vectors u = [4 1 0 0] and v = [1 0 2 1] form a base of nullspace of matrix $$ A\in M_{5,4}(R) $$ Find a reduced row echelon form of Matrix A.
Since $ n-r = dimN(A) $ we know we got two base variables and two free ones. And reduced row echelon form will look like this:
$$
\begin{bmatrix}
1 & 0 & a & b \\
0 & 1 & c & d \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
I don't know where to go from here. I am also confused about something, if I plug in [4 1 0 0] instead of x1 x2 x3 and x4 in my "potential" reduced row echelon form I would get this:
1*4 + 0*1 + a*0 + b*0 = 0
0*4 + 1*1 + c*0 + d*0 = 0
and I get 4=0 and 1=0 from that. How is that possible? (I know I am making a mistake somewhere, just don't know where.)
Best Answer
The reduced row echelon form of a matrix is unique. The fact that the two given vectors form a basis of the null space means that the reduced form of the homogeneous linear system associated to the matrix is $$ \begin{cases} x_1=4x_2+x_4\\ x_3=2x_4 \end{cases} $$ because, for $x_2=1$ and $x_4=0$ we get the first vector and with $x_2=0$ and $x_4=1$ we get the second vector. So the reduced system can be written $$ \begin{cases} x_1-4x_2-x_4=0\\ x_3-2x_4=0 \end{cases} $$ which corresponds to the matrix $$ \begin{bmatrix} 1 & -4 & 0 & -1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$
Your error is considering the third and fourth variables as free, while they are the second and fourth ones.