7. Solving Ax = 0: Pivot Variables, Special Solutions

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Recommends Patrik Bašo · 2020-05 · Mathematics Stack Exchange · on “Solving Ax = 0 , pivot variables, free varaibles, need more specific explanation”

Recently i started to lear linear algebra out of MIT OpenCourceWare. But i cant quite understand one little 'proof' (demontration) of getting $N(A)$ out of expression $Ax = 0$ by creating rref form of matrix. Given point is at minute 28:05 in this video [this video] So basically what i cant understand is that.. We have $Ax = 0$ in form of : $$ \begin{bmatrix} I & F \\ 0 & 0 \end{bmatrix}\begin{bmatrix} -F \\ I \end{bmatrix}=0 $$ Also the matrices sizes are : $ r = rank(A) \\ A^{n\times m} \\ I^{r\times r} \\ F^{r\times (n-r)} $ So i really cant grasp this idea when there is $ (n-r) \neq r$ because then we cant express second matrix $x$ (because those blocks of I and F are not of the same width...

Recommends Vishal Jain · 2020-06 · Mathematics Stack Exchange · on “query about finding null space of matrices”

I was watching gilbert strangs linear algebra lecture [this video] series and got a bit stuck when trying to reason where his use of free variables came from. At 8:17 he introduces the idea of free variables as the variables which multiply non pivot columns. What I am confused about is why we are only allowed to freely set those variables which we choose to label as free. Example used in lecture: Consider the matrix $A$ : \begin{pmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{pmatrix} To find the null space we are looking for solutions to the equations: $x_1+2x_2+2x_3+2x_4=0 $ , $2x_3+4x_4=0$ I see that we have 4 variables and 2 independant equations. This would mean that we should

Cites Elizabeth · 2017-03 · Mathematics Stack Exchange · on “Why are free variables free?”

I'm having trouble understanding this: The free variables are called free because they can take on any value; none of the equations relates any of them to each other. Why can we set free variables to an arbitrary s or t? Example: Say we have $A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ and we are looking for solutions to $Ax = 0$ . Here, we have column $1$ and $3$ as our pivot columns (and $x_1$ , $x_3$ are pivot varibles ) and column $2$ and $4$ as our free columns (and $x_2$ , $x_4$ are free variables )[1]. To find a solution vector $x =\begin{bmatrix} x_1 \\ x_2 \\x_3 \\ x_4\end{bmatrix}$ , we can set any arbitrary values to $x_2$ and $x_4$ which wil

Cites Aur Bhai · 2022-05 · Mathematics Stack Exchange · on “If entries at pivot position are zero then the column is a linear combination of the previous columns?”

In the 7th lecture of Gilbert Strang's Linear Algebra course, he does row operations on a matrix (m x n) to convert it to echelon form. He found the first k pivot elements; for k+1 th pivot element, he found that the entries of a column from row k+1 to m are zero, and from this, he concludes that this column is a linear combination of previous columns. Why or how can we figure that out? Here is the link to the video (from the point where he says that) [this video]