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Re: check if a square matrix is diagonal
*To*: mathgroup at smc.vnet.net
*Subject*: [mg115324] Re: check if a square matrix is diagonal
*From*: Bill Rowe <readnews at sbcglobal.net>
*Date*: Fri, 7 Jan 2011 04:12:40 -0500 (EST)
On 1/6/11 at 2:06 AM, xiaochu at gmail.com (benyaya) wrote:
>What I try to do is extract the diagonal, subtract if from the
>matrix, then compare the new matrix with 0. My code doesn't work out
>though, can anyone help? thanks a lot.
>checkIfDiagonalMatrix[m_] = Module[{d, mtemp},
>d = Dimensions[m];
>mtemp = DiagonalMatrix[Diagonal[m]] - m;
>If[mtemp == Table[Table[0, {i, 1, d}], {i, 1, d}],
>True,
>False]
>]
One immediate problem is Dimensions[m] returns a list.
Consequently, d cannot be used as the end point for the Table
iterator. The following will do what you want and has the same
logic as you have used
checkIfDiagonalMatrix[m_] := Module[{d, mtemp},
d = Dimensions[m];
mtemp = DiagonalMatrix[Diagonal[m]] - m;
If[mtemp == ConstantArray[0,d],
True,
False]
]
Note, I used SetDelayed (:=) not Set (=)
But I think this logic is doing more work than necessary. I
would accomplish this as:
diagonalQ[
m_] := (ArrayRules[SparseArray@m] /.
HoldPattern[{a_, a_} -> _] :> Sequence[]) == {}
If m is a diagonal matrix this should be pretty fast. However,
it m is a dense array, this might be fairly slow.
What I am doing is using ArrayRules to extract all of the
non-zero elements of m. Those are returned as rules that look
like {m,n}->number. I then apply a pattern matching rule that
deletes all cases where m and n are equal and compare that to
the empty list {}.
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