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Proof regarding direct sum of the dual space of a v-space

  1. Jan 19, 2017 #1
    (From Hoffman and Kunze, Linear Algebra: Chapter 6.7, Exercise 11.) Note that ##V_j^0## means the annihilator of the space ##V_j##. V* means the dual space of V.

    1. The problem statement, all variables and given/known data
    Let V be a vector space, Let ##W_1 , \cdots , W_k## be subspaces of V, and let
    $$V_j = W_1 + \cdots + W_{j-1} + W_{j+1} + \cdots + W_k$$
    Suppose that ##V = W_1 \oplus \cdots \oplus W_k##. Prove that the dual space V* has the direct-sum decomposition ##V^{*}= V_1^0 \oplus \cdots \oplus V_k^0##


    2. Relevant equations
    I use a portion of a theorem in the text referred to as Theorem 9 that states "If ##E_1 , \cdots , E_k## are k linear operators on V which satisfy conditions;
    (i) each ##E_i## is a projection.
    (ii) ##E_i \circ E_j = 0##, if i ≠ j
    (iii): I = ##E_1 + \cdots + E_k##, where I is the identity operator.
    then if we let ##W_i## be the range of ##E_i##, then ##V = W_1 \oplus \cdots \oplus W_k##."

    3. The attempt at a solution
    Let ##E_i## (i = 1,...,k) be a linear operator on V such that, if ##\alpha## is in V and ##\alpha = \alpha_1 + \cdots + \alpha_k## (with ##\alpha_j \in W_j##), then ##E_i(\alpha) = \alpha_i##. Then the image of each function ##E_i## is the corresponding subspace ##W_i## and the null space is the sum ##W_1 + \cdots + W_{i-1} + W_{i+1} + \cdots + W_k = V_i##.

    Let ##^tE_i## denote the transpose of ##E_i## (i.e., the linear operator from V* into V* defined by ##^tE_i(f) = f \circ E_i##), then the image of ##^tE_i## is the annihilator of the space ##V_i##. We seek to show that that V* is the direct sum of the images of each ##^tE_i## for i = 1, ... , k.

    (i): To show each transpose operator is a projection, let ##f \in V^{*}## and we have $$(^tE_j \circ ^tE_j)(f) = ^tE_j(f \circ E_j) = f \circ E_j \circ E_j = f \circ E_j^2 = f \circ E_j = ^tE_j(f)$$ Thus ##^tE_j## is a projection for j = 1, ..., k.

    (ii): We now show that ##^tE_i \circ ^tE_j = 0## for i ≠ j. Once again if ##f \in V^{*}##, we have
    ##(^tE_i \circ ^tE_j)(f) = ^tE_i(^tE_j(f)) = ^tE_i(f \circ E_j) = f \circ E_j \circ E_i = 0## where the last result follows from the fact that ##E_i \circ E_j = 0## for i ≠ j as the image of ##E_j## is in the null space of ##E_i##.

    (iii): Lastly, we must show that, $$I = ^tE_1 + \cdots + ^tE_k$$ It follows that ##I = E_1 + \cdots + E_k## as if ##\alpha \in V## and ##\alpha = \alpha_1 + \cdots + \alpha_k## (##\alpha_j \in W_j##) then ##\alpha = E_1(\alpha) + \cdots + E_k(\alpha)## for all ##\alpha \in V##. Also as ##E_i \circ E_j = E_j \circ E_i## for all i and j and each ##E_i## is diagonalizable it follows that there is a basis for V such that the matrix associated with each projection with respect to that basis is a diagonal matrix. And as the associated matrix of a transpose linear transformation is the transpose of the matrix of its associated map with respect to the same base(s), it follows that ##I = ^tE_1 + \cdots + ^tE_k##.
    Therefore, by Theorem 9, $$V^{*}= V_1^0 \oplus \cdots \oplus V_k^0$$ as was to be shown.


    A part I'm not entirely clear on is the proof of property (iii). Mainly that I'm fairly sure that if a linear transformation is the sum of some other linear transformations then the matrix associated with the sum map with respect to some pair of bases should be the sum the of the matrix representations of the summand linear transformations in the same basis but Chapter 3 was awhile ago :P
     
  2. jcsd
  3. Jan 20, 2017 #2

    fresh_42

    Staff: Mentor

    Correct me if I'm wrong, but isn't the essential part here ##W^* \cong ann(V/W)## for a subspace ##W \subseteq V## and all the rest is only a distraction? At least I think you should explicitly define the annihilator and show why ##im (E^*\, : \, W^* \rightarrow V^*) = ann (V/W)##.
     
