User Name Remember Me? Password

 Linear Algebra Linear Algebra Math Forum

 May 13th, 2018, 09:04 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Left Inverse Matrix question Let Y1 and Y2 be two left inverses of a matrix A. a) Show that if b + c = 1 then Y = bY1 + cY2 is also a left inverse of A. b) If we remove the condition that b + c = 1 is Y still a left inverse of A? I am not sure I did it correctly. Would anyone confirm my answers?? Thanks. My Answer(a): Given both Y1 and Y2 are left inverse of a matrix A Therefore $\displaystyle Both~~Y1~and~Y2 = (A^{T}A)^{-1}A^{T} ~~even~~ Y1 <> Y2 \\ Y = (1 - c) (A^{T}A)^{-1}A^{T} + c (A^{T}A)^{-1}A^{T} \\ Y = (A^{T}A)^{-1}A^{T} - c (A^{T}A)^{-1}A^{T} + c (A^{T}A)^{-1}A^{T} \\ Y = (A^{T}A)^{-1}A^{T}$ Y is also a left inverse of A. My Answer(b): No Last edited by zollen; May 13th, 2018 at 09:07 AM. May 13th, 2018, 09:23 AM #2 Senior Member   Joined: Oct 2009 Posts: 752 Thanks: 261 It is true that $Y = (A^T A)^{-1} A^T$ is a left-inverse of $A$. But it isn't the only one (in general). A left inverse is simply a matrix $Y$ such that $YA = I$. There are multiple such matrices, so you cannot assume it has the special form $(A^T A)^{-1}A^T$. Thanks from zollen May 13th, 2018, 09:29 AM   #3
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 209
Thanks: 3

Quote:
 Originally Posted by Micrm@ss It is true that $Y = (A^T A)^{-1} A^T$ is a left-inverse of $A$. But it isn't the only one (in general). A left inverse is simply a matrix $Y$ such that $YA = I$. There are multiple such matrices, so you cannot assume it has the special form $(A^T A)^{-1}A^T$.
ok, Given...
$\displaystyle Y_{1}A = I, ~so~ Y_{1} = I A^{-1} \\ Y_{2}A = I, ~so~ Y_{2} = I A^{-1} \\ YA = I, ~so~ Y = I A^{-1} \\ \\ \\ \\ Y = (1 - c) IA^{-1} + c I A^{-1} \\ Y = IA^{-1} - c IA^{-1} + c IA^{-1} \\ Y = IA^{-1}$ May 13th, 2018, 09:37 AM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,375 Thanks: 1276 you can't further evaluate $I A^{-1}$ ? I would do this as $Y=b Y_1 + c Y_2$ $YA = b Y_1 A + c Y_2 A = b I + c I = (b+c) I = I$ $\text{Yes, }(b+c=1) \Rightarrow Y \text{ is a left inverse of }A$ if $b+c \neq 1$ then it's clear that $YA = (b+c)I \neq I$ and in this case $Y$ is not a left inverse of A Thanks from zollen May 13th, 2018, 09:38 AM   #5
Senior Member

Joined: Oct 2009

Posts: 752
Thanks: 261

Quote:
 Originally Posted by zollen ok, Given... $\displaystyle Y_{1}A = I, ~so~ Y_{1} = I A^{-1} \\ Y_{2}A = I, ~so~ Y_{2} = I A^{-1} \\ YA = I, ~so~ Y = I A^{-1} \\ \\ \\ \\ Y = (1 - c) IA^{-1} + c I A^{-1} \\ Y = IA^{-1} - c IA^{-1} + c IA^{-1} \\ Y = IA^{-1}$ How about this?
No, $A$ is not necessarily invertible, so $A^{-1}$ might not exist. So $YA=I$ does not imply $Y=IA^{-1}$. Tags inverse, left, matrix, question 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 soroban New Users 0 June 21st, 2012 07:27 AM JOANA Abstract Algebra 0 March 18th, 2011 06:29 AM TsAmE Linear Algebra 3 October 29th, 2010 08:51 PM RMG46 Linear Algebra 1 July 5th, 2010 05:07 PM ced Linear Algebra 1 October 29th, 2009 04:19 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      