       Re: Working with arrays

• To: mathgroup at smc.vnet.net
• Subject: [mg131206] Re: Working with arrays
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Mon, 17 Jun 2013 06:28:12 -0400 (EDT)
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• References: <20130616091915.825D36A2D@smc.vnet.net>

```Clear[IA,xtest,\$Post];

CenterDot is not really an operator, it is a notation

IA\[CenterDot]xtest

IA\[CenterDot]xtest

IA=Array[a,{3,3}];

xtest=Array[x,3];

IA\[CenterDot]xtest

{{a[1,1],a[1,2],a[1,3]},{a[2,1],a[2,2],a[2,3]},{a[3,1],a[3,2],a[3,3]}}\[CenterDot]{x,x,x}

If you want CenterDot to carry out the operation Dot, you can use a
replacement rule

IA.xtest == Dot[IA,xtest]==
Inner[Times,IA,xtest,Plus]==
(IA\[CenterDot]xtest/.CenterDot->Dot)==
(CenterDot[IA,xtest]/.CenterDot->Dot)

True

If you want CenterDot to always act like Dot with lists then use \$Post to
automatically use the replacement rule.

\$Post=#/.CenterDot[x_List,y_List]->
Dot[x,y]&;

IA.xtest == Dot[IA,xtest]==
Inner[Times,IA,xtest,Plus]==
IA\[CenterDot]xtest==CenterDot[IA,xtest]

True

Symbolic use of CenterDot is not changed

x\[CenterDot]y

x\[CenterDot]y

Bob Hanlon

On Sun, Jun 16, 2013 at 5:19 AM, amannucci
<Anthony.J.Mannucci at jpl.nasa.gov>wrote:

> I thought I understood variables. This sequence completely mystifies me:
> Clear[lA, xtest]
> lA = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
> xtest = {4, 5, 6};
> lA\[CenterDot]xtest (* First case *)
> {{a, b}, {c, d}} . {x, y} (* Second case *)
>
> The output is:
>
> {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}\[CenterDot]{4, 5, 6}
>
> {a x + b y, c x + d y}
>
> How do I force matrix multiplication to actually occur, as in the second
> answer? Why does Mathematica do the matrix multiply in the second case but
> not the first?
>
> Thanks for any help.
>
> -Tony
>
>

```

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