How to find $|\{A \in \mathbb F _7^{5 \times 5}|A \text{ is invertible}\}|$?

$$ \begin{align} GL_n(K) & = \{ A \in K^{n \times n }| A \text{ is invertible}\} \\ & = \{ A \in K^{n \times n }| \phi_A \text{ is an isomorphism}\} \\ & = \{ A \in K^{n \times n }| rk_{\phi A} = \text{dim }K^{n \times n}\} \\ & = \{ A \in K^{n \times n }| rk A = n\} \end{align} $$

And also

$Col(A) = Im \phi A$

$rk_KA = dim_kCol(A)$

Question: How can I zip this information to get the cardinality of all possible invertible matrices in $\mathbb F _7^{5 \times 5}$?

This post seems equivalent but somehow it does not help me. A technique without determinants would be the best for me at the moment. I appreciate also solutions or help using determinants.

  • 3
    $\begingroup$ Take a look at the answers here or here. At least some of the answers are generic. I would close this as a duplicate, but need to ask more experienced moderators about what that will do to the bounty. $\endgroup$ – Jyrki Lahtonen Jun 11 '17 at 20:59

The columns must form a basis of $\Bbb F_7^5$. For the $k$th column, you can pick any of $7^5$ vectors, except those $7^{k-1}$ that are in the $(k-1)$-dimensional subspace generated by the preceding $k-1$ columns. So the count is $$(7^5-7^0)(7^5-7^1)(7^5-7^2)(7^5-7^3)(7^5-7^4) $$

  • $\begingroup$ Unfortunately I do not fully understand you explanation. Why do you subtract $7^0, 7^1, 7^2...$? Yes, you mention it has something to do with the $(k-1)$-dimensional subpace.. $\endgroup$ – jublikon Jun 7 '17 at 7:15
  • 2
    $\begingroup$ He counts the numer of matrices as follows. First he observe that any martix is given by 5 vectors (of lenth 5). A matrix is invertible if and only if these vectors are linearily independent. Now let's count for the first vector we have no restrictions except from not being 0, that is $7^5-1$. The second vector can't be in the span of the first vector, this span contains 7 elements so we have $7^5-7$. Do this 3 more times and you're done. $\endgroup$ – Yanko Jun 11 '17 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.