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Chapter 13: Problem 7

Suppose \(T\) is invertible. Show that \(\left(T^{-1}\right)^{*}=\left(T^{*}\right)^{-1}\)

Short Answer

Expert verified
In order to show that \(\left(T^{-1}\right)^{*}=\left(T^{*}\right)^{-1}\), we need to prove that the compositions \(\left(T^{-1}\right)^{*} \circ T^{*}\) and \(T^{*} \circ \left(T^{-1}\right)^{*}\) are both identity maps. By evaluating the inner products and showing that \(\left(T^{-1}\right)^{*}(T^{*}(\mathbf{v})) = \mathbf{v}\) and \(T^{*}\left(\left(T^{-1}\right)^{*}(\mathbf{u})\right) = \mathbf{u}\) for all \(\mathbf{v}\in V\) and \(\mathbf{u}\in W\), we prove that \(\left(T^{-1}\right)^{*}=\left(T^{*}\right)^{-1}\).

Step by step solution

01

Prove \(\left(T^{-1}\right)^{*} \circ T^{*}\) is the identity map

To prove that \(\left(T^{-1}\right)^{*} \circ T^{*}\) is the identity map, we can show that their composition is equal to the identity map, which means applying the composition on any vector must return the same vector. Given any vector \(\mathbf{v}\in V\), we have: \[\left(\left(T^{-1}\right)^{*} \circ T^{*}\right)(\mathbf{v}) = \left(T^{-1}\right)^{*}(T^{*}(\mathbf{v}))\] Now we need to find \(\left(T^{-1}\right)^{*}(T^{*}(\mathbf{v}))\) and show that it equals \(\mathbf{v}\).
02

Rewrite the inner product

We start by rewriting the inner product \(\langle T^{*}(\mathbf{v}),\mathbf{w}\rangle\) in terms of \(T^{-1}\), where \(\mathbf{w}\in W\): \[\langle T^{*}(\mathbf{v}),\mathbf{w}\rangle = \langle \mathbf{v}, T(\mathbf{w})\rangle\] We can now multiply both sides by \(T^{-1}\): \[\langle T^{-1}(T^{*}(\mathbf{v})),\mathbf{w}\rangle = \langle \mathbf{v}, T(T^{-1}(\mathbf{w}))\rangle\] But we know that \(T(T^{-1}(\mathbf{w})) = \mathbf{w}\), so we have: \[\langle T^{-1}(T^{*}(\mathbf{v})),\mathbf{w}\rangle = \langle \mathbf{v},\mathbf{w}\rangle\] This implies that \(T^{-1}(T^{*}(\mathbf{v}))\) is the vector that satisfies the above equality for all \(\mathbf{w}\). Therefore, we conclude that: \[\left(T^{-1}\right)^{*}(T^{*}(\mathbf{v})) = \mathbf{v}\] Thus, \(\left(T^{-1}\right)^{*} \circ T^{*}\) is the identity map.
03

Prove \(T^{*} \circ \left(T^{-1}\right)^{*}\) is the identity map

Similarly, to prove that \(T^{*} \circ \left(T^{-1}\right)^{*}\) is the identity map, we can show that their composition is equal to the identity map, which means applying the composition on any vector must return the same vector. Given any vector \(\mathbf{u}\in W\), we have: \[\left(T^{*} \circ \left(T^{-1}\right)^{*}\right)(\mathbf{u}) = T^{*}\left(\left(T^{-1}\right)^{*}(\mathbf{u})\right)\] Now we need to find \(T^{*}\left(\left(T^{-1}\right)^{*}(\mathbf{u})\right)\) and show that it equals \(\mathbf{u}\).
04

Rewrite the inner product again

We rewrite the inner product \(\langle \left(T^{-1}\right)^{*}(\mathbf{u}),\mathbf{v}\rangle\) in terms of \(T^{*}\), where \(\mathbf{v}\in V\): \[\langle \left(T^{-1}\right)^{*}(\mathbf{u}),\mathbf{v}\rangle = \langle \mathbf{u}, T^{*}(\mathbf{v})\rangle\] We can now multiply both sides by \(T\): \[\langle T\left(\left(T^{-1}\right)^{*}(\mathbf{u})\right),\mathbf{v}\rangle = \langle \mathbf{u}, T(T^{*}(\mathbf{v}))\rangle\] But we know that \(T(T^{*}(\mathbf{v})) = \mathbf{v}\), so we have: \[\langle T\left(\left(T^{-1}\right)^{*}(\mathbf{u})\right),\mathbf{v}\rangle = \langle \mathbf{u},\mathbf{v}\rangle\] This implies that \(T\left(\left(T^{-1}\right)^{*}(\mathbf{u})\right)\) is the vector that satisfies the above equality for all \(\mathbf{v}\). Therefore, we conclude that: \[T^{*}\left(\left(T^{-1}\right)^{*}(\mathbf{u})\right) = \mathbf{u}\] Thus, \(T^{*} \circ \left(T^{-1}\right)^{*}\) is the identity map. Since both \(\left(T^{-1}\right)^{*} \circ T^{*}\) and \(T^{*} \circ \left(T^{-1}\right)^{*}\) are identity maps, we can conclude that \(\left(T^{-1}\right)^{*}=\left(T^{*}\right)^{-1}\). This completes the proof.

