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What I did:

I put this into reduced row echelon form:

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

It is clear that the $r(M)=2$, because there are two independent rows.


Now for the null space, I wrote down the equations from the reduced row echelon form:

$$x-2y+2z+4t=0$$

$$z+t=0$$

I can't seem to write $x$ and $y$ separately in terms of $z$ and $t$. Any hints?

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5 Answers 5

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Write the system

$$\left\{\begin{array}{rcl} x-2y+2z+4t& =& 0 \\ z+t & = & 0\end{array}\right.$$

as

$$\left\{\begin{array}{rcl} x+2z& =& 2y-4t \\ z & = & -t\end{array}\right.$$ and solve it. You get,

$$\left\{\begin{array}{rcl} x& =& 2y-2t \\ z & = & -t\end{array}\right.$$ That is, $$(2y-2t,y,-t,t)$$ is an element of the null space for any $y,t.$ Now, look for two linearly independent vectors.

(Note that the kernel has dimension $2.$ So the system has infinitely many solutions that have to depend on two parameters.)

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  • $\begingroup$ You're welcome. $\endgroup$
    – mfl
    Oct 29, 2014 at 13:17
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Do also backwards elimination: $$ \begin{bmatrix} 1 & -2 & 2 & 4 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \to \begin{bmatrix} 1 & -2 & 0 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$ Now your equations read $$ \begin{cases} x_1=2x_2-2x_4\\ x_3=-x_4 \end{cases} $$ You get two linearly independent vectors in the null space by setting $x_2=1, x_4=0$ and $x_2=0,x_4=1$, so the basis is given by the two linearly independent vectors $$ \begin{bmatrix} 2\\ 1\\ 0\\ 0\\ \end{bmatrix} \qquad \begin{bmatrix} -2\\ 0\\ -1\\ 1 \end{bmatrix} $$ The first corresponds to $x_2=1$ and $x_4=0$, the second to $x_2=0$ and $x_4=1$.

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  • $\begingroup$ (+1)but I've heard something like : column reduction doenst preserve the null space. Thank you :) $\endgroup$
    – The Artist
    Oct 29, 2014 at 13:20
  • $\begingroup$ @TheArtist Row reduction doesn't preserve the column space. But since row reduction consists of multiplying the original matrix $A$ by an invertible matrix, if $U$ is a reduced form, then $U=FA$ for some invertible $F$. If $Ax=0$, then also $Ux=0$ and conversely. $\endgroup$
    – egreg
    Oct 29, 2014 at 13:23
  • $\begingroup$ @TheArtist I used no column reduction. $\endgroup$
    – egreg
    Oct 29, 2014 at 13:25
  • $\begingroup$ Oh ok Thank you very much :) $\endgroup$
    – The Artist
    Oct 29, 2014 at 13:36
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Just to make the answer a bit more algorithmic: a "pivot entry" is an entry which is the first non-zero entry in its row. A "pivot column" is a column containing a pivot entry. In your matrix, columns 1 and 3 are pivot columns.

Name the variables after the columns as you did (so, $x, y, z, w$). Then the ``free variables'' are the ones that don't occur in pivot columns. In your case, these are $y$ and $w$. The remaining rows of the matrix express the bound variables in terms of the free variables.

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  • $\begingroup$ (+1) Thank you very much :) This will be helpful for other future readers :) $\endgroup$
    – The Artist
    Oct 29, 2014 at 16:37
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First case: $z=t=0$, you obtain $x-2y=0$. One of solutions is the vector $(2,1,0,0)$.

Second case: $z=-t=1$, which gives you the equation $x-2y=2$, which gives you, for example, $(2,0,1,-1)$.

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  • $\begingroup$ How $z=-t=1$ ? :) $\endgroup$
    – The Artist
    Oct 29, 2014 at 13:14
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Write $z = -t$ and put it in your first equation. You shall get $x = 2y + 6t$, after simplification. Write $$(x,y,z,t)' = y(2 , 1 , 0 , 0)' + t(6, 0, -1, 1)'$$. Now consider $y$ and $t$ in some field. This shall give you all the solutions.

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