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I've been going through MIT OCW's Linear Algebra lectures, and Professor Strang demonstrated the algorithmic steps to solving the equation $Ax = 0$ to find the nullspace of $A$.

At around nineteen minutes in, he starts introducing pivot and free columns of an echelon matrix:

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

Where the second and forth columns represent free columns and the first and third pivot columns. I understand pivot columns, but then he goes onto say that the free columns correspond to free variables in the solution $x$.

Finally, he sets $x_2 = 1$ and $x_4 = 0$ and solves for that system, calling it a special solution. He then switches them around with $x_2 = 0$ and $x_4 = 1$, concluding with the solution set of:

$$x = c\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + d\begin{bmatrix}0\\0\\-2\\1\end{bmatrix}$$

I want to know why these variables are essentially "free" and are able to be whatever. My current understanding is that $1,0$ and $0,1$ are substituted because they are somehow "basis" vectors in which the linear combination would give us all solutions. But what makes these specific columns "free"? In what way are they special, giving them the ability to take any value?

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  • $\begingroup$ I think you have a typo in the last vector, as $(2,0,-2,1)$ isn't a solution. I think you meant $(0,0,-2,1)$. $\endgroup$ Jun 1, 2018 at 15:51

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I think the situation is clearest if you perform the full row reduction first; then your system is equivalent to the one given by the matrix $$\begin{pmatrix} 1 & 2 & 0 & 0 \\ 0 & 0 & 1 & 2 \end{pmatrix}.$$

This shows that the pivots are independent of one another, yet that they depend on the so-called free columns. Explicitly, the system is given by $x_1 + 2x_2 = 0$, $x_3 + 2x_4= 0$, and the dependence may be rewritten as $x_1 = -2x_2$, $x_3 = -2x_4$. Now it is transparent that $x_2,x_4$ may take arbitrary values, and this forces the pivots to take specific values given by the dependence relations. This generalizes to any homogeneous system.

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  • $\begingroup$ So essentially $x_4$ and $x_2$ are independent and $x_1$ and $x_3$ are dependent on them, so allowing them to range over all possible values (via linear combos) gives all solutions. But what prevents us from isolating $x_4$ and $x_2$ and making them dependent on $x_1$ and $x_3$? $\endgroup$
    – Andrew Li
    Jun 1, 2018 at 16:08
  • $\begingroup$ @AndrewLi There is indeed nothing preventing that: you could easily swap the roles of $x_1,x_2$ and $x_3,x_4$ and divide through by $2$ to obtain an equivalent system where the roles of "pivot" and "free" variables are interchanged. However, no matter how you rename things, the number of pivot variables (called the rank of the system) and the number of free variables (called the nullity of the system) are invariant. $\endgroup$ Jun 1, 2018 at 16:45
  • $\begingroup$ Okay, so the rank & nullity don't change meaning the nullspace doesn't, right (and thus renaming doesn't matter)? Could you gauge the correctness of my current understanding in the original question? Thanks for your time. $\endgroup$
    – Andrew Li
    Jun 1, 2018 at 17:36
  • $\begingroup$ @AndrewLi Correct. In our example, the nullspace is a plane spanned by $(-2,1,0,0)$ and $(0,0,-2,1)$. Swapping the roles of the pivots and free variables, we would obtain the relations $x_2 = -x_1/2$ and $x_4 = -x_3/2$. Now $x_1$ and $x_3$ are free to take any values, and we see that the nullspace is spanned by $(1,-1/2,0,0)$ and $(0,0,1,-1/2)$; this is exactly the same plane as before! $\endgroup$ Jun 1, 2018 at 18:15
  • $\begingroup$ Okay, thanks so much for clearing this up! $\endgroup$
    – Andrew Li
    Jun 1, 2018 at 18:17

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