How to show that a map is linear in $C^n$?
$begingroup$
Could someone tell me if I am on right way solving Problem b)?
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.
b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.
$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$
My thoughts:
$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)
$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)
linear-algebra map-projections
edited Jan 5 at 17:22
Omnomnomnom
129k794188
asked Jan 5 at 16:47
KaiKai
686
$endgroup$
|
show 5 more comments
$begingroup$
Could someone tell me if I am on right way solving Problem b)?
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.
b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.
$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$
My thoughts:
$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)
$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)
linear-algebra map-projections
edited Jan 5 at 17:22
Omnomnomnom
129k794188
asked Jan 5 at 16:47
KaiKai
686
$endgroup$
$begingroup$
What does $x_U$ mean?
$endgroup$
– José Carlos Santos
Jan 5 at 16:48
$begingroup$
Projection of x onto u where U is a subspace
$endgroup$
– Kai
Jan 5 at 16:49
$begingroup$
And how do you define that?
$endgroup$
– José Carlos Santos
Jan 5 at 16:50
$begingroup$
Your function doesn't make sense.
$endgroup$
– K. Y
Jan 5 at 16:58
$begingroup$
@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
$endgroup$
– Crostul
Jan 5 at 17:15
|
show 5 more comments
$begingroup$
Could someone tell me if I am on right way solving Problem b)?
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.
b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.
$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$
My thoughts:
$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)
$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)
linear-algebra map-projections
edited Jan 5 at 17:22
Omnomnomnom
129k794188
asked Jan 5 at 16:47
KaiKai
686
$endgroup$
Could someone tell me if I am on right way solving Problem b)?
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
a) Show that every $xinmathbb C^n$ can be written as $x=x_U+x_V$ with $x_Uin U$ and $x_Vin V$ and that this decomposition is unique.
b) Define f : $C^n$ → $C^n$, $f$($x$) := $x_U$. Show that $f$ is a linear map.
$x_u$ is a Projection of $x$ onto $U$, where $U$ is a subspace of $C^n$
My thoughts:
$f$($x$) = < $lambda $ , $x_U$> = $lambda $ . $f$($x$)
$f$($x$) = ($x$+ $y$) = $x_u$ + $y_u$= $f$($x$) + $f$($y$)
linear-algebra map-projections
linear-algebra map-projections
edited Jan 5 at 17:22
Omnomnomnom
129k794188
asked Jan 5 at 16:47
KaiKai
686
edited Jan 5 at 17:22
Omnomnomnom
129k794188
asked Jan 5 at 16:47
KaiKai
686
edited Jan 5 at 17:22
Omnomnomnom
129k794188
edited Jan 5 at 17:22
Omnomnomnom
129k794188
edited Jan 5 at 17:22
Omnomnomnom
129k794188
129k794188
asked Jan 5 at 16:47
KaiKai
686
asked Jan 5 at 16:47
KaiKai
686
asked Jan 5 at 16:47
KaiKai
686
686
$begingroup$
What does $x_U$ mean?
$endgroup$
– José Carlos Santos
Jan 5 at 16:48
$begingroup$
Projection of x onto u where U is a subspace
$endgroup$
– Kai
Jan 5 at 16:49
$begingroup$
And how do you define that?
$endgroup$
– José Carlos Santos
Jan 5 at 16:50
$begingroup$
Your function doesn't make sense.
$endgroup$
– K. Y
Jan 5 at 16:58
$begingroup$
@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
$endgroup$
– Crostul
Jan 5 at 17:15
|
show 5 more comments
$begingroup$
What does $x_U$ mean?
$endgroup$
– José Carlos Santos
Jan 5 at 16:48
$begingroup$
Projection of x onto u where U is a subspace
$endgroup$
– Kai
Jan 5 at 16:49
$begingroup$
And how do you define that?
$endgroup$
– José Carlos Santos
Jan 5 at 16:50
$begingroup$
Your function doesn't make sense.
