2
$\begingroup$

\begin{bmatrix} 1 & 2 & 1 & 1 \\ 2 & 4 & 3 & 5\\ \end{bmatrix}

When I reduce this matrix to reduced row echelon form and set Ax=0, I get

\begin{bmatrix} 1 & 2 & 0 & -2 \\ 0 & 0 & 1 & 3 \\ \end{bmatrix}


I've found the vectors forming the basis of the nullspace of A: $(2,−1,0,0),(2,0,−3,1)$ What would the other two vectors be? $(1,0,0,0)$ and $(0,1,0,0)$?

$\endgroup$
  • $\begingroup$ I approved the edit which removed the $[X]$ as you put it in linear algebra, i just want to tell you that $\mathbb{R}[X]$ is the ring of polynomials over $\mathbb{R}$ $\endgroup$ – Dominic Michaelis Mar 7 '13 at 19:52
  • $\begingroup$ My mistake, sorry. $\endgroup$ – Mathlete Mar 7 '13 at 19:52
1
$\begingroup$

I have no idea from where from where the $-2$ and the $2$ in the reduced echolon form comes.

The idea is at first find two linear independet vectors which are in the null space. (for example by the reduced echolon form), afterward add some vectors for a basis.

$\endgroup$
  • $\begingroup$ Sorry that was an error on my part. When the matrix is in RREF, do I use the 3 or the -2 in the fourth column? $\endgroup$ – Mathlete Mar 7 '13 at 19:59
  • $\begingroup$ for what you want to use it ? remember that the columns of a matrix are the images of the unit vectors, so $(2,-1,0,0)$ will be in the kernel $\endgroup$ – Dominic Michaelis Mar 7 '13 at 20:11
  • $\begingroup$ I've found the vectors forming the basis of the nullspace of A: $(2,-1,0,0), (2,0,-3,1)$ How do I find the other two vectors? $\endgroup$ – Mathlete Mar 7 '13 at 20:38
  • 1
    $\begingroup$ Oh it's good you are asking, you need some linear independet vectors, you can always fill them up with unit vectors, you just need to chose the right ones. the first one is $(1,0,0,0)$. the vector $(0,1,0,0)$ won't work, you could choose $(0,0,1,0)$ or $(0,0,0,1)$ that doesn't make a difference $\endgroup$ – Dominic Michaelis Mar 7 '13 at 20:49
  • 1
    $\begingroup$ after choosing $(1,0,0,0)$ you have the first vector or the null spcae $(2,-1,0,0)$, if you take the difference of $(2,0,0,0)$ and $(2,-1,0,0)$ you get $(0,1,0,0)$ which is the second unit vector and so this one isn't linear independent anymore $\endgroup$ – Dominic Michaelis Mar 7 '13 at 20:55
0
$\begingroup$

Here is a general procedure for completing the basis of a space. Suppose you already have a few vectors that are the basis of a subspace (it does not matter what the subspace is). Let these vectors be $$v_1, v_2, v_3, \cdots v_k$$ Now pick any basis for the space (columns/rows of identity) is an obvious choice. Let these vectors be $$u_1, u_2, u_3 \cdots u_n$$ Now run the gram-schmdit process on the $k+n$ vectors ordered as $v$'s followed by $u$'s, i.e. $$ [v_1, v_2 \cdots v_k, u_1, u_2, \cdots u_n] \rightarrow [w_1, w_2, \cdots w_k, w_{k+1}, \cdots w_n] ~~\hbox{Using Gram-Schmidt} $$

Now $w_1, w_2 \cdots w_k$ is an orthonormal basis for the original space spanned by $v_1, \cdots v_k$ and the rest complete the space (actually forms a basis for the orthogonal complemnt

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.