Prove every subspace of $V$ is part of a direct sum equal to $V$
$begingroup$
In Linear Algebra Done Right, the book proves that every subspace of $V$ is part of a direct sum equal to $V$. I generally follow the proof, but do not understand some points.
Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then there is a subspace $W$ of $V$ such that $V = Ubigoplus W$.
Proof
Because $V$ is finite-dimensional, so is $U$ (proved in 2.26. I am OK). Thus there is a basis $u_1,...,u_m$ of $U$. Of course $u_1,...u_m$ is a linearly independent list of vectors in $V$ (I am OK). Hence this list can be extended to a basis $u_1,...,u_m, w_1,...,w_n$ of $V$ (I am not OK. Does it mean $w_1,...w_n$ can only be vectors that extend to the basis of $V$? If so, why the title of the proof said EVERY subspace of $V$ is part of a direct sum equal to $V$).
To prove $V = Ubigoplus W$, we need only show that $$V = U+W text{ and } U cap W = {0}$$ (I am OK).
To prove the first equation above, suppose $v in V$. Then, because the list $u_1,...,u_m, w_1,...w_n$ spans $V$, there exist $a_1,...a_m, b_1,...b_m in mathbb{F}$ such that $$v=a_1u_1+...+a_mu_m+b_1w_1+...+b_nw_n$$.
In other words, we have $v=u+w$, wheere $u in U$ and $w in W$ are defined as above. Thus $v in U + W$, completing the proof that $V = U + W$. (I am also OK.)
To show that $U cap W = {0}$, suppose $v in U cap W$. Then there exist scalars $a_1,...,a_m,b_1,...b_n in mathbb{F}$ such that
$$v=a_1v_1+...+a_mv_m = b_1w_1+...+b_nw_n$$. Thus
$$a_1u_ + ...+a_mu_m - b_1w_1-...-b_nw_n = 0$$
Because $u_1,...u_m,w_1,...,w_n$ is linearly independent, this implies that $a_1=...=a_m=b_1=...=b_n = 0$. Thus $v=0$, completing the proof that $U cap W = {0}$.
Is the proof means for every subspace $U$ of finite-dimensional $V$, we can find a $W$ that is the direct sum of $V$?
linear-algebra
asked Jan 1 at 23:01
JOHN JOHN
1638
$endgroup$
add a comment |
$begingroup$
In Linear Algebra Done Right, the book proves that every subspace of $V$ is part of a direct sum equal to $V$. I generally follow the proof, but do not understand some points.
Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then there is a subspace $W$ of $V$ such that $V = Ubigoplus W$.
Proof
Because $V$ is finite-dimensional, so is $U$ (proved in 2.26. I am OK). Thus there is a basis $u_1,...,u_m$ of $U$. Of course $u_1,...u_m$ is a linearly independent list of vectors in $V$ (I am OK). Hence this list can be extended to a basis $u_1,...,u_m, w_1,...,w_n$ of $V$ (I am not OK. Does it mean $w_1,...w_n$ can only be vectors that extend to the basis of $V$? If so, why the title of the proof said EVERY subspace of $V$ is part of a direct sum equal to $V$).
To prove $V = Ubigoplus W$, we need only show that $$V = U+W text{ and } U cap W = {0}$$ (I am OK).
To prove the first equation above, suppose $v in V$. Then, because the list $u_1,...,u_m, w_1,...w_n$ spans $V$, there exist $a_1,...a_m, b_1,...b_m in mathbb{F}$ such that $$v=a_1u_1+...+a_mu_m+b_1w_1+...+b_nw_n$$.
In other words, we have $v=u+w$, wheere $u in U$ and $w in W$ are defined as above. Thus $v in U + W$, completing the proof that $V = U + W$. (I am also OK.)
