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May 13th, 2018, 10:04 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2  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 10:07 AM. 
May 13th, 2018, 10:23 AM  #2 
Senior Member Joined: Oct 2009 Posts: 608 Thanks: 186 
It is true that $Y = (A^T A)^{1} A^T$ is a leftinverse 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 $(A^T A)^{1}A^T$.

May 13th, 2018, 10:29 AM  #3  
Senior Member Joined: Jan 2017 From: Toronto Posts: 200 Thanks: 2  Quote:
$\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?  
May 13th, 2018, 10:37 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,169 Thanks: 1141 
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 
May 13th, 2018, 10:38 AM  #5 
Senior Member Joined: Oct 2009 Posts: 608 Thanks: 186  No, $A$ is not necessarily invertible, so $A^{1}$ might not exist. So $YA=I$ does not imply $Y=IA^{1}$.


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