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November 17th, 2018, 09:37 PM   #1
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basis for left null space

How can we find the basis for left null space?

specifically for these matrices A( 1 2 5 and B=(1 2 5
2 7 10 ) 2 4 10)
what I did was :

A transpose 1 2
2 7
5 10 reduced it and I obtained: 1 0
0 1
0 0

after here I made x1= 0
x2=0
x3=0
so the left null space basis is (0) is that so?

for matrix B transpose reduced = 1 2
0 0
0 0
then x1= -2x2
x2= 0 and
x3=0
then the basis is ............
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November 18th, 2018, 11:27 PM   #2
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This isn't correct. If I understand your post

$A= \begin{pmatrix}1 &2 &5\\2 &4 &10 \end{pmatrix}$

and you want to find the basis for $v = \begin{pmatrix}v_1\\v_2 \end{pmatrix}$ such that

$v^T A=0$

$A^T v = 0$

$A^T=\begin{pmatrix}1 &2 \\ 2 &4 \\ 5 &10 \end{pmatrix}$

this clearly reduces to

$\begin{pmatrix}1 &2 \\ 0 &0 \\ 0 &0 \end{pmatrix}$

and we're left with $v_1 + 2v_2 = 0 \Rightarrow v_2 = -\dfrac 1 2 v_1$

This in turn gives us a basis vector of $v_b = \begin{pmatrix}2 \\-1 \end{pmatrix}$ for the Null space of $A$

Similarly I think $B = \begin{pmatrix}1 &2 &5 \\ 2 &7 &10\end{pmatrix}$

$B^T = \begin{pmatrix}1 &2 \\2 &7 \\5 &10\end{pmatrix}$

This reduces to

$\begin{pmatrix}1 &2 \\0 &3 \\0 &0 \end{pmatrix}$

which results in a basis vector of $v_b = \begin{pmatrix}0 \\ 0\end{pmatrix}$
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