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Key Concepts

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

Adjoint Operator
The concept of the adjoint operator is crucial in understanding how linear transformations behave with respect to inner products. When you have a linear transformation like \(T\), the adjoint of \(T\), denoted as \(T^{*}\), is another operator that closely relates to \(T\) through the inner products involved. Put simply, for any vectors \(\mathbf{v}\) and \(\mathbf{w}\), the equation \(\langle T(\mathbf{v}), \mathbf{w} \rangle = \langle \mathbf{v}, T^{*}(\mathbf{w}) \rangle\) must hold. The adjoint operator helps in simplifying calculations and proofs, especially in complex vector spaces.
Matrix Inverse
The matrix inverse, represented as \(T^{-1}\) for an operator \(T\), is a fundamental concept representing the reversal of a transformation. When applied to invertible linear transformations, the inverse ensures that applying \(T\) followed by \(T^{-1}\) (and vice-versa) results in the identity action鈥攅ffectively doing nothing to the vector. This property, \(T \circ T^{-1} = I\) where \(I\) is the identity map, is essential in solving systems of linear equations and ensuring transformations can be reversed.
Inner Product
An inner product is a mathematical tool used to generalize the concept of angle and length in vector spaces. For vectors \(\mathbf{v}\) and \(\mathbf{w}\), the inner product, denoted as \(\langle \mathbf{v}, \mathbf{w} \rangle\), is a scalar that provides a measure of their interaction. Inner products have properties like being commutative (\(\langle \mathbf{v}, \mathbf{w} \rangle = \overline{\langle \mathbf{w}, \mathbf{v} \rangle}\)) and positive-definite (\(\langle \mathbf{v}, \mathbf{v} \rangle \geq 0\)). These properties are useful in various applications, including projecting vectors and defining orthogonal vectors.
Identity Map
The identity map is a simple yet essential concept in linear algebra. It's the transformation that leaves every element unchanged, defined as \(I(\mathbf{v}) = \mathbf{v}\) for any vector \(\mathbf{v}\). In proofs, the identity map is often used to demonstrate that a certain composition of maps reverses itself鈥攔eturning to the starting point. Verifying that \(T^{-1} \circ T = I\) and \(T \circ T^{-1} = I\) confirms an operator is invertible and highlights the importance of returning a vector to its original form.

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Most popular questions from this chapter

Find the adjoint of: (a) \(A=\left[\begin{array}{ll}5-2 i & 3+7 i \\ 4-6 i & 8+3 i\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{rr}3 & 5 i \\ i & -2 i\end{array}\right]\) (c) \(C=\left[\begin{array}{ll}1 & 1 \\ 2 & 3\end{array}\right]\)

Find the adjoint of \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) defined by $$F(x, y, z)=(3 x+4 y-5 z, 2 x-6 y+7 z, 5 x-9 y+z)$$

Determine which of the following matrices is normal: (a) \(A=\left[\begin{array}{ll}1 & i \\ 0 & 1\end{array}\right]\) (b) \(\quad B=\left[\begin{array}{cc}1 & i \\ 1 & 2+i\end{array}\right]\)

Show that there exists an orthonormal basis \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) of \(V\) consisting of eigenvectors of \(T\) if and only if there exist orthogonal projections \(E_{1}, \ldots, E_{r}\) and scalars \(\lambda_{1}, \ldots, \lambda_{r}\) such that (i) \(T=\lambda_{1} E_{1}+\cdots+\lambda_{r} E_{r}\) (ii) \(\quad E_{1}+\cdots+E_{r}=I\) (iii) \(\quad E_{i} E_{j}=0 \quad\) for \(\quad i \neq j\)

Let \(V\) be the vector space of polynomials over \(\mathbf{R}\) with inner product defined by $$\langle f, g\rangle=\int_{0}^{1} f(t) g(t) d t$$ Give an example of a linear functional \(\phi\) on \(V\) for which Theorem 13.3 does not hold-that is, for which there is no polynomial \(h(t)\) such that \(\phi(f)=\langle f, h\rangle\) for every \(f \in V\).
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