A question about Span in Linear Algebra The Next CEO of Stack OverflowA question about subspacesFind an orthogonal vector under the constraints describedFind orthonormal basis of $mathbbR^3$ with a given span of two basis vectorsName of outer “product” of two vectors with exponentiation instead of multiplicationInternal Direct Sum Linear Algebra Proof QuestionLinear transformation linear algebralinear algebra span of four unknown vectors.Prove that every element in $V$ is on the form $vecv$linear system of equations and the intersection of two spansIntersection of two subspaces is the span of the vector b
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A question about Span in Linear Algebra
The Next CEO of Stack OverflowA question about subspacesFind an orthogonal vector under the constraints describedFind orthonormal basis of $mathbbR^3$ with a given span of two basis vectorsName of outer “product” of two vectors with exponentiation instead of multiplicationInternal Direct Sum Linear Algebra Proof QuestionLinear transformation linear algebralinear algebra span of four unknown vectors.Prove that every element in $V$ is on the form $vecv$linear system of equations and the intersection of two spansIntersection of two subspaces is the span of the vector b
$begingroup$
So I have this question about Span that I'm not sure how to do:
Let $DeclareMathOperatorSpanSpanu_1 =beginbmatrix
1\
2\
3
endbmatrix,
u_2 =beginbmatrix
2\
3\
4
endbmatrix,
v_1 =beginbmatrix
1\
1\
2
endbmatrix,
v_2 =beginbmatrix
2\
2\
3
end bmatrix$, in the real vector space $mathbbR^3$.
Let $U=Span(u_1,u_2)$ and $V=Span(v_1, v_2)$.
Show that $Ucap V$ is equal to $Span(v)$ where $v = beginbmatrix
1\
1\
1
endbmatrix$.
So I have already shown that all elements in $V$ can be written as $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ (1)
and that $beginbmatrix
1 & 2\
2 & 3\
3 & 4
endbmatrix
x = beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ only has a solution if $alpha = beta$ (2). I have also shown that $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ is equal to $betabeginbmatrix
1\
1\
1
endbmatrix = beta v$ when $alpha = beta$. (3)
I know that I have to show that $Ucap V ⊆ Span(v)$ and that $Span(v) ⊆ Ucap V$.
My answer so far:
$Ucap V ⊆ Span(v):$
Since we know that (3) and (2) is true then every element in $V$ can be written as a linear combination of $v$ which means that $V=Span(v)$. Since $Ucap V⊆ V Rightarrow Ucap V ⊆ Span(v)$.
However I'm not sure how to show that $Span(v) ⊆Ucap V$. Any help would be appreciated!
Edit:
With help from the comments I think I have found the answer:
Since $Span(v)=alpha in mathbbR$ then $alpha v=alpha cdot (−gamma)cdot v_1 + beta cdot v_2$, where $beta=−gamma$ and $alpha cdot v=alpha cdot (−gamma)cdot u_1 + alpha cdot delta cdot u_2$ where $delta=−gamma$ so $Span(v)⊆Ucap V$.
Are both these arguments correct?
linear-algebra
asked Mar 27 at 19:02
NikolajNikolaj
578
$endgroup$
add a comment |
$begingroup$
So I have this question about Span that I'm not sure how to do:
Let $DeclareMathOperatorSpanSpanu_1 =beginbmatrix
1\
2\
3
endbmatrix,
u_2 =beginbmatrix
2\
3\
4
endbmatrix,
v_1 =beginbmatrix
1\
1\
2
endbmatrix,
v_2 =beginbmatrix
2\
2\
3
end bmatrix$, in the real vector space $mathbbR^3$.
Let $U=Span(u_1,u_2)$ and $V=Span(v_1, v_2)$.
Show that $Ucap V$ is equal to $Span(v)$ where $v = beginbmatrix
1\
1\
1
endbmatrix$.
So I have already shown that all elements in $V$ can be written as $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ (1)
and that $beginbmatrix
1 & 2\
2 & 3\
3 & 4
endbmatrix
x = beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ only has a solution if $alpha = beta$ (2). I have also shown that $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ is equal to $betabeginbmatrix
1\
1\
1
endbmatrix = beta v$ when $alpha = beta$. (3)
I know that I have to show that $Ucap V ⊆ Span(v)$ and that $Span(v) ⊆ Ucap V$.
My answer so far:
$Ucap V ⊆ Span(v):$
Since we know that (3) and (2) is true then every element in $V$ can be written as a linear combination of $v$ which means that $V=Span(v)$. Since $Ucap V⊆ V Rightarrow Ucap V ⊆ Span(v)$.
However I'm not sure how to show that $Span(v) ⊆Ucap V$. Any help would be appreciated!
Edit:
With help from the comments I think I have found the answer:
Since $Span(v)=alpha in mathbbR$ then $alpha v=alpha cdot (−gamma)cdot v_1 + beta cdot v_2$, where $beta=−gamma$ and $alpha cdot v=alpha cdot (−gamma)cdot u_1 + alpha cdot delta cdot u_2$ where $delta=−gamma$ so $Span(v)⊆Ucap V$.
