I want to make sure I'm understanding what the matrix of a linear transformation says about its null space and range.

It's clear for me with rows (as this is how Gaussian elimination seems to be applied in most examples)

For an $m \times n$ matrix: After applying Gaussian elimination to the rows turning it into RRE form, if any row(s) can be row reduced to a row of zeros, then the number non-zero rows is the row rank, (i.e. dimension of the range).

$n$, the column count of the matrix, minus this number, is the dimension of the null space.

How does this work for columns?

For example, the matrix: $\begin{bmatrix} 1 & 2 & 0 & 1 & 0\\ 2 & 4 & 1 & 0 & 0\\ 3 & 6 & 0 & 0 & 1 \\ 4 & 8 & 0 & 0 & 0 \\ \end{bmatrix}$

Where $c_i$ denotes the column, it's clear that $-2c_1 + c_2 + 0c_3 + 0 c_4 + 0 c_5=0$. But on first sight, it's not that arithmetically easy to do row reduction. So I would prefer to use any conclusion by looking at columns. In this case, what can we say about this matrix?


Not sure if this is what you are asking, but the number of linearly independent columns is also equal to the rank of the matrix. So, for the case above, from inspection it appears that the rank is 4, because columns 3,4,5 are obviously linearly independent (that's 3 so far), column 2 is independent from 3,4,5 (because there is no way to multiply 0 to get 8 in the fourth row). This adds one more to the rank, so we have rank=4. Lastly, columns 1 and 2 are multiples of each other, so the rank does not increase. Therefore, the rank=4 by inspection, and the dimension of the null-space is then n-4 = 5-4 = 1. I hope this helps.

  • $\begingroup$ The number of linearly independent columns is also equal to the rank of the matrix. How can you show this to be true? $\endgroup$
    – Snowball
    Nov 30 '19 at 18:54
  • $\begingroup$ It's a known property of matrices in linear algebra: mathworld.wolfram.com/MatrixRank.html. You can check wikipedia for proof examples if you wish: en.wikipedia.org/wiki/… among other resources. I hope this helps. $\endgroup$
    – ad2004
    Nov 30 '19 at 19:00

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