I was reading some facts about the determinant and refreshed my memory with the fact that the determinant of the $ n\times n $ matrix can be defined as $ \det(A)=\sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^{n} a_{i, \sigma_i} $.

Now,I would like to know this:

How to prove that for every real number $\alpha$ there exist sequence of matrices $ A_1(\alpha),A_2(\alpha),...,A_n(\alpha),... $ ($ A_i $ is $ i \times i $ matrix) such that we have $ \det(A_i(\alpha))=\alpha $ for every $ i \in \mathbb N $?

In other words I would like to know how it can be proven from the above stated definition of determinant that every real number is the value of the determinant of matrix of any number rows and columns.

I am aware that this can easily be proven by using other properties of the determinant but would like to know is there an easy way to prove this by using only the above stated definition.

  • 3
    $\begingroup$ You can take the diagonal matrix with entries $1, 1, \dots, 1, 1, \alpha$. $\endgroup$ – David Feb 1 '16 at 18:03

Consider diagonal matrices with entries $\alpha$, $1,\ldots,1$ in the main diagonal. Then the only term in your sum is the one coming from the identity permutation.

  • $\begingroup$ Is the determinant of diagonal matrix product of the elements on the diagonal? $\endgroup$ – Farewell Feb 1 '16 at 18:04
  • $\begingroup$ Yes, it is the product. $\endgroup$ – John B Feb 1 '16 at 18:14

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