Decompose a finite dimensional vector space into a direct sum of $1$-dimensional sub-spaces.Can infinite-dimensional vector spaces be decomposed into direct sum of its subspaces?Infinite dimensional vector space, and infinite dimensional subspaces.Dimension of Direct sum of same Vector SpacesProof regarding the direct sum of subspacesLinear Algebra Proof with one-dimensional subspacesDirect sum of vector spaces and dimensiondimension of a direct sum is the sum of the separate dimensionssub spaces of vector spaces problemsA confusion of direct Sum in finite dimensional vector spaceEvery nontrivial subspace of a finite-dimensional vector space has a basis
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Decompose a finite dimensional vector space into a direct sum of $1$-dimensional sub-spaces.
Can infinite-dimensional vector spaces be decomposed into direct sum of its subspaces?Infinite dimensional vector space, and infinite dimensional subspaces.Dimension of Direct sum of same Vector SpacesProof regarding the direct sum of subspacesLinear Algebra Proof with one-dimensional subspacesDirect sum of vector spaces and dimensiondimension of a direct sum is the sum of the separate dimensionssub spaces of vector spaces problemsA confusion of direct Sum in finite dimensional vector spaceEvery nontrivial subspace of a finite-dimensional vector space has a basis
$begingroup$
Suppose $V$ is finite-dimensional, with $dim V= n$, for some $ngeq 1$.
Prove that there exist $1$-dimensional sub-spaces $U_1, ...., U_n$ of $V$ such that $V = U_1 oplus... oplus U_n$.
linear-algebra
edited Mar 13 at 18:43
Yanko
7,8801830
asked Mar 13 at 18:37
noobiskonoobisko
715
$endgroup$
add a comment |
$begingroup$
Suppose $V$ is finite-dimensional, with $dim V= n$, for some $ngeq 1$.
Prove that there exist $1$-dimensional sub-spaces $U_1, ...., U_n$ of $V$ such that $V = U_1 oplus... oplus U_n$.
linear-algebra
edited Mar 13 at 18:43
Yanko
7,8801830
asked Mar 13 at 18:37
noobiskonoobisko
715
$endgroup$
add a comment |
$begingroup$
Suppose $V$ is finite-dimensional, with $dim V= n$, for some $ngeq 1$.
Prove that there exist $1$-dimensional sub-spaces $U_1, ...., U_n$ of $V$ such that $V = U_1 oplus... oplus U_n$.
linear-algebra
edited Mar 13 at 18:43
Yanko
7,8801830
asked Mar 13 at 18:37
noobiskonoobisko
715
$endgroup$
Suppose $V$ is finite-dimensional, with $dim V= n$, for some $ngeq 1$.
Prove that there exist $1$-dimensional sub-spaces $U_1, ...., U_n$ of $V$ such that $V = U_1 oplus... oplus U_n$.
linear-algebra
linear-algebra
edited Mar 13 at 18:43
Yanko
7,8801830
asked Mar 13 at 18:37
noobiskonoobisko
715
edited Mar 13 at 18:43
Yanko
7,8801830
asked Mar 13 at 18:37
noobiskonoobisko
715
edited Mar 13 at 18:43
Yanko
7,8801830
edited Mar 13 at 18:43
Yanko
7,8801830
edited Mar 13 at 18:43
Yanko
7,8801830
7,8801830
asked Mar 13 at 18:37
noobiskonoobisko
715
asked Mar 13 at 18:37
noobiskonoobisko
715
asked Mar 13 at 18:37
noobiskonoobisko
715
715
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is given that $dim V = n$ so you can find a basis $v_1,...,v_n$.
Now let $U_i:=textspan v_i$ for $1leq i leq n$ and you're done.
answered Mar 13 at 18:41
YankoYanko
7,8801830
$endgroup$
$begingroup$
So from what i first understood, 1-dimensional means n=1. Therefore $U_1 := spanu_1$ so $V=U_1$...is that correct or do i not understand something.
$endgroup$
– noobisko
Mar 13 at 18:44
$begingroup$
@noobisko $n=dim V$. If a vector space $U$ is $1$-dimensional that means that $dim U = 1$. In my answer $U_i$ is spanned by one non-zero vector (by $v_i$) and so it has dimension $1$.
$endgroup$
– Yanko
Mar 13 at 18:45
add a comment |
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1 Answer
1
active
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1 Answer
1
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oldest
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active
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active
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votes
$begingroup$
It is given that $dim V = n$ so you can find a basis $v_1,...,v_n$.
