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 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
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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
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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 $Y=IA^{-1}$.

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