# Creation of one column matrix

I want to create an one column matrix with the 1st element to be -1, and all the other elements to be zero. However, I want to be able to make a matrix of this kind with arbitrary dimension, that is, an arbitrary number of rows.

Here is one way to do it.

makeArray[n_Integer] :=
Module[{a = ConstantArray[{0}, n]}, a[[1, 1]] = -1; a]

makeArray


{{-1}, {0}, {0}, {0}}

There almost certainly more elegant ways to doing this, but this is the first thing that came to my mind.

– DK13
Jul 5 '13 at 15:28
• I'm glad to help, but if you really like my answer, click on the accept button (check mark on the left of this answer). Jul 5 '13 at 15:35
• @m_goldberg I always encourage users to wait longer before accepting an answer as I'm sure you've read before. It certainly doesn't hurt to point out that functionality as it's not entirely obvious, but unless you have strong feelings to the contrary I think it best not to encourage fast Accepts. Jul 5 '13 at 18:07
• @Mr.Wizard. I understand what you are saying and agree that I was hasty here. I will be more careful not to discourage potential additional answers in the future. Jul 5 '13 at 20:27

Try this

coln[n_] := Join[{{-1}}, Table[{0}, {i, n}]];
(*n is the dimension*)
coln


gives you {{-1}, {0}, {0}, {0}, {0}, {0}}

If you want to make a Column or Matrix with only one finite element (say x) at a position (say i for a column or i,j for a matrix) the KroneckerDelta will be a better option.

col[x_, i_, n_] := Table[{x KroneckerDelta[i, j]}, {j, n}];
row[x_, i_, n_] := Table[x KroneckerDelta[i, j], {j, n}];
mat[x_, i_, j_, n_] := Table[x KroneckerDelta[i, p] KroneckerDelta[j, q], {p, n}, {q, n}];


then

mat[0.5,2,2,3]


will give you a matrix with 0.5 at (2,2) and zero everywhere else. Same for column and row.

You can also use sparse matrix to press in particular indices.Check this following example,

Normal[SparseArray[{1 -> -1}, RandomInteger[{2, 9}]]]


Output:{-1, 0, 0, 0, 0, 0, 0, 0}

You can put it in function to take it to more than 9 as number of rows.

Edit: As pointed out the snippet above will produce one row,so this minor improvisation will do it.

Normal[{SparseArray[{1 -> -1}, RandomInteger[{2, 9}]]}] // Transpose


Output:{{-1}, {0}, {0}, {0}, {0}, {0}}

• SparseArray seems a nice way to go, but this produces a row vector, not a one-column, 2D matrix Jul 5 '13 at 23:46
• You can use this: Normal[SparseArray[{{1, 1} -> -1}, {10, 1}]] for 10 elements. Jul 6 '13 at 0:07
Transpose@DiagonalMatrix[{-1}, 0, {1, 5}]


=> {{-1}, {0}, {0}, {0}, {0}}

Edit

ArrayFlatten@{{-1}, {ConstantArray[{0}, 4]}}


=> {{-1}, {0}, {0}, {0}, {0}}

Edit

Probably the most transparent and concise version is this:

-Thread[{UnitVector[5, 1]}]

(* ==> {{-1}, {0}, {0}, {0}, {0}} *)


The first argument of UnitVector is the dimension, and the second argument is the position where the 1 appears. Obviously, this is exactly the kind of problem UnitVector was made for.

End Edit

Another possibility:

-{KroneckerDelta[1, #]} & /@ Range

(* ==> {{-1}, {0}, {0}, {0}, {0}} *)


This has the advantage of being easily generalized to setting any other element to -1. For example, to do it with the third element:

-{KroneckerDelta[3, #]} & /@ Range

(* ==> {{0}, {0}, {-1}, {0}, {0}} *)

• Very nice. I was (also) not aware of you usage of Thread. (I have almost always used Transpose for this purpose). Thread@DiagonalMatrix[{-1}, 0, {1, 5}] == Transpose@ DiagonalMatrix[{-1}, 0, {1, 5}] == -Thread[{UnitVector[5, 1]}] == -Transpose@{UnitVector[5, 1]} Jul 6 '13 at 8:41