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Finite matrices only?
→‎Symmetric Case: new section
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The form in which the theorem is given could logically apply to infinite matrices also; but the details of the proof are for matrices of "the same dimension n". [[Special:Contributions/78.32.103.197|78.32.103.197]] ([[User talk:78.32.103.197|talk]]) 20:33, 24 May 2009 (UTC)
The form in which the theorem is given could logically apply to infinite matrices also; but the details of the proof are for matrices of "the same dimension n". [[Special:Contributions/78.32.103.197|78.32.103.197]] ([[User talk:78.32.103.197|talk]]) 20:33, 24 May 2009 (UTC)

== Symmetric Case ==

If the matrix (whose derivative is being computed) is a symmetric structure, does this affect the result? Because a change in the (i,j)th term will necessarily be felt in the (j,i)th element.... therefore the derivative must be "simultaneous". Right?

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Simplifying notation

How about using a specific notation for the cofactors in the proof, such as Cij, as in the cofactor and the Laplace expansion articles? I think it would make the proof more readable. The relation to the adjugate matrix could be stated at the beginning or/and at the end of the proof.

Eroblar 17:40, 1 September 2007 (UTC)[reply]

What field?

Is is not clear what kind of numbers the entries of the matrix may be for the formula to hold. Does is also hold for complex matrices? Would be great if someone could add this information to the statement of the theorem. —Preceding unsigned comment added by 128.232.241.65 (talk) 14:44, 26 October 2007 (UTC)[reply]

Finite matrices only?

The form in which the theorem is given could logically apply to infinite matrices also; but the details of the proof are for matrices of "the same dimension n". 78.32.103.197 (talk) 20:33, 24 May 2009 (UTC)[reply]

Symmetric Case

If the matrix (whose derivative is being computed) is a symmetric structure, does this affect the result? Because a change in the (i,j)th term will necessarily be felt in the (j,i)th element.... therefore the derivative must be "simultaneous". Right?