To show that $U cap W = {0}$, suppose $v in U cap W$. Then there exist scalars $a_1,...,a_m,b_1,...b_n in mathbb{F}$ such that
$$v=a_1v_1+...+a_mv_m = b_1w_1+...+b_nw_n$$. Thus
$$a_1u_ + ...+a_mu_m - b_1w_1-...-b_nw_n = 0$$
Because $u_1,...u_m,w_1,...,w_n$ is linearly independent, this implies that $a_1=...=a_m=b_1=...=b_n = 0$. Thus $v=0$, completing the proof that $U cap W = {0}$.
Is the proof means for every subspace $U$ of finite-dimensional $V$, we can find a $W$ that is the direct sum of $V$?
linear-algebra
asked Jan 1 at 23:01
JOHN JOHN
1638
$endgroup$
$begingroup$
It means that every subspace has a complementary subspace.
$endgroup$
– Bernard
Jan 1 at 23:09
$begingroup$
I'm not entirely sure what you are confused about but it seems like you are under the impression that $V$ only has one basis. Consider for example $mathbb{R} ^2$. We have the standard basis ${(1,0),(0,1)}$ but we can replace $(1,0)$ with $(1,1)$ and it is still a basis. In fact replacing $(1,0)$ with anything in the form $(a,b)$ where $aneq 0$ will result in a basis. ${(1,2),(1,3)}$ is another example of a basis.
$endgroup$
– Fortox
Jan 1 at 23:13
add a comment |
$begingroup$
In Linear Algebra Done Right, the book proves that every subspace of $V$ is part of a direct sum equal to $V$. I generally follow the proof, but do not understand some points.
Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then there is a subspace $W$ of $V$ such that $V = Ubigoplus W$.
Proof
Because $V$ is finite-dimensional, so is $U$ (proved in 2.26. I am OK). Thus there is a basis $u_1,...,u_m$ of $U$. Of course $u_1,...u_m$ is a linearly independent list of vectors in $V$ (I am OK). Hence this list can be extended to a basis $u_1,...,u_m, w_1,...,w_n$ of $V$ (I am not OK. Does it mean $w_1,...w_n$ can only be vectors that extend to the basis of $V$? If so, why the title of the proof said EVERY subspace of $V$ is part of a direct sum equal to $V$).
To prove $V = Ubigoplus W$, we need only show that $$V = U+W text{ and } U cap W = {0}$$ (I am OK).
To prove the first equation above, suppose $v in V$. Then, because the list $u_1,...,u_m, w_1,...w_n$ spans $V$, there exist $a_1,...a_m, b_1,...b_m in mathbb{F}$ such that $$v=a_1u_1+...+a_mu_m+b_1w_1+...+b_nw_n$$.
In other words, we have $v=u+w$, wheere $u in U$ and $w in W$ are defined as above. Thus $v in U + W$, completing the proof that $V = U + W$. (I am also OK.)
To show that $U cap W = {0}$, suppose $v in U cap W$. Then there exist scalars $a_1,...,a_m,b_1,...b_n in mathbb{F}$ such that
$$v=a_1v_1+...+a_mv_m = b_1w_1+...+b_nw_n$$. Thus
$$a_1u_ + ...+a_mu_m - b_1w_1-...-b_nw_n = 0$$
Because $u_1,...u_m,w_1,...,w_n$ is linearly independent, this implies that $a_1=...=a_m=b_1=...=b_n = 0$. Thus $v=0$, completing the proof that $U cap W = {0}$.
Is the proof means for every subspace $U$ of finite-dimensional $V$, we can find a $W$ that is the direct sum of $V$?
linear-algebra
asked Jan 1 at 23:01
JOHN JOHN
1638
$endgroup$
In Linear Algebra Done Right, the book proves that every subspace of $V$ is part of a direct sum equal to $V$. I generally follow the proof, but do not understand some points.
Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$. Then there is a subspace $W$ of $V$ such that $V = Ubigoplus W$.
Proof
Because $V$ is finite-dimensional, so is $U$ (proved in 2.26. I am OK). Thus there is a basis $u_1,...,u_m$ of $U$. Of course $u_1,...u_m$ is a linearly independent list of vectors in $V$ (I am OK). Hence this list can be extended to a basis $u_1,...,u_m, w_1,...,w_n$ of $V$ (I am not OK. Does it mean $w_1,...w_n$ can only be vectors that extend to the basis of $V$? If so, why the title of the proof said EVERY subspace of $V$ is part of a direct sum equal to $V$).
