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I have the following matrix :

$$A= \begin{bmatrix} 0 &1 &2 &3 &4 \\ 0 &0 &0 &1 &2\\ 0 &0 &0 &0 &0\\ \end{bmatrix} $$

So here the 1st pivot is missing but the 2nd one is there. Can we conclude that the column space is the 4th column and dimension of the column space would be 1? Or do we need the first pivot in order to come up with such conclusions.

Or does the first element of the first row (which is 1) become the first pivot? since we need dependent variables, that are dependent on free variables? Or does the first row completely go to free variables? Help me out please.

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    $\begingroup$ I would still call the $1$ in the top left the "first pivot" because it is a pivot, and it is the first one to appear. While the first pivot can and will often be found in the first column, as this example shows, that is not always the case. The locations of the pivots in the REF implies which columns of the original matrix can be used to form a maximal independent set that spans the column space. Here, the second and fourth columns both have pivots and the matrix is already in REF (albeit not RREF). $\endgroup$
    – JMoravitz
    Jul 19, 2016 at 21:39

1 Answer 1

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Hint:

The column space is defined as the span of the column vectors, so the column space is:

$$\mathrm{span}\left(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \\ 0 \end{bmatrix}\right)$$

What does having a zero vector included in the list do to that span? If you remove the zero vector, will the span remain the same? (Hint: Given any scalar $a$, what is: $a*\begin{bmatrix} 0 \\ 0 \\0 \end{bmatrix}$?)

Which elements can be removed to form a basis of the span? (another way to ask this question: which vectors in that spanning list can you write as linear combinations of other vectors).

By removing the linearly dependent vectors, you will have a basis of the column space. I claim that this basis will be of length 2 and thus the column space has dimension 2. Do you see why that is?

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