What is the null space of an inconsistent matrix?

Question

ganeshie8 · Answer

Do you mean "inconsistent matrix `system` " ?

anonymous · Answer

yes

ganeshie8 · Answer

consistency is defined for system of eqns, not for matrices. how does it matter to null solution ?

ganeshie8 · Answer

you simply set your right hand side to 0 vector and solve the matrix equation ?

anonymous · Answer

i did that. I got an inconsistent matrix system after I reduced it to row echelon form

anonymous · Answer

can I determine a null space from an inconsistent system?

ganeshie8 · Answer

could you show your work ? maybe take a screenshot and attach if psble

anonymous · Answer

attachment

ikram002p · Answer

so u got aome answer with respect to some variable t , sure u can assume t any number to get initial solution hmm idk if it called null or what lol but yes u can assume t=1 as an answer ( usually )

ganeshie8 · Answer

$$\large 

\left[\begin{array} ~1&i&-2\3&4i&-5\-1&-3t&i\end{array}\right]
\left[\begin{array}~x_1\x_2\x_3\end{array}\right] = 
\left[\begin{array}~0\0\0\end{array}\right]

$$

anonymous · Answer

that is suppose to be a -3i? in the third row?

ganeshie8 · Answer

$$\large 

\left[\begin{array} ~1&i&-2\3&4i&-5\-1&-3i&i\end{array}\right]
\left[\begin{array}~x_1\x_2\x_3\end{array}\right] = 
\left[\begin{array}~0\0\0\end{array}\right]

$$

correct ?

anonymous · Answer

yes

ganeshie8 · Answer

First of all, notice that the system is always consisten cuz $x_1=x_2=x_3=0$ always satisfies the system

ganeshie8 · Answer

A homogeneous system is never inconsistent, there always the trivial solution : $x_i = 0$

ganeshie8 · Answer

after row reducing do you get :

$$\large 

\left[\begin{array} ~1&i&-2\0&i&~1\0&0&~i\end{array}\right]
\left[\begin{array}~x_1\x_2\x_3\end{array}\right] = 
\left[\begin{array}~0\0\0\end{array}\right]

$$
?

anonymous · Answer

-3 instead of -2, 1 instead of i in the 2nd row, and 1/i in the second row

anonymous · Answer

$$\left[\begin{matrix}1 & 0 & -3 \ 0 & 1 & 1/i\ 0 &0&i  \end{matrix}\right]$$

ganeshie8 · Answer

Okay don't divide $i$ in the second row as you don't know what values it can take.

anonymous · Answer

it is not a variable. it is the complex number i

ganeshie8 · Answer

Oh oh oh.. so the rank is 3 here ?

ganeshie8 · Answer

that just means the nullspace just contains the 0 vector, right ?

anonymous · Answer

I'm not sure. but I can't solve for any variables since $$i \neq 0$$

ganeshie8 · Answer

you want to solve  $$\large   \rm   i x_3 = 0$$

right ?

ganeshie8 · Answer

since $\rm  i$ is not 0,  $\rm  x_3 $ has to be 0

anonymous · Answer

no, the bottom row has to be $$\left[\begin{matrix}0 & 0 & 0\end{matrix}\right]$$ to a system to solve.

anonymous · Answer

to have a system to solve, but if there is a value in the bottom row, it is to be interpreted as $$i =0$$

anonymous · Answer

so this is a inconsistent system by definition

ganeshie8 · Answer

I think you're talking about `augmented matrix` which is not we are dealing with here.

anonymous · Answer

oh ok

ganeshie8 · Answer

Our goal here is to find all the possible combinations of columns that yield the 0 vector.

There exists a trivial solution always since when you don't take any columns, you get a 0 vector.

ganeshie8 · Answer

$$\large 

\left[\begin{array} ~1&i&-2\0&i&~1\0&0&~i\end{array}\right]
\left[\begin{array}~x_1\x_2\x_3\end{array}\right] = 
\left[\begin{array}~0\0\0\end{array}\right]

$$

gives you below system of equations :

$$
\begin{array} 
&&&&ix_3&=&0\
&&ix_2+&x_3&=&0\
x_1&+&ix_2-&2x_3&=&0\

\end{array}

$$

anonymous · Answer

alright

ganeshie8 · Answer

use wolfram to double check http://www.wolframalpha.com/input/?i=null+space+%7B%7B1%2C0%2C-3%7D%2C%7B0%2Ci%2C1%7D%2C%7B0%2C0%2Ci%7D%7D

anonymous · Answer

okay. i appreciate the help

ganeshie8 · Answer

are you really sure, $i$ is imaginary and not just a variable like $t$ ?

anonymous · Answer

it doesn't specify