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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 $v_b = \begin{pmatrix}0 \\ 0\end{pmatrix}$ Tags basis, left, null, space Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gaussrelatz Algebra 4 October 5th, 2013 05:01 AM fysampy Linear Algebra 0 September 8th, 2012 07:35 PM mbradar2 Linear Algebra 6 May 4th, 2011 11:40 AM Sambit Linear Algebra 0 November 1st, 2010 04:22 AM wontonsoup Linear Algebra 2 May 26th, 2009 08:21 PM

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