say how you go from $U \oplus U$ to $U$.

By now, I think you know one thing I am going to say about expression 2. If you say the map is $A \oplus B$, it is a map $U \oplus U \rightarrow U \oplus V$. Also, you have two different uses of the symbol $T$ in your definition. Removing the domain/codomain information, you have written \begin{align*} T:u \mapsto T(u \oplus B(u)) \text{.} \end{align*} The second "$T$" is the one from the top of your post. The first "$T$" is the one you are defining. It seems very unlikely that these two things are actually the same thing, so using the same letter for both is confusing. (If, miraculously, it were to turn out that these were the same thing, you would explain in detail why that were so.)

What you are wanting to do in expression 2 is duplicate your input. This is again a job for composition. First, we duplicate the input, then we act on it. \begin{align*} d &: U \rightarrow U \oplus U : u \mapsto u \oplus u \text{.} \end{align*} Then $(A \oplus B) \circ d : U \rightarrow U \oplus V$. Since in your expression 2, $A \oplus B$ is a map with domain $U \oplus U$, if you want a map that starts at $U$ and incorporates $A \oplus B$, you will have to interpose some process that turns as element of $U$ into an element of $U \oplus U$. To summarize,

  • The direct sum of two operators is the map from the direct sum of the domains to the direct sum of their codomains.
  • The object $U \oplus U$ is not generally the same as $U$. If you wish to treat them as if they are, you must explain how to do so and/or why it is the case that this happens to be true. Generally, though, you specify a function that maps from one to the other (in whichever direction you want to go) and apply function composition as needed.
  • When defining a new symbol, that symbol should not appear in the definition of the symbol. (There may be a couple of constructions where this is not true. You shouldn't have reached one yet. Practice not writing circular definitions. You may eventually recognize one of the few appropriate relaxations of this rule.) New symbols should be different from previous symbols.
  • $\begingroup$ Thanks for your instructions. Upvoted and edited. For both of the case (1) and (2), is it ok to define (by myself in my homework) that $A\oplus B =T$? $\endgroup$
    – High GPA
    Jul 16, 2017 at 0:30
  • 1
    $\begingroup$ @HighGPA : I'd usually write that in the other order, "$T = A \oplus B$". Also, your edited expression 2 has a variation of the same defect: "$u \mapsto T(u \oplus v)$", but there is no introduction of $v$. Using the parts available, I would expect $u \mapsto T(u \oplus B(u))$. $\endgroup$ Jul 16, 2017 at 6:07
  • $\begingroup$ I believe you are right, thank you. For a final question, regarding the expression (1), if I write $T=A\oplus B$, would the readers naturally think that $T:U\oplus V\to U\oplus V$ rather than my definition $T:U\oplus V \to U$? $\endgroup$
    – High GPA
    Jul 16, 2017 at 22:00
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    $\begingroup$ @HighGPA : I suspect a reader might stumble on the mixture of function assignment and definition. Also, once you specify $A: U \rightarrow U$ and $B: V \rightarrow U$, you have forced $A \oplus B: U \oplus V \rightarrow U \oplus U$. Since $U \oplus U \neq U$, there is a bug here as well. I think I see where you want to go. I'll edit my answer. $\endgroup$ Jul 19, 2017 at 21:31

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