How does the implication true? Suppose $V$ is $n$ dimensional vector space. We know that $$v_j=\sum_{k=1}^{n}v_{kj}e_{k}.$$ Then, $$\alpha^{i}(v_j)=v_{ij}.$$ Can you please help me? What is the meaning of the symbol $\det[\alpha^{i}(v_j)]$? How it is defined?
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Each $\alpha^i$ is a linear function which eats a vector $v_j$ and returns a number $\alpha^i(v_j)$. You put these guys in a matrix $(\alpha^i(v_j))_{i,j=1}^n$ and compute its determinant. Your last equal sign uses that by definition, the determinant of a matrix $A = (a_{ij})_{i,j=1}^n$ is $$\det A = \sum_{\sigma \in S_n}{\rm sgn}(\sigma) a_{1\sigma(1)}\cdots a_{n\sigma(n)}.$$In tensor calculus, it is customary in some situations to write the row index $i$ upwards as $a^i_{\;j} = \alpha^i(v_j)$ and such, so this shouldn't cause any confusion.
You can think of it as being the components of a matrix, so for example: \begin{equation} \mathrm{det}(a^i(v_j)) = \mathrm{det}\begin{pmatrix} a^1(v_1) & \cdots & a^1(v_k)\\ \vdots & \ddots & \vdots\\ a^k(v_1) & \cdots & a^k(v_k) \end{pmatrix} \end{equation} This shows us that the vectors $(v_1,\dots v_k)$ are projected on the space spanned by $(a^1,\dots a^k)$ and we compute the corresponding volume, since in general, one of the sets is made up out of unit basis vectors. To see why the determinant shows up you might want to look up the formula for the determinant based on permutations.