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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.)

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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.

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  • $\begingroup$ 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? $\endgroup$ – Jay Sep 19 '14 at 14:41
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    $\begingroup$ @Jay The set of matrices having the same RREF is not a subspace of the vector space of matrices. $\endgroup$ – egreg Sep 19 '14 at 14:45
  • $\begingroup$ 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. $\endgroup$ – Jay Sep 19 '14 at 15:06
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    $\begingroup$ @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). $\endgroup$ – egreg Sep 19 '14 at 15:08
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    $\begingroup$ @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, $(4-2)\cdot 5=10$. $\endgroup$ – egreg Sep 19 '14 at 15:21

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