Now let $U_i:=textspan v_i$ for $1leq i leq n$ and you're done.
answered Mar 13 at 18:41
YankoYanko
7,8801830
$endgroup$
$begingroup$
So from what i first understood, 1-dimensional means n=1. Therefore $U_1 := spanu_1$ so $V=U_1$...is that correct or do i not understand something.
$endgroup$
– noobisko
Mar 13 at 18:44
$begingroup$
@noobisko $n=dim V$. If a vector space $U$ is $1$-dimensional that means that $dim U = 1$. In my answer $U_i$ is spanned by one non-zero vector (by $v_i$) and so it has dimension $1$.
$endgroup$
– Yanko
Mar 13 at 18:45
add a comment |
$begingroup$
It is given that $dim V = n$ so you can find a basis $v_1,...,v_n$.
Now let $U_i:=textspan v_i$ for $1leq i leq n$ and you're done.
answered Mar 13 at 18:41
YankoYanko
7,8801830
$endgroup$
$begingroup$
So from what i first understood, 1-dimensional means n=1. Therefore $U_1 := spanu_1$ so $V=U_1$...is that correct or do i not understand something.
$endgroup$
– noobisko
Mar 13 at 18:44
$begingroup$
@noobisko $n=dim V$. If a vector space $U$ is $1$-dimensional that means that $dim U = 1$. In my answer $U_i$ is spanned by one non-zero vector (by $v_i$) and so it has dimension $1$.
$endgroup$
– Yanko
Mar 13 at 18:45
add a comment |
$begingroup$
It is given that $dim V = n$ so you can find a basis $v_1,...,v_n$.
Now let $U_i:=textspan v_i$ for $1leq i leq n$ and you're done.
answered Mar 13 at 18:41
YankoYanko
7,8801830
$endgroup$
It is given that $dim V = n$ so you can find a basis $v_1,...,v_n$.
Now let $U_i:=textspan v_i$ for $1leq i leq n$ and you're done.
answered Mar 13 at 18:41
YankoYanko
7,8801830
answered Mar 13 at 18:41
YankoYanko
7,8801830
answered Mar 13 at 18:41
YankoYanko
7,8801830
answered Mar 13 at 18:41
YankoYanko
7,8801830
7,8801830
$begingroup$
So from what i first understood, 1-dimensional means n=1. Therefore $U_1 := spanu_1$ so $V=U_1$...is that correct or do i not understand something.
$endgroup$
– noobisko
Mar 13 at 18:44
$begingroup$
@noobisko $n=dim V$. If a vector space $U$ is $1$-dimensional that means that $dim U = 1$. In my answer $U_i$ is spanned by one non-zero vector (by $v_i$) and so it has dimension $1$.
$endgroup$
– Yanko
Mar 13 at 18:45
add a comment |
$begingroup$
So from what i first understood, 1-dimensional means n=1. Therefore $U_1 := spanu_1$ so $V=U_1$...is that correct or do i not understand something.
$endgroup$
– noobisko
Mar 13 at 18:44
$begingroup$
@noobisko $n=dim V$. If a vector space $U$ is $1$-dimensional that means that $dim U = 1$. In my answer $U_i$ is spanned by one non-zero vector (by $v_i$) and so it has dimension $1$.
$endgroup$
– Yanko
Mar 13 at 18:45
$begingroup$
So from what i first understood, 1-dimensional means n=1. Therefore $U_1 := spanu_1$ so $V=U_1$...is that correct or do i not understand something.
$endgroup$
– noobisko
Mar 13 at 18:44
$begingroup$
So from what i first understood, 1-dimensional means n=1. Therefore $U_1 := spanu_1$ so $V=U_1$...is that correct or do i not understand something.
$endgroup$
– noobisko
Mar 13 at 18:44
$begingroup$
@noobisko $n=dim V$. If a vector space $U$ is $1$-dimensional that means that $dim U = 1$. In my answer $U_i$ is spanned by one non-zero vector (by $v_i$) and so it has dimension $1$.
$endgroup$
– Yanko
Mar 13 at 18:45
$begingroup$
@noobisko $n=dim V$. If a vector space $U$ is $1$-dimensional that means that $dim U = 1$. In my answer $U_i$ is spanned by one non-zero vector (by $v_i$) and so it has dimension $1$.
$endgroup$
– Yanko
Mar 13 at 18:45
add a comment |
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