How to replace the diagonal entries of a square matrix with entries from a vectore of equal length?

I have an nxn matrix M, and a vector v of length n. I want to swap the diagonal of M with the vector v, without a for loop. Is that possible? Thanks!

Answer 1

[John D'Errico]

M = M - diag(diag(M)) + diag(v)

or...

M = M + diag(v - diag(M))

In either case, I killed the diagonal, then added the diagonal I want back in.

Or, you can simply replace those elements. A lazy way to do that is:

M(find(eye(size(M))) = v;

There are lots of simple variations on the latter theme of course.

n = size(M,1);
M(1:(n+1):end) = v;

For example:

M = rand(5)
M =
      0.45054      0.82582      0.10665      0.86869      0.43141
     0.083821      0.53834       0.9619     0.084436      0.91065
      0.22898      0.99613    0.0046342      0.39978      0.18185
      0.91334     0.078176      0.77491      0.25987       0.2638
      0.15238      0.44268       0.8173      0.80007      0.14554
v = 1:5;
n = size(M,1);

M(1:(n+1):end) = v
M =
            1      0.82582      0.10665      0.86869      0.43141
     0.083821            2       0.9619     0.084436      0.91065
      0.22898      0.99613            3      0.39978      0.18185
      0.91334     0.078176      0.77491            4       0.2638
      0.15238      0.44268       0.8173      0.80007            5

[merhisam]

Thank you very much!

[merhisam]

Let me pick your brain a little further. What if I want to do the same thing for the #1 diagonal instead (which is of length n-1)?

[John D'Errico]

Being fundamentally lazy, a slightly modified diag solution would not work in that case? (Of course it will.) diag does allow two arguments, the default being zero, which implies the main diagonal.

If you want to be more constructive, think about why this solution works for the main case.

M(1:(n+1):end) = v;

How are elements stored in memory in MATLAB?

reshape(1:25,5,5)
ans =
   1     6    11    16    21
   2     7    12    17    22
   3     8    13    18    23
   4     9    14    19    24
   5    10    15    20    25

MATLAB stores each column in turn. So a vector of elements like

1:(n+1):end

when used as an index into the square matrix m takes elements

1,m+2,2*m+3,...

You should see that those are the diagonal elements of a square matrix of order n.

So now, what if you wanted to replace the first super diagonal of a square matrix of order n? You would start with element n+1.

M = rand(4);
n = 4;
M((n+1):(n+1):end) = 1:3

M =
    0.13607            1      0.51325      0.12332
    0.86929      0.85303            2      0.18391
     0.5797      0.62206     0.075967            3
    0.54986      0.35095      0.23992      0.41727

The point is to understand and then visualize how elements are stored in memory.

[merhisam]

Thanks again :)

[Shuvashish Roy]

awesome explanation.

[Oliver Johnson]

Is there a way of using repmat function on this to create a 100 x 100 grid counting from 1 to 100 diagonally?

[Julio César Sayago Cabrera]

1000 Thanks, so simple and so good. I had a while thinking about it and couldn't come up with a solution. Many times the most simple way of do it is the best of them all

[Tom]

How do you go about changing the diagonal above and below the central diagonal?

[Uday]

Please read John D'Errico's detailed explanation above on why

M(1:(n+1):end) = v;

works again.

[John D'Errico]

M = rand(5)

M = 5x5

0.5108 0.7342 0.5212 0.0691 0.2529 0.4613 0.0433 0.2281 0.6686 0.3280 0.6924 0.1114 0.6931 0.0803 0.8222 0.9802 0.3333 0.5494 0.9603 0.1489 0.4885 0.3833 0.0137 0.5150 0.5014

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% replace the super-diagonal with the numbers 1:4, using diag
M = M + diag((1:4)' - diag(M,1),1) 

M = 5x5

0.5108 1.0000 0.5212 0.0691 0.2529 0.4613 0.0433 2.0000 0.6686 0.3280 0.6924 0.1114 0.6931 3.0000 0.8222 0.9802 0.3333 0.5494 0.9603 4.0000 0.4885 0.3833 0.0137 0.5150 0.5014

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% replace the sub-diagonal, with the vector [2 3 5 7], using indexing
n = size(M,1);
M(2:n+1:end) = [2 3 5 7]

M = 5x5

0.5108 1.0000 0.5212 0.0691 0.2529 2.0000 0.0433 2.0000 0.6686 0.3280 0.6924 3.0000 0.6931 3.0000 0.8222 0.9802 0.3333 5.0000 0.9603 4.0000 0.4885 0.3833 0.0137 7.0000 0.5014

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Answer 2

[Fangjun Jiang]

a=rand(5);
a(boolean(eye(5)))=1:5

[merhisam]

Very neat! Thank you!