Is it true that if $b bot im(A)$ then $b bot im(A^T)$
$begingroup$
Is it true that:
If $b bot im(A)$ then $b bot im(A^T)$ ?
I think vaguely remembering LA class I would say it is not true.
But this seems like this is what is being implied by scribe notes for linear regression
linear-algebra regression
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
$endgroup$
add a comment |
$begingroup$
Is it true that:
If $b bot im(A)$ then $b bot im(A^T)$ ?
I think vaguely remembering LA class I would say it is not true.
But this seems like this is what is being implied by scribe notes for linear regression
linear-algebra regression
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
$endgroup$
2
$begingroup$
What's actually being used here is: if $b perp im(A)$ then $b in ker(A^T)$. Which is true.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 23:01
$begingroup$
@DanielSchepler thank you kind man
$endgroup$
– YohanRoth
Dec 7 '18 at 23:25
add a comment |
$begingroup$
Is it true that:
If $b bot im(A)$ then $b bot im(A^T)$ ?
I think vaguely remembering LA class I would say it is not true.
But this seems like this is what is being implied by scribe notes for linear regression
linear-algebra regression
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
$endgroup$
Is it true that:
If $b bot im(A)$ then $b bot im(A^T)$ ?
I think vaguely remembering LA class I would say it is not true.
But this seems like this is what is being implied by scribe notes for linear regression
linear-algebra regression
linear-algebra regression
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
asked Dec 7 '18 at 22:44
YohanRothYohanRoth
6361714
6361714
2
$begingroup$
What's actually being used here is: if $b perp im(A)$ then $b in ker(A^T)$. Which is true.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 23:01
$begingroup$
@DanielSchepler thank you kind man
$endgroup$
– YohanRoth
Dec 7 '18 at 23:25
add a comment |
2
$begingroup$
What's actually being used here is: if $b perp im(A)$ then $b in ker(A^T)$. Which is true.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 23:01
$begingroup$
@DanielSchepler thank you kind man
$endgroup$
– YohanRoth
Dec 7 '18 at 23:25
2
2
$begingroup$
What's actually being used here is: if $b perp im(A)$ then $b in ker(A^T)$. Which is true.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 23:01
$begingroup$
What's actually being used here is: if $b perp im(A)$ then $b in ker(A^T)$. Which is true.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 23:01
$begingroup$
@DanielSchepler thank you kind man
$endgroup$
– YohanRoth
Dec 7 '18 at 23:25
$begingroup$
@DanielSchepler thank you kind man
$endgroup$
– YohanRoth
Dec 7 '18 at 23:25
add a comment |
1 Answer
1
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$begingroup$
The answer is no.
Take for example the matrix
$$ A = pmatrix{1 & 2 \ -1 & -2}.$$
The image of $A$ is $Im(A)= text{span} pmatrix{1 \ -1}$
Take $$b = pmatrix{1 \ 1}.$$ Vector $b$ belongs to $Im(A)^{perp}$. However, since $Im(A^{T}) = text{span} pmatrix{1 \ 2}$, $bnotin Im(A^T)^{perp}$.
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The answer is no.
Take for example the matrix
$$ A = pmatrix{1 & 2 \ -1 & -2}.$$
The image of $A$ is $Im(A)= text{span} pmatrix{1 \ -1}$
Take $$b = pmatrix{1 \ 1}.$$ Vector $b$ belongs to $Im(A)^{perp}$. However, since $Im(A^{T}) = text{span} pmatrix{1 \ 2}$, $bnotin Im(A^T)^{perp}$.
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
$endgroup$
add a comment |
$begingroup$
The answer is no.
Take for example the matrix
$$ A = pmatrix{1 & 2 \ -1 & -2}.$$
The image of $A$ is $Im(A)= text{span} pmatrix{1 \ -1}$
Take $$b = pmatrix{1 \ 1}.$$ Vector $b$ belongs to $Im(A)^{perp}$. However, since $Im(A^{T}) = text{span} pmatrix{1 \ 2}$, $bnotin Im(A^T)^{perp}$.
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
$endgroup$
add a comment |
$begingroup$
The answer is no.
Take for example the matrix
$$ A = pmatrix{1 & 2 \ -1 & -2}.$$
The image of $A$ is $Im(A)= text{span} pmatrix{1 \ -1}$
Take $$b = pmatrix{1 \ 1}.$$ Vector $b$ belongs to $Im(A)^{perp}$. However, since $Im(A^{T}) = text{span} pmatrix{1 \ 2}$, $bnotin Im(A^T)^{perp}$.
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
$endgroup$
The answer is no.
Take for example the matrix
$$ A = pmatrix{1 & 2 \ -1 & -2}.$$
The image of $A$ is $Im(A)= text{span} pmatrix{1 \ -1}$
Take $$b = pmatrix{1 \ 1}.$$ Vector $b$ belongs to $Im(A)^{perp}$. However, since $Im(A^{T}) = text{span} pmatrix{1 \ 2}$, $bnotin Im(A^T)^{perp}$.
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
answered Dec 7 '18 at 22:56
dallonsidallonsi
1187
1187
add a comment |
add a comment |
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$begingroup$
What's actually being used here is: if $b perp im(A)$ then $b in ker(A^T)$. Which is true.
$endgroup$
– Daniel Schepler
Dec 7 '18 at 23:01
$begingroup$
@DanielSchepler thank you kind man
$endgroup$
– YohanRoth
Dec 7 '18 at 23:25