# How to find the index of the minimum value in a matrix.

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Micaiah Barletta on 19 Jan 2022
Answered: Steven Lord on 20 Jan 2022
Hi everyone,
So I have a 9x21 matrix, V, and I am trying to find the index (i,j) of the minimum (closest to zero) value in that matrix. I have tried using several forms of the min function, but it keeps returning multiple indices for the the minimum values in each column. I only want one index (i,j), that of the minimum value in the ENTIRE matrix.
How would I go about doing this? Thanks!
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### Answers (3)

Stephen23 on 19 Jan 2022
Edited: Stephen23 on 19 Jan 2022
Where M is your matrix:
[R,C] = find(M==min(M(:)))
or
[~,X] = min(M(:)); % or [~,X] = min(M,[],'all','linear');
[R,C] = ind2sub(size(M),X)
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Image Analyst on 19 Jan 2022
Try
% Sample data between -2.5 and +2.5
M = 5 * rand(5, 10) - 2.5
M = 5×10
2.2558 2.0285 0.7944 2.3343 0.7202 0.8739 -0.0449 -1.9158 -1.0654 -1.8531 -0.9877 1.7554 2.4745 -2.4179 0.1576 1.4184 0.2105 1.1018 2.1986 -2.3464 0.0450 2.3143 -0.8332 2.0228 0.1946 -0.5806 -0.5764 1.6439 -1.6989 1.2254 -1.1251 -0.8863 0.5182 -0.3079 -2.0536 -0.3036 1.9644 -1.4574 -0.9964 1.8352 -0.9463 -0.9253 0.9574 -0.8925 0.3358 1.1363 -2.4485 -2.0140 0.0612 0.1545
% Find value closest to 0:
minValue = min(abs(M(:)))
minValue = 0.0449
[r, c] = find(abs(M) == minValue)
r = 1
c = 7

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Yusuf Suer Erdem on 19 Jan 2022
Hi, did you try it with this way. It gives the index which has the smallest number.
a=[8 2 5 1;2 -2 4 0;9 5 4 8];
[r,c]=find(a==min(a(:)))
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Yusuf Suer Erdem on 20 Jan 2022
Hi, I strived a lot on your codes but I finally achieved it. The codes which are below works. I am glad if you accept my answer;
clc; clear;
a = rand(9,21);
b= 0;
differences = abs(a-b)
minDiff = min(differences);
closestValue = min(minDiff);
[r c]=find(a==closestValue)

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Steven Lord on 20 Jan 2022
Since you're using release R2021b (according to the information listed on the right side of this page) you can use 'all' as the dimension input and specify both the ComparisonMethod parameter and the 'linear' option.
M = randn(6)
M = 6×6
0.9423 0.3336 0.1760 1.0120 1.5974 -0.7014 -2.2171 -0.4520 -0.2327 -0.8261 -0.9182 -0.8860 0.7431 -0.7179 -0.2381 0.0671 0.9823 -1.0860 1.6596 -0.9416 -0.0795 -0.2284 -1.1704 -0.8441 -0.4919 -0.3952 0.6310 2.4187 -0.0775 0.2292 0.1780 0.3546 -0.3958 0.1260 -0.0577 0.7462
[minValue, minIndex] = min(M, [], 'all', 'linear', 'ComparisonMethod', 'abs')
minValue = -0.0577
minIndex = 30
For this particular random matrix, the entry with the smallest absolute value is the element with linear index 30.
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