# If we know nullspace of matrix, how to find reduced row echelon form of that matrix?

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.)

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.

• Thanks a lot, I also got one more question. If we have a subspace of all matrices who have rref that you pointed above, what is dimension of such subspace?
– Jay
Sep 19 '14 at 14:41
• @Jay The set of matrices having the same RREF is not a subspace of the vector space of matrices. Sep 19 '14 at 14:45
• I am sorry, I made a mistake. If we have subspace of all matrices who have same nullspace ({u,v}) like my matrix above, what is the dimension of such subspace? I converted matrices to linear operators and proved it is a subspace (I think) and I thought I have to solve the dimension problem using RREF.
– Jay
Sep 19 '14 at 15:06
• @Jay That's not a subspace either; if $A$ has $U$ as null space, then also $-A$ has, but $A+(-A)$ has a different null space. I think this warrants a fresh question on the site, but you probably mean "The set of matrices whose null space contains a fixed subspace” (this is indeed a subspace of the vector space of $5\times4$ matrices). Sep 19 '14 at 15:08
• @Jay If the kernel of a linear map $f\colon V\to W$ contains a subspace $U$, then it induces a linear map $\bar{f}\colon V/U\to W$. Conversely a linear map $g\colon V/U\to W$ can be composed with the canonical projection. So your subspace has the same dimension as $\operatorname{Hom}(V/U,W)$. In the present case, $(4-2)\cdot 5=10$. Sep 19 '14 at 15:21

I believe that the accepted solution does not have full generality and will note work if, say, the first two columns of the matrix are linearly dependent.

If the columns of matrix $$N$$ represent the null space, then the RREF can be reconstructed by the following matlab command

rref(transpose(null(transpose(N))))


See this lesson on Lemma for a discussion of the problem (but not the above solution).

First write your null space basis as a matrix:

$$N = \begin{bmatrix} 4 & 1 \\ (1) & 0 \\ 0 & 2 \\ 0 & (1) \end{bmatrix}$$

Then modify the basis (without changing the nullspace) so that each vector's "last" element is a "1" and in a different row from all other basis vectors. That is already done in your problem.

Then pad the matrix with zero columns to put the pivots in the right place:

$$N_2 = \begin{bmatrix} 0 & 4 & 0 & 1 \\ 0 & (1) & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & (1) \end{bmatrix}$$

Then subtract the matrix from the identity:

$$I - N_2 = \begin{bmatrix} 1 & -4 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

That's the RRE for the matrix. The reverse of this can be used to find the null space in a fairly easy way.