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November 17th, 2018, 08:37 PM  #1 
Senior Member Joined: Apr 2017 From: New York Posts: 155 Thanks: 6  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 ............ 
November 18th, 2018, 10:27 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 
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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basis, left, null, space 
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