  4. Jan 20, 2017 #3
    Perhaps, but other than a quick wikipedia article read just now I haven't learned anything about quotient spaces so I'm not entirely sure what you're pointing out. As for the second part, would defining the annihilator of one of the subspaces ##V_i## as ##\{ f \in V^* | f(E_i(\alpha)) = 0 for all \alpha \in V_i \}## suffice? And as the image of ##^tE_i## is the space of linear functionals of the form ##f \circ E_i## for some ##f \in V^*## and ##E_i = 0## for all ##\alpha \in V_i## it follows that ##(f \circ E_i)(\alpha) = 0## too for all ##\alpha \in V_i## and hence ##Im(^tE_i) = ann(V_i)##.
     
    Last edited: Jan 20, 2017
  5. Jan 20, 2017 #4
    Woops, for the last part to show ##ann(V_i) \subseteq Im(^tE_i)## let ##g \in ann(V_i)##, ##\alpha \in V## and suppose ##\alpha = \alpha_1 + \cdots + \alpha_k## with ##\alpha_i \in W_i## for i = 1, . . . , k. As ##^tE_i## is a projection we must show that ##g(E_i(\alpha)) = g(\alpha)##. We have, $$g(E_i(\alpha)) = g(E_i(\alpha_1 + \cdots + alpha_k)) = g(E_i(\alpha_i)) = g(\alpha_i) = g(\alpha)$$
    with the last equality following from the fact that g is an element in the annihilator of ##V_i##. Thus ##ann(V_i) \subseteq Im(^tE_i)## and hence ##Im(^tE_i) = ann(V_i)##.
     
  6. Jan 22, 2017 #5

    fresh_42

    Staff: Mentor

    I have really difficulties to grasp the complexity of the solution you suggested.
    Beside this, haven't you confused ##V_i## and ##W_i\,## in the definition of the annihilator?
    You defined ##E_i\, : \,V \twoheadrightarrow W_i## as the projection onto the subspace ##W_i## and the annihilator of ##V_i## as
    $$ \begin{align*} V_i^0 &= ann(V_i) \\ & = ann(W_1+\ldots +W_{i-1}+W_{i+1}+\ldots +W_k) \\ & \stackrel{(1)}{=} \{f \in V^* \,\vert \, f \circ E_i\,(V_i) = 0 \} \\ & = \{f \in V^* \,\vert \, f \circ E_i\,(W_1+ \ldots + W_{i-1} + W_{i+1}+\ldots +W_k) = 0 \} \\ & = \{ f \in V^* \,\vert \, f(0)=0 \} \\ & = V^* \end{align*} $$
    What is left, if we have the corrected definition of the annihilator (see ##(1)##)? I think the actual work is hidden in the way you defined the annihilator. What says the book - I mean literally? If it has another definition, then you should show that both are equivalent. Maybe you should also use the embeddings ##\iota_i \, : \, W_i \hookrightarrow V##.
     
    Last edited: Jan 22, 2017
  7. Jan 22, 2017 #6
    If you mean the definition of the annihilator the book gives it as, (I wasn't sure what exactly you meant by explicitly defining it)
    "If V is a vector space over the field F and S is a subset of V, the annihilator of S is the set ##S^0## of linear functionals f on V such that ##f(\alpha) = 0## for all ##\alpha \in S##." I'm not sure what you mean by using "the embeddings ##\iota_i \, : \, W_i \hookrightarrow V##". I don't know what an embedding is.
     
  8. Jan 22, 2017 #7

    fresh_42

    Staff: Mentor

    Fine. So ##V_i^0=ann(V_i)=\{\,f \in V^*\,\vert \,f(V_i)=f(W_1+\ldots +W_{i-1}+W_{i+1}+\ldots +W_k)=0\, \}##.

    Now we want to show ##V^* = V_1^0 \oplus \ldots \oplus V_k^0##. Why couldn't it be done directly?

    Simply show ##V^* \subseteq V_1^0 + \ldots + V_k^0 \subseteq V^*## and ##V_i^0 \cap V_j^0 = \{0\}## for ##i \neq j##. You could consider linear mappings ##f_i = f|_{W_i}\in W_i^*##.

    Remark: The embeddings would only add a little more precision to it. You have projections ##E_i \, : \,V \rightarrow W_i## and ##W_i \subseteq V = W_1 \oplus \ldots \oplus W_k##. The corresponding embedding is just the linear map ##W_i \rightarrow V## that maps ##\alpha_i \mapsto (0,\ldots,0,\alpha_i,0,\ldots 0)## which formally makes the vector space ##W_i## a subspace of ##V## again. The difference between a single line and and a line as the part of a plane, e.g. the ##x-##axis in a ##(x,y)## coordinate plane if you like. Taking the ##\alpha_i## out of the context as part of ##\alpha## somehow "forgets" it is living in ##V## and to embed it again in ##V## formally adds this information again. If it's confusing, just leave it.
     
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