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Chapter 11: Problem 34

Suppose \(T_{1}: U \rightarrow V\) and \(T_{2}: V \rightarrow W\) are linear. Prove that \(\left(T_{2} \circ T_{1}\right)^{t}=T_{1}^{t} \circ T_{2}^{t}\).

Short Answer

Expert verified
We define the transpose of a linear transformation and show the relationship between the transpose of the composition and the composition of the transposes. Using the property of the transpose, we proved the given identity: \(\left(T_{2} \circ T_{1}\right)^{t}=T_{1}^{t} \circ T_{2}^{t}\)

Step by step solution

01

Define the transpose of a linear transformation

The transpose of a linear transformation \(T: U \rightarrow V\) is a linear transformation \(T^t: V^* \rightarrow U^*\), where \(V^*\) and \(U^*\) are the dual spaces of \(V\) and \(U\), respectively. The transpose of a linear transformation \(T\), denoted as \(T^t\), has the property that, for any vector \(u \in U\) and any linear functional \(\phi \in V^*\), the following equation holds: - \((T^t\phi)(u) = \phi(Tu)\)
02

Show the relationship between the transpose of the composition and the composition of the transposes

Let \(u \in U\), \(\psi \in W^*\), and \(\phi = T_2^t(\psi)\). Then, we have: - \((T_2^t \circ T_1^t \psi)(u) = \psi((T_2 \circ T_1)u)\) To prove our identity, we will show that: - \((T_2^t \circ T_1^t \psi)(u) = (T_1^t \psi)(u)\) - \((T_1^t \circ T_2^t \psi)(u) = \psi(T_2(T_1u))\)
03

Prove the given identity

We have the equation: - \((T_2^t \circ T_1^t \psi)(u) = \psi((T_2 \circ T_1)u)\) Now, we can write the left side of the equation as: - \((T_2^t \circ T_1^t \psi)(u) = (T_2^t(T_1^t\psi))(u)\) Using the property of the transpose, we get: - \((T_2^t(T_1^t\psi))(u) = T_1^t\psi(Tu) = \psi(T_2(T_1 u))\) So, we have: - \((T_2^t \circ T_1^t \psi)(u) = \psi(T_2(T_1 u))\) Since \((T_2 \circ T_1)(u) = T_2(T_1 u)\), we get: - \((T_2^t \circ T_1^t \psi)(u) = \psi((T_2 \circ T_1) u)\) Thus, - \(\left(T_{2} \circ T_{1}\right)^{t}=T_{1}^{t} \circ T_{2}^{t}\) This completes the proof.

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Most popular questions from this chapter

Let \(V\) be the vector space of polynomials of degree \(\leq 2\). Let \(a, b, c \in K\) be distinct scalars. Let \(\phi_{a}, \phi_{b}, \phi_{c}\) be the linear functionals defined by \(\phi_{a}(f(t))=f(a), \phi_{b}(f(t))=f(b), \phi_{c}(f(t))=f(c) .\) Show that \(\left\\{\phi_{a}, \phi_{b}, \phi_{c}\right\\}\) is linearly independent, and find the basis \(\left\\{f_{1}(t), f_{2}(t), f_{3}(t)\right\\}\) of \(V\) that is its dual.

Suppose \(T: V \rightarrow U\) is linear and \(u \in U .\) Prove that \(u \in \operatorname{Im} T\) or there exists \(\phi \in V^{*}\) such that \(T^{t}(\phi)=0\) and \(\phi(u)=1\).

Let \(V\) be a vector space over \(\mathbf{R}\). Let \(\phi_{1}, \phi_{2} \in V^{*}\) and suppose \(\sigma: V \rightarrow \mathbf{R},\) defined by \(\sigma(v)=\phi_{1}(v) \phi_{2}(v)\) also belongs to \(V^{*}\). Show that either \(\phi_{1}=\mathbf{0}\) or \(\phi_{2}=\mathbf{0}\).

Find the basis \(\left\\{\phi_{1}, \phi_{2}, \phi_{3}\right\\}\) that is dual to the following basis of \(\mathbf{R}^{3}\) : \\[ \left\\{v_{1}=(1,-1,3), \quad v_{2}=(0,1,-1), \quad v_{3}=(0,3,-2)\right\\} \\]

Prove Theorem 11.3. Let \(\left\\{v_{i}\right\\}\) and \(\left\\{w_{i}\right\\}\) be bases of \(V\) and let \(\left\\{\phi_{i}\right\\}\) and \(\left\\{\sigma_{i}\right\\}\) be the respective dual bases in \(V^{*}\). Let \(P\) be the change-of-basis matrix from \(\left\\{v_{i}\right\\}\) to \(\left\\{w_{i}\right\\}\). Then \(\left(P^{-1}\right)^{T}\) is the change-of- basis matrix from \(\left\\{\phi_{i}\right\\}\) to \(\left\\{\sigma_{i}\right\\}\). Suppose, for \(i=1, \ldots, n\), $$ w_{i}=a_{i 1} v_{1}+a_{i 2} v_{2}+\cdots+a_{i n} v_{n} \quad \text { and } \quad \sigma_{i}=b_{i 1} \phi_{1}+b_{i 2} \phi_{2}+\cdots+a_{i n} v_{n} $$ Then \(P=\left[a_{i j}\right]\) and \(Q=\left[b_{i j}\right]\). We seek to prove that \(Q=\left(P^{-1}\right)^{T}\). Let \(R_{i}\) denote the ith row of \(Q\) and let \(C_{j}\) denote the \(j\) th column of \(P^{T}\). Then $$ R_{i}=\left(b_{i 1}, b_{i 2}, \ldots, b_{i n}\right) \quad \text { and } \quad C_{j}=\left(a_{j 1}, a_{j 2}, \ldots, a_{j h}\right)^{T} $$By definition of the dual basis, $$ \begin{aligned} \sigma_{i}\left(w_{j}\right) &=\left(b_{i 1} \phi_{1}+b_{i 2} \phi_{2}+\cdots+b_{i n} \phi_{n}\right)\left(a_{j 1} v_{1}+a_{j 2} v_{2}+\cdots+a_{j n} v_{n}\right) \\ &=b_{i 1} a_{j 1}+b_{i 2} a_{j 2}+\cdots+b_{i n} a_{j n}=R_{i} C_{j}=\delta_{i j} \end{aligned} $$ where \(\delta_{i j}\) is the Kronecker delta. Thus, $$ Q P^{T}=\left[R_{i} C_{j}\right]=\left[\delta_{i j}\right]=I $$ Therefore, \(Q=\left(P^{T}\right)^{-1}=\left(P^{-1}\right)^{T}\), as claimed.
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