Understanding basis algorithm result

Question

I've a matrix ${\bf A}$ defined as

A = \begin{pmatrix} 1 & -2 & 0 & 3 & 7\\ 2 & 1 & -3 & 1 & 1\\ \end{pmatrix}

And ${W_1}$ is the solution space associated to the homogenous system ${\bf A}$. I am asked to find a basis for ${W_1}$.

I re-wrote the matrix as linear combination of ${c_1}$, i.e.:

\begin{align} x = c_1 + 2 c_2 \\ y = -2c_1+ c_2 \\ z = -3c_2 \\ v = 3c_1 + c_2 \\ w = 7c_1 + c_2 \\ \end{align}

I use these equations to build another matrix, I solve it and then i reach to:

\begin{align} x - \frac{2}{3}y = c_1 \\ x - \frac{-1}{3}z = c_2 \\ 2x + y \frac{-5}{3}z = 0 \\ x - y - v = 0 \\ 3x - 2y - 2 = 0 \end{align}

So, the last three equations are the general ones for $W_1$ and from there I can describe the basis, is this right ?

Answer 1

A standard Gaussian elimination leads to a reduced row echelon form: \begin{align} \begin{pmatrix} 1 & -2 & 0 & 3 & 7\\ 2 & 1 & -3 & 1 & 1 \end{pmatrix} &\to \begin{pmatrix} 1 & -2 & 0 & 3 & 7\\ 0 & 5 & -3 & -5 & -13 \end{pmatrix} &&R_2\gets R_2-2R_1 \\ &\to \begin{pmatrix} 1 & -2 & 0 & 3 & 7\\ 0 & 1 & -3/5 & -1 & -13/5 \end{pmatrix} &&R_2\gets \frac{1}{5}R_2 \\ &\to \begin{pmatrix} 1 & 0 & -6/5 & 1 & 9/5\\ 0 & 1 & -3/5 & -1 & -13/5 \end{pmatrix} &&R_2\gets \frac{1}{5}R_2 \end{align} This means the system is equivalent to $$ \begin{cases} x_1=\dfrac{6}{5}x_3-x_4-\dfrac{9}{5}x_5 \\[6px] x_2=\dfrac{3}{5}x_3+x_4+\dfrac{13}{5}x_5 \end{cases} $$ Now it's just a matter of choosing values for $x_3$, $x_4$ and $x_5$ to obtain linearly independent vectors; the choices, in order to also remove denominators, can be $$ x_3=5,\quad x_4=0,\quad x_5=0\\ x_3=0,\quad x_4=1,\quad x_5=0\\ x_3=0,\quad x_4=0,\quad x_5=5 $$ leading to the three vectors $$ \begin{pmatrix}6\\3\\5\\0\\0\end{pmatrix},\quad \begin{pmatrix}-1\\1\\0\\1\\0\end{pmatrix},\quad \begin{pmatrix}-9\\13\\0\\0\\5\end{pmatrix}. $$ These three vectors form a basis for $W_1$.

Answer 2

I can't really follow your reasoning; your equations connecting $x,y,z,v,w$ and $c_1,c_2$ seem inconsistent, and there's that odd appearance of $2$.

One standard way of a getting a basis for $\mathrm{Null}(A)$, i.e. the solution space for $A\mathbf{x}=\mathbf{0}$, is to find the reduced row echelon form of $A$ and just read off the parameterised equations. I'll let you do the calculations for your matrix, but here's a similar example:

$A=\begin{bmatrix}1&2&3&4\\1&4&1&6\end{bmatrix}$, RRE form of $A$ is $\begin{bmatrix}1&0&5&2\\0&1&-1&1\end{bmatrix}$.

The equations from the RRE matrix are $\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}-5x_3&-2x_4\\x_3&-x_4\\x_3\\&x_4\end{bmatrix}=x_2\begin{bmatrix}-5\\1\\1\\0\end{bmatrix}+x_4\begin{bmatrix}-2\\-1\\0\\1\end{bmatrix}$