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Direct Sums of Vector Spaces Thursday 3 November 2005 Lectures for Part a of Oxford FHS in Mathematics and Joint Schools

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Direct Sums of Vector Spaces Thursday 3 November 2005 Lectures for Part a of Oxford FHS in Mathematics and Joint Schools Linear Algebra 2: Direct sums of vector spaces Thursday 3 November 2005 Lectures for Part A of Oxford FHS in Mathematics and Joint Schools • Direct sums of vector spaces • Projection operators • Idempotent transformations • Two theorems • Direct sums and partitions of the identity Important note: Throughout this lecture F is a field and V is a vector space over F . 0 Direct sum decompositions, I Definition: Let U, W be subspaces of V . Then V is said to be the direct sum of U and W , and we write V = U ⊕ W , if V = U + W and U ∩ W = {0}. Lemma: Let U, W be subspaces of V . Then V = U ⊕ W if and only if for every v ∈ V there exist unique vectors u ∈ U and w ∈ W such that v = u + w. Proof. 1 Projection operators Suppose that V = U ⊕ W . Define P : V → V as follows. For v ∈ V write v = u + w where u ∈ U and w ∈ W : then define P (v) := u. Observations: (1) P is well-defined; (2) P is linear; (3) ImP = U, KerP = W ; (4) P 2 = P . Proofs. Terminology: P is called the projection of V onto U along W . 2 Notes on projection operators Note 1. Suppose that V is finite-dimensional. Choose a basis u1, . , ur for U and a basis w1, . , wm for U. Then the matrix of P with respect to the basis u1, . , ur, w1, . , wm of V is Ir 0 . 0 0 Note 2. If P is the projection onto U along W then I − P is the projection onto W along U. 3 Idempotent operators: a theorem Terminology: An operator T such that T 2 = T is said to be idempotent. Theorem. Every idempotent operator is a projection operator. Proof. 4 A theorem about projections Theorem. Let P : V → V be the projection onto U along W . Let T : V → V be a linear transformation. Then PT = TP if and only if U and W are T -invariant (that is TU 6 U and TW 6 W ). Proof. 5 Direct sum decompositions, II Definition: V is said to be direct sum of subspaces U1, ..., Uk , and we write V = U1 ⊕ · · · ⊕ Uk , if for every v ∈ V there exist unique vectors ui ∈ Ui for 1 6 i 6 k such that v = u1 + ··· + uk . Note: U1 ⊕ U2 ⊕ · · · ⊕ Uk = (··· ((U1 ⊕ U2) ⊕ U3) ⊕ · · · ⊕ Uk). Note: If Ui V for 1 i k then V = U1 ⊕ · · · ⊕ Uk if and only 6 6 6 X if V = U1 + U2 + ··· + Uk and Ur ∩ Ui = {0} for 1 6 r 6 k. i6=r It is NOT sufficient that Ui ∩ Uj = {0} whenever i 6= j. Note: If V = U1 ⊕ U2 ⊕ · · · ⊕ Uk and Bi is a basis of Ui then k X B1∪B2∪· · ·∪Bk is a basis of V . In particular, dimV = dimUi . i=1 6 Partitions of the identity 2 Let P1,...,Pk be linear mappings V → V such that Pi = Pi for all i and PiPj = 0 whenever i 6= j. If P1 + ··· + Pk = I then {P1,...,Pk} is known as a partition of the identity on V . Example: If P is a projection then {P, I − P } is a partition of the identity. Theorem. Suppose that V = U1 ⊕· · ·⊕Uk. Let Pi be the projec- M tion of V onto Ui along Uj. Then {P1,...,Pk} is a partition j6=i of the identity on V . Conversely, if {P1,...,Pk} is a partition of the identity on V and Ui := ImPi then V = U1 ⊕ · · · ⊕ Uk . Proof. 7.
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