Re: check if a square matrix is diagonal
- To: mathgroup at smc.vnet.net
- Subject: [mg115351] Re: check if a square matrix is diagonal
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 8 Jan 2011 03:40:18 -0500 (EST)
Why not simply:
diagonalQ[m_] := m == DiagonalMatrix@Diagonal@m
?
Bobby
On Fri, 07 Jan 2011 03:12:40 -0600, Bill Rowe <readnews at sbcglobal.net>
wrote:
> 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 {}.
>
>
--
DrMajorBob at yahoo.com