To prove $V = Ubigoplus W$, we need only show that $$V = U+W text{ and } U cap W = {0}$$ (I am OK).
To prove the first equation above, suppose $v in V$. Then, because the list $u_1,...,u_m, w_1,...w_n$ spans $V$, there exist $a_1,...a_m, b_1,...b_m in mathbb{F}$ such that $$v=a_1u_1+...+a_mu_m+b_1w_1+...+b_nw_n$$.
In other words, we have $v=u+w$, wheere $u in U$ and $w in W$ are defined as above. Thus $v in U + W$, completing the proof that $V = U + W$. (I am also OK.)
To show that $U cap W = {0}$, suppose $v in U cap W$. Then there exist scalars $a_1,...,a_m,b_1,...b_n in mathbb{F}$ such that
$$v=a_1v_1+...+a_mv_m = b_1w_1+...+b_nw_n$$. Thus
$$a_1u_ + ...+a_mu_m - b_1w_1-...-b_nw_n = 0$$
Because $u_1,...u_m,w_1,...,w_n$ is linearly independent, this implies that $a_1=...=a_m=b_1=...=b_n = 0$. Thus $v=0$, completing the proof that $U cap W = {0}$.
Is the proof means for every subspace $U$ of finite-dimensional $V$, we can find a $W$ that is the direct sum of $V$?
linear-algebra
linear-algebra
asked Jan 1 at 23:01
JOHN JOHN
1638
asked Jan 1 at 23:01
JOHN JOHN
1638
asked Jan 1 at 23:01
JOHN JOHN
1638
asked Jan 1 at 23:01
JOHN JOHN
1638
asked Jan 1 at 23:01
JOHN JOHN
1638
1638
$begingroup$
It means that every subspace has a complementary subspace.
$endgroup$
– Bernard
Jan 1 at 23:09
$begingroup$
I'm not entirely sure what you are confused about but it seems like you are under the impression that $V$ only has one basis. Consider for example $mathbb{R} ^2$. We have the standard basis ${(1,0),(0,1)}$ but we can replace $(1,0)$ with $(1,1)$ and it is still a basis. In fact replacing $(1,0)$ with anything in the form $(a,b)$ where $aneq 0$ will result in a basis. ${(1,2),(1,3)}$ is another example of a basis.
$endgroup$
– Fortox
Jan 1 at 23:13
add a comment |
$begingroup$
It means that every subspace has a complementary subspace.
$endgroup$
– Bernard
Jan 1 at 23:09
$begingroup$
I'm not entirely sure what you are confused about but it seems like you are under the impression that $V$ only has one basis. Consider for example $mathbb{R} ^2$. We have the standard basis ${(1,0),(0,1)}$ but we can replace $(1,0)$ with $(1,1)$ and it is still a basis. In fact replacing $(1,0)$ with anything in the form $(a,b)$ where $aneq 0$ will result in a basis. ${(1,2),(1,3)}$ is another example of a basis.
$endgroup$
– Fortox
Jan 1 at 23:13
$begingroup$
It means that every subspace has a complementary subspace.
$endgroup$
– Bernard
Jan 1 at 23:09
$begingroup$
It means that every subspace has a complementary subspace.
$endgroup$
– Bernard
Jan 1 at 23:09
$begingroup$
I'm not entirely sure what you are confused about but it seems like you are under the impression that $V$ only has one basis. Consider for example $mathbb{R} ^2$. We have the standard basis ${(1,0),(0,1)}$ but we can replace $(1,0)$ with $(1,1)$ and it is still a basis. In fact replacing $(1,0)$ with anything in the form $(a,b)$ where $aneq 0$ will result in a basis. ${(1,2),(1,3)}$ is another example of a basis.
