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May 13th, 2018, 09:04 AM   #1
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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.
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May 13th, 2018, 09:23 AM   #2
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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 $(A^T A)^{-1}A^T$.
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May 13th, 2018, 09:29 AM   #3
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Quote:
Originally Posted by Micrm@ss View Post
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}
$

How about this?
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May 13th, 2018, 09:37 AM   #4
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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
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May 13th, 2018, 09:38 AM   #5
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Quote:
Originally Posted by zollen View Post
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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