$endgroup$
– K. Y
Jan 5 at 16:58
$begingroup$
@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
$endgroup$
– Crostul
Jan 5 at 17:15
$begingroup$
What does $x_U$ mean?
$endgroup$
– José Carlos Santos
Jan 5 at 16:48
$begingroup$
What does $x_U$ mean?
$endgroup$
– José Carlos Santos
Jan 5 at 16:48
$begingroup$
Projection of x onto u where U is a subspace
$endgroup$
– Kai
Jan 5 at 16:49
$begingroup$
Projection of x onto u where U is a subspace
$endgroup$
– Kai
Jan 5 at 16:49
$begingroup$
And how do you define that?
$endgroup$
– José Carlos Santos
Jan 5 at 16:50
$begingroup$
And how do you define that?
$endgroup$
– José Carlos Santos
Jan 5 at 16:50
$begingroup$
Your function doesn't make sense.
$endgroup$
– K. Y
Jan 5 at 16:58
$begingroup$
Your function doesn't make sense.
$endgroup$
– K. Y
Jan 5 at 16:58
$begingroup$
@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
$endgroup$
– Crostul
Jan 5 at 17:15
$begingroup$
@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
$endgroup$
– Crostul
Jan 5 at 17:15
|
show 5 more comments
1 Answer
1
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oldest
votes
$begingroup$
You did not prove a).
Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.
Your proof of b) is not correct (although you probably had the right idea).
Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that
$$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
$endgroup$
add a comment |
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$begingroup$
You did not prove a).
Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.
Your proof of b) is not correct (although you probably had the right idea).
Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that
$$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
$endgroup$
add a comment |
$begingroup$
You did not prove a).
Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.
Your proof of b) is not correct (although you probably had the right idea).
Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that
$$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
$endgroup$
add a comment |
$begingroup$
You did not prove a).
Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.
Your proof of b) is not correct (although you probably had the right idea).
Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that
$$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
$endgroup$
You did not prove a).
Since $mathbb{C}^n = U + V = { x_U + x_V mid x_U in U, x_V in V}$, we know that each $xinmathbb C^n$ can be written as $x=x_U + x_V$ with $x_Uin U$ and $x_V in V$. Assume we have another such decomposition $x = x'_U + x'_V$. Then $x_U - x'_U = x'_V - x_V$. The left hand side is an element of $U$, the right hand side an element of $V$. Since $U cap V = {0}$, we see that $x_U - x'_U = x'_V - x_V = 0$. This proves uniqueness.
Your proof of b) is not correct (although you probably had the right idea).
Let $x,y in mathbb{C}^n$ and let $x = x_U + x_V, y = y_U +y_V$ be their unique decompositions. Then $lambda x + mu y = (lambda x_u + mu y_u) + (lambda x_V + mu y_V)$ is a decomposition of $lambda x + mu y$ and by uniqueness we see that
$$f(lambda x + mu y) = (lambda x + mu y)_U = lambda x_u + mu y_u = lambda f(x) + mu f(y) .$$
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
edited Jan 5 at 17:59
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
answered Jan 5 at 17:40
Paul FrostPaul Frost
12.5k31035
12.5k31035
add a comment |
add a comment |
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$begingroup$
What does $x_U$ mean?
$endgroup$
– José Carlos Santos
Jan 5 at 16:48
$begingroup$
Projection of x onto u where U is a subspace
$endgroup$
– Kai
Jan 5 at 16:49
$begingroup$
And how do you define that?
$endgroup$
– José Carlos Santos
Jan 5 at 16:50
$begingroup$
Your function doesn't make sense.
$endgroup$
– K. Y
Jan 5 at 16:58
$begingroup$
@K.Y That function makes sense. The decomposition as sum of an element of $U$ and an element of $V$ is unique. This defines two projections over $U$ and $V$.
$endgroup$
– Crostul
Jan 5 at 17:15