$endgroup$
– Fortox
Jan 1 at 23:13
$begingroup$
I'm not entirely sure what you are confused about but it seems like you are under the impression that $V$ only has one basis. Consider for example $mathbb{R} ^2$. We have the standard basis ${(1,0),(0,1)}$ but we can replace $(1,0)$ with $(1,1)$ and it is still a basis. In fact replacing $(1,0)$ with anything in the form $(a,b)$ where $aneq 0$ will result in a basis. ${(1,2),(1,3)}$ is another example of a basis.
$endgroup$
– Fortox
Jan 1 at 23:13
add a comment |
1 Answer
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You have a subspace $U$ with basis $u_1,...,u_m$.
Extend the basis to a basis of $V$ by adding vectors $w_1,...,w_n$. There is some freedom to choose the $w_k$ but they must be linearly indepdendent and the collection must span $V$.
Let $W=operatorname{sp} {w_k}$.
Since the whole collection spans $V$, we must have $V = U +W$. If $uin U, win W$ and
$u+w = 0$, we must have $u=w=0$ since the whole collection is linearly independent.
Note that $W$ is not unique.
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
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add a comment |
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$begingroup$
You have a subspace $U$ with basis $u_1,...,u_m$.
Extend the basis to a basis of $V$ by adding vectors $w_1,...,w_n$. There is some freedom to choose the $w_k$ but they must be linearly indepdendent and the collection must span $V$.
Let $W=operatorname{sp} {w_k}$.
Since the whole collection spans $V$, we must have $V = U +W$. If $uin U, win W$ and
$u+w = 0$, we must have $u=w=0$ since the whole collection is linearly independent.
Note that $W$ is not unique.
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
$endgroup$
add a comment |
$begingroup$
You have a subspace $U$ with basis $u_1,...,u_m$.
Extend the basis to a basis of $V$ by adding vectors $w_1,...,w_n$. There is some freedom to choose the $w_k$ but they must be linearly indepdendent and the collection must span $V$.
Let $W=operatorname{sp} {w_k}$.
Since the whole collection spans $V$, we must have $V = U +W$. If $uin U, win W$ and
$u+w = 0$, we must have $u=w=0$ since the whole collection is linearly independent.
Note that $W$ is not unique.
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
$endgroup$
add a comment |
$begingroup$
You have a subspace $U$ with basis $u_1,...,u_m$.
Extend the basis to a basis of $V$ by adding vectors $w_1,...,w_n$. There is some freedom to choose the $w_k$ but they must be linearly indepdendent and the collection must span $V$.
Let $W=operatorname{sp} {w_k}$.
Since the whole collection spans $V$, we must have $V = U +W$. If $uin U, win W$ and
$u+w = 0$, we must have $u=w=0$ since the whole collection is linearly independent.
Note that $W$ is not unique.
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
$endgroup$
You have a subspace $U$ with basis $u_1,...,u_m$.
Extend the basis to a basis of $V$ by adding vectors $w_1,...,w_n$. There is some freedom to choose the $w_k$ but they must be linearly indepdendent and the collection must span $V$.
Let $W=operatorname{sp} {w_k}$.
Since the whole collection spans $V$, we must have $V = U +W$. If $uin U, win W$ and
$u+w = 0$, we must have $u=w=0$ since the whole collection is linearly independent.
Note that $W$ is not unique.
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
answered Jan 1 at 23:09
copper.hatcopper.hat
126k559160
126k559160
add a comment |
add a comment |
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$begingroup$
It means that every subspace has a complementary subspace.
$endgroup$
– Bernard
Jan 1 at 23:09
$begingroup$
I'm not entirely sure what you are confused about but it seems like you are under the impression that $V$ only has one basis. Consider for example $mathbb{R} ^2$. We have the standard basis ${(1,0),(0,1)}$ but we can replace $(1,0)$ with $(1,1)$ and it is still a basis. In fact replacing $(1,0)$ with anything in the form $(a,b)$ where $aneq 0$ will result in a basis. ${(1,2),(1,3)}$ is another example of a basis.
$endgroup$
– Fortox
Jan 1 at 23:13