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Given a matrix:

$$ A_{m,n} = \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{pmatrix} $$

$a_{i,j}$ is a signless integral and bounded. And $b_{i,j}$ is the same. Is there any similarity function between $a_{i,j}$ and $b_{i,j}$. Such that $f(a_{i,j})=f(b_{i,j})$ if and only if $a_{i,j}=b_{i,j}$.

For example, the rank of matrices can identify a class of matrices, not a single matrix.

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    What exactly do you mean by "identity" here? One can see a complex in different ways, e.g. a matrix can be both symmetric and orthogonal, or Hessenberg and banded, or any number of other properties...2012-02-14
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    I don't understand. On the simplest interpretation of your question, the way to detect the identity is to see whether $AB=BA=B$ for all matrices $B$.2012-02-14
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    I think by "identity" the OP means "how can I tell if some matrix is identical to a given matrix $A$?" I don't really understand what kind of answer the OP is expecting other than "compare their entries."2012-02-14
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    You could pack all the $a_{ij}$'s into a bit string...2012-02-14
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    @Rahul Narain, Thankk you for your answer. Could you give me an reason. Then it would be a matrix of 1*mn. It is differetn with the original matrix.2012-02-14
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    I think the poster is asking whether there is some smaller piece of information that can be derived from a matrix, $f(M)$, such that $f(M)=f(N)$ if and only if $M=N$; i.e., some simpler *equality-invariant* than the entire matrix. I doubt it.2012-02-14
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    If "$a_{i,j}$ is a signless integral and bounded" means the entries are nonnegative integers with some known bound $B$, then $f(M) = \sum_i \sum_j a_{i,j} B^{i+mj}$ would do. This could conveniently be written as $f(M) = u^T M v$ where $u^T = [B, B^2, \ldots, B^m]$ and $v = [B^m, B^{2m}, \ldots, B^{nm}]^T$.2012-02-14
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    Following Robert, maybe hashing is really what you are looking for: http://en.wikipedia.org/wiki/Hash_function ?2012-02-14
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    @Arturo Magidin, Thank you, exactly right2012-02-23

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