Are both these arguments correct?
linear-algebra
asked Mar 27 at 19:02
NikolajNikolaj
578
$endgroup$
$begingroup$
It could be helpful to know the field underlying the vector space. Is it a vector space over real numbers?
$endgroup$
– NazimJ
Mar 27 at 20:02
$begingroup$
It is in the real vectorspace $mathbbR^3$. I have added it.
$endgroup$
– Nikolaj
Mar 27 at 20:17
add a comment |
$begingroup$
So I have this question about Span that I'm not sure how to do:
Let $DeclareMathOperatorSpanSpanu_1 =beginbmatrix
1\
2\
3
endbmatrix,
u_2 =beginbmatrix
2\
3\
4
endbmatrix,
v_1 =beginbmatrix
1\
1\
2
endbmatrix,
v_2 =beginbmatrix
2\
2\
3
end bmatrix$, in the real vector space $mathbbR^3$.
Let $U=Span(u_1,u_2)$ and $V=Span(v_1, v_2)$.
Show that $Ucap V$ is equal to $Span(v)$ where $v = beginbmatrix
1\
1\
1
endbmatrix$.
So I have already shown that all elements in $V$ can be written as $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ (1)
and that $beginbmatrix
1 & 2\
2 & 3\
3 & 4
endbmatrix
x = beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ only has a solution if $alpha = beta$ (2). I have also shown that $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ is equal to $betabeginbmatrix
1\
1\
1
endbmatrix = beta v$ when $alpha = beta$. (3)
I know that I have to show that $Ucap V ⊆ Span(v)$ and that $Span(v) ⊆ Ucap V$.
My answer so far:
$Ucap V ⊆ Span(v):$
Since we know that (3) and (2) is true then every element in $V$ can be written as a linear combination of $v$ which means that $V=Span(v)$. Since $Ucap V⊆ V Rightarrow Ucap V ⊆ Span(v)$.
However I'm not sure how to show that $Span(v) ⊆Ucap V$. Any help would be appreciated!
Edit:
With help from the comments I think I have found the answer:
Since $Span(v)=alpha in mathbbR$ then $alpha v=alpha cdot (−gamma)cdot v_1 + beta cdot v_2$, where $beta=−gamma$ and $alpha cdot v=alpha cdot (−gamma)cdot u_1 + alpha cdot delta cdot u_2$ where $delta=−gamma$ so $Span(v)⊆Ucap V$.
Are both these arguments correct?
linear-algebra
asked Mar 27 at 19:02
NikolajNikolaj
578
$endgroup$
So I have this question about Span that I'm not sure how to do:
Let $DeclareMathOperatorSpanSpanu_1 =beginbmatrix
1\
2\
3
endbmatrix,
u_2 =beginbmatrix
2\
3\
4
endbmatrix,
v_1 =beginbmatrix
1\
1\
2
endbmatrix,
v_2 =beginbmatrix
2\
2\
3
end bmatrix$, in the real vector space $mathbbR^3$.
Let $U=Span(u_1,u_2)$ and $V=Span(v_1, v_2)$.
Show that $Ucap V$ is equal to $Span(v)$ where $v = beginbmatrix
1\
1\
1
endbmatrix$.
So I have already shown that all elements in $V$ can be written as $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ (1)
and that $beginbmatrix
1 & 2\
2 & 3\
3 & 4
endbmatrix
x = beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ only has a solution if $alpha = beta$ (2). I have also shown that $beginbmatrix
alpha +2beta\
alpha + 2beta\
2alpha + 3beta
endbmatrix$ is equal to $betabeginbmatrix
1\
1\
1
endbmatrix = beta v$ when $alpha = beta$. (3)
I know that I have to show that $Ucap V ⊆ Span(v)$ and that $Span(v) ⊆ Ucap V$.
My answer so far:
$Ucap V ⊆ Span(v):$
Since we know that (3) and (2) is true then every element in $V$ can be written as a linear combination of $v$ which means that $V=Span(v)$. Since $Ucap V⊆ V Rightarrow Ucap V ⊆ Span(v)$.
However I'm not sure how to show that $Span(v) ⊆Ucap V$. Any help would be appreciated!
Edit:
With help from the comments I think I have found the answer:
Since $Span(v)=alpha in mathbbR$ then $alpha v=alpha cdot (−gamma)cdot v_1 + beta cdot v_2$, where $beta=−gamma$ and $alpha cdot v=alpha cdot (−gamma)cdot u_1 + alpha cdot delta cdot u_2$ where $delta=−gamma$ so $Span(v)⊆Ucap V$.
