(Change of basis) Let $e_1^{\ast }\dots .e_n^{\ast }$ be an orthonormal basis of $V^n$ , so that (1.114) and (1.115) hold for an arbitrary vector $v$ . In particular, we can write $e_j=a_{1j}{e}_1^{\ast }+\cdots +a_{nj}{e}_n^{\ast }\text{. }$ $a_{i j}=\mathbf{e}_{j} \cdot \mathbf{e}_{i}^{*}, \quad i=1, \ldots, n . j=1, \ldots, n .$a)Showthat,with$A = \left(a_{ij}\right)$.$col\left(v_{1}^{ \ast },…,v_{n}^{ \ast }\right) = Acol\left(v_{1}…,v_{n}\right)\text{. }$ThusthematrixAprovidesthelinkbetweenthecomponentsof$v$ with respect to the two orthonormal bases. [Hint: Write $\mathbf{v}=v_{1} \mathbf{e}_{1}+\cdots+v_{n} \mathbf{e}_{n}=v_{1}^{*} \mathbf{e}_{1}^{*}+\cdots+v_{n}^{*} \mathbf{e}_{n}^{*}$ and dot both sides with $e_i^{\ast }$ for $i=1,\dots ,n$ .] b) Show that the $j$ th column of $A$ gives the components of $e_j$ with respect to the basis $e_1^{\ast },\dots ,e_n^{\ast }$ and the $i$ th row of $A$ gives the components of $e_i^{\ast }$ with respect to the basis $e_1,\dots ,e_n$ . c) Show that $A$ is an orthogonal matrix (Section 1.13).

Step 1: Understand the given information We are given that $e_1^{\ast },\dots ,e_n^{\ast }$ and $e_1,\dots ,e_n$ are orthonormal basis of a vector space $V^n$. We have also given the relationship between these two bases: $e_j=a_{1j}{e}_1^{\ast }+\cdots +a_{nj}{e}_n^{\ast }$ and $a_{ij}=e_j\cdot e_i^{\ast },i=1,\dots ,n.j=1,\dots ,n.$ Step 2: Express the components of a vector with respect to different bases Using the hint, we write $v$ as: $v=v_1{e}_1+\cdots +v_n{e}_n=v_1^{\ast }{e}_1^{\ast }+\cdots +v_n^{\ast }{e}_n^{\ast }.$ To find the relationship between the components $v_i$ and $v_i^{\ast }$, we dot both sides of the equation with $e_i^{\ast }$: $v\cdot e_i^{\ast }=\sum _{k=1}^n{v}_k(e_k\cdot e_i^{\ast })=\sum _{k=1}^n{v}_k{a}_{ik}=v_1^{\ast }{e}_1^{\ast }\cdot e_i^{\ast }+\cdots +v_n^{\ast }{e}_n^{\ast }\cdot e_i^{\ast }=v_i^{\ast }.$ Now, we can write this relationship in matrix form: $\left(\begin{array}{c}{v}_1^{\ast }\\ ⋮\\ v_n^{\ast }\end{array}\right)=A\left(\begin{array}{c}{v}_1\\ ⋮\\ v_n\end{array}\right).$ So, the relationship between the components of $v$ with respect to the two different bases can be expressed as: $col\left(v_1^{\ast },\dots ,v_n^{\ast }\right)=Acol\left(v_1,\dots ,v_n\right).$ Step 3: Interpret the columns and rows of A The $j$th column of A is given by $\left(\begin{array}{c}{a}_{1j}\\ ⋮\\ a_{nj}\end{array}\right)$. We know that $e_j=a_{1j}{e}_1^{\ast }+\cdots +a_{nj}{e}_n^{\ast }.$ So, the $j$th column of A represents the components of $e_j$ with respect to the basis $e_1^{\ast },\dots ,e_n^{\ast }$. Similarly, the $i$th row of A represents the components of $e_i^{\ast }$ with respect to the basis $e_1,\dots ,e_n$. Step 4: Prove that A is an orthogonal matrix To prove A is an orthogonal matrix, we need to show that $A^{T}A=I$. Let's consider the elements of the product $A^{T}A$. We have: $(A^{T}A{)}_{ij}=\sum _{k=1}^n(A^T{)}_{ik}{A}_{kj}=\sum _{k=1}^n{a}_{ki}{a}_{kj}.$ Using the definition of the components of A, we have: $(A^{T}A{)}_{ij}=\sum _{k=1}^n(e_k\cdot e_i^{\ast })(e_k\cdot e_j^{\ast }).$ Now, using the properties of the dot product, we can rewrite this expression as $(A^{T}A{)}_{ij}=\left(\sum _{k=1}^n{e}_k\cdot e_i^{\ast }\right)\left(\sum _{k=1}^n{e}_k\cdot e_j^{\ast }\right)=(e_i^{\ast }\cdot e_i^{\ast })(e_j^{\ast }\cdot e_j^{\ast })={\delta }_{ij}.$ where ${\delta }_{ij}$ is the Kronecker delta function, which is equal to 1 when $i=j$ and 0 otherwise. Therefore, $A^{T}A=I$ and A is an orthogonal matrix.