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Q :Let $\left\lbrace v_i \right\rbrace^n_{i=1}$ and $\left\lbrace w_i \right\rbrace^n_{i=1}$ be basis of V and also let $\left\lbrace \phi_i \right\rbrace^n_{i=1}$ and $\left\lbrace \sigma_i \right\rbrace^n_{i=1}$ be basis of $V^{\ast}$ and dual basis to $\left\lbrace v_i \right\rbrace^n_{i=1}$ and $\left\lbrace w_i \right\rbrace^n_{i=1}$ respectively. Suppose, P is the change of basis matrix from $\left\lbrace v_i \right\rbrace^n_{i=1}$ to $\left\lbrace w_i \right\rbrace^n_{i=1}$ , then show that $$(P^{-1})^T\text{ is the change of basis matrix from }\left\lbrace \phi_i \right\rbrace^n_{i=1} \text{to } \left\lbrace \sigma_i \right\rbrace^n_{i=1}$$

My approach :By hypothesis $\exists P,\text{ a transition matrix from }\left\lbrace v_i \right\rbrace^n_{i=1}$ to $\left\lbrace w_i \right\rbrace^n_{i=1}$. That is $w_1=a_{11}v_1+a_{12}v_2+\cdots+a_{1n}v_n$
$w_2=a_{21}v_1+a_{22}v_2+\cdots+a_{2n}v_n$
$\vdots$
$w_n=a_{n1}v_1+a_{n2}v_2+\cdots+a_{nn}v_n$ $$P=\begin{pmatrix}a_{11} & a_{21}&\cdots&a_{n1}\\\ \vdots & \vdots&&\vdots\\\ a_{n1}& a_{2n}&\cdots& a_{nn} \end{pmatrix}\\$$ Again similar way, $\exists Q,\text{ a transition matrix from }\left\lbrace \phi_i \right\rbrace^n_{i=1}$ to $\left\lbrace \sigma_i \right\rbrace^n_{i=1}$. That is
$\sigma_1=b_{11}\phi_1+b_{12}\phi_2+\cdots+b_{1n}\phi_n$
$\sigma_2=b_{21}\phi_1+b_{22}\phi_2+\cdots+b_{2n}\phi_n$
$\vdots$
$\sigma_n=b_{n1}\phi_1+b_{n2}\phi_2+\cdots+b_{nn}\phi_n$ $$Q=\begin{pmatrix}b_{11} & b_{21}&\cdots&b_{n1}\\\ \vdots & \vdots&&\vdots\\\ b_{n1}& b_{2n}&\cdots& b_{nn} \end{pmatrix}\\$$Now by definition of dual basis we know,$\sigma_i(w_j)=\delta_{ij} = \begin{cases} 1, i=j \\ 0, i\neq j \end{cases}$. Now \begin{align*} \sigma_i(w_j)&=(b_{i1}\phi_1+\cdots+b_{in}\phi_n)(a_{j1}v_1+\cdots+a_{jn}v_n)\\ &=b_{i1}\phi_1(a_{j1}v_1+\cdots+a_{jn}v_n)+\cdots+b_{in}\phi_n(a_{j1}v_1+\cdots+a_{jn}v_n) \\ &=b_{i1}a_{j1}+\cdots+b_{in}a_{jn} \end{align*} But here I stuck. I am not found anything to link up with P to Q. Besides, please give me some mathematical notation which help me to reduce my solution size if any. Any hint or solution will be appreciated .
Thanks in advance .

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I'll use your notation for the basis vectors, but for the coefficients, I'll write them as $P_{ij}$ and $Q_{ij}$ rather than $a_{ij}$ and $b_{ij}$. The key observation which allows you to link the two matrices is that for any $f \in V^*$, we can expand it as \begin{equation} f = \sum_{i=1}^n f(v_i) \cdot \phi_i \end{equation} (i.e the $b_{ij}$'s which you wrote can actually be written in terms of $\sigma_j$). To see why this is true, temporarily call the RHS $T$. Then, for every $j$, (by definition of dual basis) it is immediate that $T(v_j) = f(v_j)$. So we have shown that $f$ and $T$ agree on the basis $\{v_1, \dots, v_n\}$; hence they are equal everywhere. Another result we need is that $P^{-1}$ is the transition matrix from $\{w_1, \dots , w_n\}$ to $\{v_1, \dots, v_n\}$. Now, we have a few lines of computation, and the only thing to be careful of is the indices.

For every $1 \leq j \leq n$, we have that \begin{align} \sigma_j &= \sum_{i=1}^n \sigma_j(v_i) \cdot \phi_i \\ &= \sum_{i=1}^n \sigma_j \left(\sum_{k=1}^n \left( P^{-1}\right)_{ik}\, w_k\right) \cdot \phi_i \\ &= \sum_{i=1}^n \left( P^{-1}\right)_{ij} \cdot \phi_i \end{align} The last equality is because the $\sigma$'s are dual to the $w$'s. On the other hand, since we said $Q$ is the transition matrix from the $\phi$'s to $\sigma$'s, this means, for every $1 \leq j \leq n$, we have \begin{equation} \sigma_j = \sum_{i=1}^n Q_{ji} \cdot \phi_i. \end{equation}

Notice how the indices of $i$ and $j$ in the two summations are reversed. Thus, it follows that $Q = \left( P^{-1}\right)^t$.

(You may be interested to know that this question is in fact a special case of a much more general theorem regarding the matrix representation of a linear map with respect to given bases and the matrix representation of the dual map with respect to the corresponding dual bases. If you want to learn more about this, I highly recommend reading Linear Algebra, by Friedberg, Insel and Spence; in particular Section 2.6 on dual spaces, and Theorem 2.25 in that section.)

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  • $\begingroup$ Thanks for a nice explanation and I will definitely look your suggested book. Thanks again @peek-a-boo $\endgroup$ – emonhossain May 22 at 12:05

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