I have a matrix $A=\{a_{ij}\}$ of dimensions $n$ rows by $m$ columns, and I would like to multiply each row of the matrix by a constant $b_i$, $i=1,\ldots,n$, and store the result in the same variable. In linear-algebra speak, I basically would like to do $A=\mathrm{diag}(b_1,\ldots,b_n) A$. However, I would like to avoid having to create a diagonal matrix off the vector $(b_1,\ldots,b_n)$, because $n$ is large and I wouldn't want to use any more RAM. Is there a different and less memory demanding way to do $A=\mathrm{diag}(b_1,\ldots,b_n) A$?

My approach is to go over the rows of $A$ and multiply each by the respective element of the vector $(b_1,\ldots,b_n)$. May be there is a more elegant way to do this? By "elegant" I mean faster yet not too memory-demanding.

m = 3;
n = 5;
A = Table[RandomInteger[{0, 5}], {i, m}, {j, n}];
b = Table[RandomInteger[{5, 10}], {i, m}];

(* Standard Linear-Algebraic Solution *)

(* Less memory demanding solution *)
For[i = 1, i <= m, i += 1,
  A[[i, All]] = b[[i]]*A[[i, All]];
  • 2
    $\begingroup$ Use a SparseArray[] for the diagonal matrix: SparseArray[Band[{1, 1}] -> b].A. However, due to listability, even a Hadamard product suffices: b A. $\endgroup$ Oct 28, 2015 at 0:35

2 Answers 2


This is also a kind of inner product, hence:

amat = Array[a, {3, 5}];
bvec = Array[b, 3];
ans = Transpose[Inner[Times, bvec, amat, List]]

enter image description here

Or, numerically:

amat = RandomInteger[{-5, 5}, {3, 5}];
bvec = RandomInteger[{-5, 5}, 3];
ans = Transpose[Inner[Times, bvec, amat, List]];

enter image description here

  • $\begingroup$ Good answer. But, I just wanted to know that if I want multiplication e.g. a[1,1].b[1]. I have a[1,1]=2 and b[1]=2, then a[1,1].b[1]=4. But I am getting in my case where b[1]-> a[1][1], as a[1][1]a[1][1]. No multiplication but like tensor product. $\endgroup$
    – L.K.
    Sep 26, 2016 at 15:13
  • $\begingroup$ You probably have old definitions. Try quitting the kernel and starting again. $\endgroup$
    – bill s
    Sep 26, 2016 at 16:30

Oh, yes, there is a more elegant way!

Mathematica does element-by-element multiplication.

I think this does what you want:

A = Array[a, {3, 5}]

 (* {{a[1, 1], a[1, 2], a[1, 3], a[1, 4], a[1, 5]}, {a[2, 1], 
  a[2, 2], a[2, 3], a[2, 4], a[2, 5]}, {a[3, 1], a[3, 2], a[3, 3], 
  a[3, 4], a[3, 5]}} *)

 bvec = Array[b, {3}]

 (* {b[1], b[2], b[3]} *)

 bvec * A

 (* {{a[1, 1] b[1], a[1, 2] b[1], a[1, 3] b[1], a[1, 4] b[1], 
  a[1, 5] b[1]}, {a[2, 1] b[2], a[2, 2] b[2], a[2, 3] b[2], 
  a[2, 4] b[2], a[2, 5] b[2]}, {a[3, 1] b[3], a[3, 2] b[3], 
  a[3, 3] b[3], a[3, 4] b[3], a[3, 5] b[3]}} *)

I have used symbolic quantities in the test vector and matrix so you can see what is going on.

As a general rule, if you have a nested For loop in Mathematica, you are missing something about its capabilities.

  • $\begingroup$ Thanks. What about reciprocating each element of a vector, is there a specific command for that as well? $\endgroup$
    – Alex
    Oct 28, 2015 at 1:01
  • $\begingroup$ @Alex: 1/myvector? $\endgroup$
    – Verbeia
    Oct 28, 2015 at 1:22
  • $\begingroup$ @Alex: all the basic arithmetic functions, and most of the mathematical functions, are Listable, which lets you avoid having to construct elementwise loops. $\endgroup$ Oct 28, 2015 at 1:33

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