Rank on symbolic matrices

5 ビュー (過去 30 日間)
Ohad Shapira
Ohad Shapira 2021 年 4 月 4 日
編集済み: Bruno Luong 2021 年 4 月 7 日
Hey all,
I'm trying to figure out the rank of several (sym) matrices that I am working on and the built-in 'Rank' function gives me different answer.
As I saw in the help, It's not reliable and I can't trust it ("rank returns an incorrect result because the outputs of intermediate steps are not simplified").
Is there a different way to get the rank of my matrices?
TIA
  9 件のコメント
Bruno Luong
Bruno Luong 2021 年 4 月 5 日
According to https://www.mathworks.com/help/symbolic/rank.html it seems you cannot use RANK until a numerical value is plugged-in.
Seem like very limited usefulness to me.
That reminds me why I don't like using computer symbolic calculation.
Ohad Shapira
Ohad Shapira 2021 年 4 月 5 日
Thanks, most of my problems are solved in this forum.
BTW, your scenario:

サインインしてコメントする。

回答 (1 件)

Aditya Patil
Aditya Patil 2021 年 4 月 7 日
Rank does not take identities satisfied by functions into account. As a workaround, substitute values into variables, and then calculate rank. See Rank Function Does Not Simplify Symbolic Calculations more further details.
  2 件のコメント
Ohad Shapira
Ohad Shapira 2021 年 4 月 7 日
Thank you,
I saw this and I'm trying to understand what can I do to "fix" the rank function that it will return the right answer.
What can I do to verify it?
Bruno Luong
Bruno Luong 2021 年 4 月 7 日
編集済み: Bruno Luong 2021 年 4 月 7 日
May be you can select p arbitrary combinations of values of your variables and substitute in then calculation of the rank after substitution.
If the ranks obtained after substitution give the identical result then it is ikely this holds true for almost all values.
I would suggest select p as the size of the matrix (n) + 1 or larger, because the determinant is a polynomial of order n, threfore has n+1 DOFs.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeNumber Theory についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by