Steven Dale Cutkosky's Class notes on linear algebra [Lecture notes] PDF

By Steven Dale Cutkosky

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Let Vr = Span{u1 , . . , ur } and Vi = Span{u1 , . . , ur , vr+1 , . . , vi } for r + 1 ≤ i ≤ n. Then Vi = Span{u1 , . . , ui } for r + 1 ≤ i ≤ n. 4, we have that ui ∈ (V i−1 )⊥ for r + 1 ≤ i ≤ n. Thus {u1 , . . , un } are an orthonormal set of vectors which form a basis of V . 6. Suppose that W is a subspace of V . Then V = W W ⊥. Proof. First construct, using Gram Schmidt, an orthonormal basis {w1 , . . , ws } of W . Now apply Gram Schmidt again to extend this to an orthonormal basis {w1 , .

T Since [(v)β β ] = [(v)β , A(w)β ] for all v, w ∈ V , we have that A = A . 1. Let y = Φ−1 (x) ∈ V . (L(y))β = A(y)β = Ax = λx = (λy)β . Thus L(y) = λy. Since y is nonzero, as x is nonzero, we have that y is an eigenvector of L. 3. Suppose that V is a finite dimensional real vector space with an inner product. Suppose that L : V → V is a symmetric linear map. Then V has an orthonormal basis consisting of eigenvectors of L. Proof. We prove the Theorem by induction on n = dim V . 2 there exists an eigenvector v ∈ V for L.

The above Lemma motivates the following definition. 7. L : V → V is an isometry if LL∗ = I, where I is the identity map of V. If V is a real inner product space, then an isometry is called an orthogonal transformation (or real unitary). If V is an Hermitian inner product space, then an isometry is called a unitary transformation. 8. Let {v1 , . . , vn } be a an orthonormal basis of V . Then L is an isometry if and only if {Lv1 , . . , Lvn } is an orthonormal basis of V . 9. If W is an L-invariant subspace of V and L = L∗ , then W ⊥ is an L-invariant subspace of V .

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Class notes on linear algebra [Lecture notes] by Steven Dale Cutkosky


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