Guided Proof Let W be a subspace of the innerproduct space V. Prove that the setW = {v

QUESTION:

Guided Proof Let W be a subspace of the inner product space V. Prove that the set \(W^{\perp}=\{\mathbf{v} \in V:\langle\mathbf{v}, \mathbf{w}\rangle=0\) for all \(\mathbf{w} \in W\}\) is a subspace of V.

Getting Started: To prove that \(W^{\perp}\) is a subspace of V, you must show that \(W^{\perp}\) is nonempty and that the closure conditions for a subspace hold (Theorem 4.5).

(i) Find a vector in \(W^{\perp}\) to conclude that it is nonempty.

(ii) To show the closure of \(W^{\perp}\) under addition, you need to show that \(\left\langle\mathbf{v}_{1}+\mathbf{v}_{2}, \mathbf{w}\right\rangle=0\) for all \(\mathbf{w} \in W\) and for any \(\mathbf{v}_{1}, \mathbf{v}_{2} \in W^{\perp}\). Use the properties of inner products and the fact that \(\left\langle\mathbf{v}_{1}, \mathbf{w}\right\rangle\) and \(\left\langle\mathbf{v}_{2}, \mathbf{w}\right\rangle\) are both zero to show this.

(iii) To show closure under multiplication by a scalar, proceed as in part (ii). Use the properties of inner products and the condition of belonging to \(W^{\perp}\).

Text Transcription:

W^{perp} = v in V: langle v, w rangle = 0

w in W

W^{perp}

langle v_1 + v_2, w rangle = 0

v_1, v_2 in W^{perp}

langle v_1, w rangle

langle v_2, w rangle