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Code is changed on 22.07.2025, Now it also works for Complex Number.
For wrong or incomplete solution, please submit the feedback form.
So, I will try my best to improve it soon.
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Solution
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Solution provided by AtoZmath.com
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Column Space calculator
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 1. `[[8,-6,2],[-6,7,-4],[2,-4,3]]`
2. `[[6,-2,2],[-2,3,-1],[2,-1,3]]`
3. `[[3,2,4],[2,0,2],[4,2,3]]`
4. `[[1,1,1],[-1,-3,-3],[2,4,4]]`
5. `[[2,3],[4,10]]`
6. `[[5,1],[4,2]]`
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Example1. Find Column Space ...
`[[1,-2,0,3,-4],[3,2,8,1,4],[2,3,7,2,3],[-1,2,0,4,-3]]`Solution: | `1` | `-2` | `0` | `3` | `-4` | | | `3` | `2` | `8` | `1` | `4` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
Now, reduce the matrix to row echelon form interchanging rows `R_1 harr R_2` | = | | `3` | `2` | `8` | `1` | `4` | | | `1` | `-2` | `0` | `3` | `-4` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
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`R_2 larr R_2-1/3xx R_1` | = | | `3` | `2` | `8` | `1` | `4` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `2` | `3` | `7` | `2` | `3` | | | `-1` | `2` | `0` | `4` | `-3` | |
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`R_3 larr R_3-2/3xx R_1` | = | | `3` | `2` | `8` | `1` | `4` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `5/3` | `5/3` | `4/3` | `1/3` | | | `-1` | `2` | `0` | `4` | `-3` | |
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`R_4 larr R_4+1/3xx R_1` | = | | `3` | `2` | `8` | `1` | `4` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `5/3` | `5/3` | `4/3` | `1/3` | | | `0` | `8/3` | `8/3` | `13/3` | `-5/3` | |
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`R_3 larr R_3+5/8xx R_2` | = | | `3` | `2` | `8` | `1` | `4` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `0` | `0` | `3` | `-3` | | | `0` | `8/3` | `8/3` | `13/3` | `-5/3` | |
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`R_4 larr R_4+ R_2` | = | | `3` | `2` | `8` | `1` | `4` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `0` | `0` | `3` | `-3` | | | `0` | `0` | `0` | `7` | `-7` | |
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interchanging rows `R_3 harr R_4` | = | | `3` | `2` | `8` | `1` | `4` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `0` | `0` | `7` | `-7` | | | `0` | `0` | `0` | `3` | `-3` | |
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`R_4 larr R_4-3/7xx R_3` | = | | `3` | `2` | `8` | `1` | `4` | | | `0` | `-8/3` | `-8/3` | `8/3` | `-16/3` | | | `0` | `0` | `0` | `7` | `-7` | | | `0` | `0` | `0` | `0` | `0` | |
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The rank of a matrix is the number of non all-zeros rows `:. Rank = 3` Column Space : The matrix has 3 pivots and Pivots are in the columns 1,2 and 4. We know that these columns in the original matrix define the column space of the matrix. `:.` The Column Space is `[[1],[3],[2],[-1]],[[-2],[2],[3],[2]],[[3],[1],[2],[4]]` |
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