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Chapter 2: Problem 12

Let \(V\) be a finite-dimensional vector space with the ordered basis \(\beta\). Prove that \(\psi(\beta)=\beta^{* *}\), where \(\psi\) is defined in Theorem \(2.26\).

Short Answer

Expert verified
The proof is carried out in four steps. Step 1: The linear transformation \(\psi: V \to V^{**}\), is given by \(\psi(v)(f) = f(v)\). Step 2: Given an ordered basis \(\beta = \{v_1,v_2,\dots,v_n\}\) of a finite-dimensional vector space \(V\), the dual basis \(\beta^* = \{f_1,f_2,\dots,f_n\}\) of the dual space \(V^*\) is defined with the property \(f_i(v_j) = \delta_{ij}\). Step 3: The double-dual basis \(\beta^{**} = \{f_1^*,f_2^*,\dots,f_n^*\}\) of the double-dual space \(V^{**}\) is defined with the property \(f_i^*(f_j) = \delta^{*}_{ij}\). Step 4: Upon applying the function \(\psi\) to each basis element \(\beta\), we obtain \(\psi(v_i) = f_i^{**}\) for all \(i\). It follows that \((\psi(v_i))(f_j) = f_i^{**}(f_j)\), which simplifies to \(\delta_{ij}\) as per the definition of the dual basis, verifying \(\psi(\beta) = \beta^{**}\).

Step by step solution

01

Recall Theorem 2.26 and the definition of \(\psi\)

In Theorem 2.26, we have a linear transformation \(\psi: V \to V^{**}\) defined as \[\psi: V \to V^{**},\quad \psi(v)(f) = f(v),\] where \(V^{**}\) denotes the double-dual space of the original vector space \(V\). The function \(\psi\) maps each vector \(v\) in V to a corresponding functional \(\psi(v)\) in \(V^{**}\). The goal is to show that for a basis \(\beta\) of V, the ordered basis \(\psi(\beta) = \beta^{**}\).
02

Find the dual 螔asis

For any ordered basis \(\beta = \{v_1,v_2,\dots,v_n\}\) of a finite-dimensional vector space \(V\), we can find the dual basis \(\beta^* = \{f_1,f_2,\dots,f_n\}\) of the dual space \(V^*\). The dual basis satisfies \(f_i(v_j) = \delta_{ij}\), where \(\delta_{ij}\) is the Kronecker delta: \[\delta_{ij} = \begin{cases} 1 & \text{if } i=j,\\ 0 & \text{if } i \neq j. \end{cases}\]
03

Find the double-dual 螔asis

Next, we need to find the double-dual basis \(\beta^{**} = \{f_1^*,f_2^*,\dots,f_n^*\}\) of the double-dual space \(V^{**}\). The double-dual basis can be constructed from the dual basis \(\beta^*\). Each double-dual basis element \(f_i^*\) satisfies the following property: \[f_i^*(f_j) = \delta^{*}_{ij},\] where the Kronecker delta \(\delta^{*}_{ij}\) is defined similarly to \(\delta_{ij}\).
04

Apply the function \(\psi\) to the ordered basis of \(V\)

Now we will apply the function \(\psi\) to each element of the ordered basis \(\beta\): \[\psi(v_i) = f_i^{**}, \quad i = 1,\dots,n.\] We will show that this application of \(\psi\) gives us the double-dual basis \(\beta^{**}\) of \(V^{**}\). To prove that, we need to show that for all \(i, j = 1,\dots,n\), we have: \[(\psi(v_i))(f_j) = f_i^{**}(f_j).\] By the definition of \(\psi\) (from Step 1), we have \[(\psi(v_i))(f_j) = f_j(v_i).\] According to the property of the dual basis (from Step 2), this expression simplifies to: \[f_j(v_i) = \delta_{ij}.\] Finally, recall that the double-dual basis \(f_i^*\) from Step 3 satisfies \[f_i^*(f_j) = \delta^{*}_{ij}.\] Since \(\delta^{*}_{ij} = \delta_{ij}\), we have proved that the application of \(\psi\) to the ordered basis \(\beta\) gives us the double-dual basis \(\beta^{**}\): \[\psi(\beta) = \beta^{**}.\]

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Most popular questions from this chapter

Assume the notation in Theorem \(2.13 .\) (a) Suppose that \(z\) is a (column) vector in \(\mathrm{F}^{p}\). Use Theorem 2.13(b) to prove that \(B z\) is a linear combination of the columns of \(B\). In particular, if \(z=\left(a_{1}, a_{2}, \ldots, a_{p}\right)^{t}\), then show that $$ B z=\sum_{j=1}^{p} a_{j} v_{j} . $$ (b) Extend (a) to prove that column \(j\) of \(A B\) is a linear combination of the columns of \(A\) with the coefficients in the linear combination being the entries of column \(j\) of \(B\). (c) For any row vector \(w \in \mathrm{F}^{m}\), prove that \(w A\) is a linear combination of the rows of \(A\) with the coefficients in the linear combination being the coordinates of \(w\). Hint: Use properties of the transpose operation applied to (a). (d) Prove the analogous result to (b) about rows: Row \(i\) of \(A B\) is a linear combination of the rows of \(B\) with the coefficients in the linear combination being the entries of row \(i\) of \(A\).

Prove Theorem 2.1 (i) and \((\mathrm{v})\) : (i) \((A+B)+C=A+(B+C),(\mathrm{v}) k(A+B)=k A+k B\) Suppose \(A=\left[a_{i j}\right], B=\left[b_{i j}\right], C=\left[c_{i j}\right] .\) The proof reduces to showing that corresponding \(i j\) -entries in each side of each matrix equation are equal. [We prove only \((\mathrm{i}) \text { and }(\mathrm{v}),\) because the other parts of Theorem 2.1 are proved similarly.]

Find the diagonal and trace of each matrix: (a) \(A=\left[\begin{array}{rrr}1 & 3 & 6 \\ 2 & -5 & 8 \\ 4 & -2 & 9\end{array}\right]\) (b) \(B=\left[\begin{array}{rrr}2 & 4 & 8 \\ 3 & -7 & 9 \\ -5 & 0 & 2\end{array}\right]\) (c) \(\quad C=\left[\begin{array}{rrr}1 & 2 & -3 \\ 4 & -5 & 6\end{array}\right]\).

Let \(V\) be a finite-dimensional vector space, and let \(T: V \rightarrow V\) be linear. (a) If \(\operatorname{rank}(T)=\operatorname{rank}\left(T^{2}\right)\), prove that \(R(T) \cap N(T)=\\{0\\}\). Deduce that $\mathrm{V}=\mathrm{R}(\mathrm{T}) \oplus \mathrm{N}(\mathrm{T})$ (see the exercises of Section 1.3). (b) Prove that $\mathrm{V}=\mathrm{R}\left(\mathrm{T}^{k}\right) \oplus \mathrm{N}\left(\mathrm{T}^{k}\right)\( for some positive integer \)k$.

Find a \(3 \times 3\) orthogonal matrix \(P\) whose first two rows are multiples of (a) (1,2,3) and (0,-2,3) (b) \(\quad(1,3,1)\) and (1,0,-1)
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