       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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