/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Let \(\mathrm{V}\) be a vector s... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 18

Let \(\mathrm{V}\) be a vector space over \(F\), where \(F=R\) or \(F=C\), and let \(\mathrm{W}\) be an inner product space over \(F\) with inner product \(\langle\cdot, \cdot\rangle .\) If \(\mathrm{T}: \mathrm{V} \rightarrow \mathrm{W}\) is linear, prove that \(\langle x, y\rangle^{\prime}=\langle\mathrm{T}(x), \mathrm{T}(y)\rangle\) defines an inner product on \(\mathrm{V}\) if and only if \(\mathrm{T}\) is one-to-one.

Short Answer

Expert verified
In conclusion, we have proven that the function \(\langle.,.\rangle' = \langle T(x), T(y) \rangle\) defines an inner product on V if and only if T is one-to-one by showing that the properties of conjugate symmetry, linearity in the first argument, and positive definiteness are satisfied if and only if T is one-to-one.

Step by step solution

01

Conjugate Symmetry

First, let's show that <.,.>' has conjugate symmetry if T is one-to-one. Given x, y 鈭� V, consider the following: \(' = = ^* = '\) Since W is an inner product space, = * and conjugate symmetry holds for <.,.>' if T is one-to-one. Now, assume that conjugate symmetry holds for <.,.>'. This doesn't affect the one-to-one property of T because manipulations are done in the inner product space W.
02

Linearity in the First Argument

Next, let's show that <.,.>' has linearity in the first argument if T is one-to-one. Given x鈧�, x鈧�, y 鈭� V and scalars a, b 鈭� F, consider the following: \(' = = \) Since T is linear, = a * + b * = a * ' + b * ', and linearity in the first argument holds for <.,.>' if T is one-to-one. Now, assume linearity in the first argument holds for <.,.>'. This doesn't affect the one-to-one property of T because manipulations are done in the inner product space W.
03

Positive Definite

Finally, let's show that <.,.>' is positive definite if T is one-to-one. If x 鈭� V, consider the following: \(' = 鈮� 0\) (since W is an inner product space) Now we need to show that ' = 0 if and only if x = 0. If T is one-to-one, then T(x) = 0 if and only if x = 0. So, ' = = 0 if and only if x = 0, and <.,.>' is positive definite if T is one-to-one. Now, assume that <.,.>' is positive definite. If ' = 0, then = 0. Since W is an inner product space, T(x) = 0 if and only if x = 0, and T is one-to-one.
04

Conclusion

We have shown that the function <.,.>'= defines an inner product on V if and only if T is one-to-one, satisfying the properties of conjugate symmetry, linearity in the first argument, and positive definiteness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conjugate Symmetry
In an inner product space, the property of conjugate symmetry is quite fundamental. It's about how swapping the order of the vectors in the inner product results in a conjugate. For the inner product \('\), defined as \(\), conjugate symmetry is expressed using the formula: \(' = '^*\). Here, \(^*\) denotes the complex conjugate.

For the space \(W\), the given property is already ensured by \(\langle T(x), T(y) \rangle = \langle T(y), T(x) \rangle^*\). When \(T\) is one-to-one, this is directly translated into the space \(V\) because each \(x\) and \(y\) are uniquely paired by the transformation \(T\). This unique pairing ensures that we don't lose the conjugate symmetry property when mapping these vectors. This consideration allows us to conclude that conjugate symmetry for \('\) holds true when \(T\) is bijective over its range.

In essence, the transformation \(T\) has to maintain this symmetry from \(W\) to \(V\) as long as it's one-to-one, associating each vector to a unique result in the inner product space.
Linearity
Linearity in the context of inner products refers to the property where the inner product behaves nicely with scalar multiplication and vector addition. Specifically, if \(f(x)\) and \(f(y)\) are linear transformations of vectors \(x\) and \(y\), their linear combination should also yield a consistent transformation.

For the expression \(' = \), linearity dictates that this should transform to a linear combination like \(a * ' + b * '\) when \(a, b\) are scalars. This happens because \(T\) is a linear map and thereby distributes over addition of inputs and scalar multiplication, transforming linearly. The rule \( = a \cdot + b \cdot \) precisely reflects this behavior.

So, when \(T\) is one-to-one, linearity within \(<.,.>'\) is preserved because the linear properties of the original vectors are maintained after transformation and mapping back, ensuring coherent and consistent vector transformation in \(V\) as in \(W\).
Positive Definiteness
Positive definiteness is a crucial property of inner product spaces, ensuring that the inner product of any vector with itself is non-negative and is zero only when the vector itself is the zero vector. This ensures that the space acts "positively" regarding length and angle.

For \(' = \), if the transformation \(T\) is one-to-one, then \( \geq 0\) because \(W\) is intrinsically an inner product space. The implication here is that \('\) achieves the value of zero exclusively when \(x = 0\).

This is because \(T(x) = 0\) implies \(x = 0\) in a one-to-one mapping, meaning no other value of \(x\) can map to zero except for the zero vector. This tightly preserves the positive definiteness of \(<.,.>'\) in space \(V\), underpinning a unique relationship between the space \(V\) and the transformed space \(W\). This uniqueness ensures distance and angle preservation when \(T\) is an injective linear transformation.

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Most popular questions from this chapter

Suppose that \(A\) and \(B\) are diagonalizable matrices. Prove or disprove that \(A\) is similar to \(B\) if and only if \(A\) and \(B\) are unitarily equivalent.

Let \(V=C([-1,1])\) with the inner product \(\langle f, g\rangle=\int_{-1}^{1} f(t) g(t) d t\), and let \(\mathrm{W}\) be the subspace \(\mathrm{P}_{2}(R)\), viewed as a space of functions. Use the orthonormal basis obtained in Example 5 to compute the "best" (closest) second-degree polynomial approximation of the function \(h(t)=\) \(e^{t}\) on the interval \([-1,1]\).

Let \(V\) and \(W\) be vector spaces over the same field, and let \(T: V \rightarrow W\) be a linear transformation. For any \(H \in \mathcal{B}(\mathrm{W})\), define \(\widehat{\mathrm{T}}(H): \mathrm{V} \times \mathrm{V} \rightarrow F\) by \(\widehat{\mathbf{T}}(H)(x, y)=H(\mathbf{T}(x), \mathbf{T}(y))\) for all \(x, y \in \mathbf{V}\). Prove the following results. (a) If \(H \in \mathcal{B}(\mathrm{W})\), then \(\widehat{\mathrm{T}}(H) \in \mathcal{B}(\mathrm{V})\). (b) \(\widehat{\mathrm{T}}: \mathcal{B}(\mathrm{W}) \rightarrow \mathcal{B}(\mathrm{V})\) is a linear transformation. (c) If \(\mathrm{T}\) is an isomorphism, then so is \(\widehat{\mathrm{T}}\).

Suppose that \(\beta\) and \(\gamma\) are ordered bases for an \(n\)-dimensional real [complex] inner product space V. Prove that if \(Q\) is an orthogonal [unitary] \(n \times n\) matrix that changes \(\gamma\)-coordinates into \(\beta\)-coordinates, then \(\beta\) is orthonormal if and only if \(\gamma\) is orthonormal.

Prove that if \(\left\\{w_{1}, w_{2}, \ldots, w_{n}\right\\}\) is an orthogonal set of nonzero vectors, then the vectors \(v_{1}, v_{2}, \ldots, v_{n}\) derived from the Gram-Schmidt process satisfy \(v_{i}=w_{i}\) for \(i=1,2, \ldots, n\). Hint: Use mathematical induction.
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