Are both these arguments correct?
linear-algebra
linear-algebra
asked Mar 27 at 19:02
NikolajNikolaj
578
asked Mar 27 at 19:02
NikolajNikolaj
578
edited Mar 27 at 23:09
Nikolaj
asked Mar 27 at 19:02
NikolajNikolaj
578
asked Mar 27 at 19:02
NikolajNikolaj
578
asked Mar 27 at 19:02
NikolajNikolaj
578
578
$begingroup$
It could be helpful to know the field underlying the vector space. Is it a vector space over real numbers?
$endgroup$
– NazimJ
Mar 27 at 20:02
$begingroup$
It is in the real vectorspace $mathbbR^3$. I have added it.
$endgroup$
– Nikolaj
Mar 27 at 20:17
add a comment |
$begingroup$
It could be helpful to know the field underlying the vector space. Is it a vector space over real numbers?
$endgroup$
– NazimJ
Mar 27 at 20:02
$begingroup$
It is in the real vectorspace $mathbbR^3$. I have added it.
$endgroup$
– Nikolaj
Mar 27 at 20:17
$begingroup$
It could be helpful to know the field underlying the vector space. Is it a vector space over real numbers?
$endgroup$
– NazimJ
Mar 27 at 20:02
$begingroup$
It could be helpful to know the field underlying the vector space. Is it a vector space over real numbers?
$endgroup$
– NazimJ
Mar 27 at 20:02
$begingroup$
It is in the real vectorspace $mathbbR^3$. I have added it.
$endgroup$
– Nikolaj
Mar 27 at 20:17
$begingroup$
It is in the real vectorspace $mathbbR^3$. I have added it.
$endgroup$
– Nikolaj
Mar 27 at 20:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Can you show that $v in U cap V$?
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
$endgroup$
1
$begingroup$
@Javi It's not a critique or a clarification request. It's a hint meant to guide OP to the answer and is fairly standard practice on MSE. See math.meta.stackexchange.com/questions/28969/…
$endgroup$
– Michael Biro
Mar 27 at 21:01
$begingroup$
yes $v = -u_1 + u_2 = -v_1 + v_2$ (egreg also showed this. Not sure how to proceed from here though.
$endgroup$
– Nikolaj
Mar 27 at 22:24
1
$begingroup$
Ahh so since $Span(v)=alpha v$ then $alpha v = alpha cdot (-gamma) cdot v_1 + beta cdot v_2,$ where $beta = - gamma$ and $alpha v = alpha cdot (-lambda) cdot u_1 + alpha cdot delta cdot u_2$ where $delta = -lambda$ so $Span(v) ⊆ Ucap V$
$endgroup$
– Nikolaj
Mar 27 at 22:56
$begingroup$
@Nikolaj That's right! More generally, every linear combination of vectors in a span is a vector in the span.
$endgroup$
– Michael Biro
Mar 28 at 0:38
add a comment |
$begingroup$
Suppose a vector belongs to the intersection, so it can be written as
$$
alphabeginbmatrix 1 \ 2 \ 3 endbmatrix
+
betabeginbmatrix 2 \ 3 \ 4 endbmatrix
=
gammabeginbmatrix 1 \ 1 \ 2 endbmatrix
+
deltabeginbmatrix 2 \ 2 \ 3 end bmatrix
$$
Let's then look at the null space of the matrix
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
$$
A standard elimination yields
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & -1 & 1 & 2 \
0 & -2 & 1 & 3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & -1 & -1
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 2 & 0 & -1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
$$
Thus we get $alpha=-delta$, $beta=delta$, $gamma=-delta$.
Choosing $delta=1$ and $gamma=-1$, we get the vector
$$
v=-u_1+u_2=-v_1+v_2=beginbmatrix 1 \ 1 \ 1 endbmatrixin Ucap V
$$
The equations imply that the dimension of $Ucap V$ is $1$, so $Ucap V=operatornameSpanv$. If you want a confirm of this, the elimination implies that the rank of $[u_1 u_2 v_1 v_2]$, which equals the rank of $[u_1 u_2 -v_1 -v_2]$ is $3$ and Grassmann's formula yields $dim(Ucap V)=dim U+dim V-dim(U+V)=2+2-3=1$.
answered Mar 27 at 21:34
egregegreg
185k1486206
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Can you show that $v in U cap V$?
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
$endgroup$
1
$begingroup$
@Javi It's not a critique or a clarification request. It's a hint meant to guide OP to the answer and is fairly standard practice on MSE. See math.meta.stackexchange.com/questions/28969/…
$endgroup$
– Michael Biro
Mar 27 at 21:01
$begingroup$
yes $v = -u_1 + u_2 = -v_1 + v_2$ (egreg also showed this. Not sure how to proceed from here though.
$endgroup$
– Nikolaj
Mar 27 at 22:24
1
$begingroup$
Ahh so since $Span(v)=alpha v$ then $alpha v = alpha cdot (-gamma) cdot v_1 + beta cdot v_2,$ where $beta = - gamma$ and $alpha v = alpha cdot (-lambda) cdot u_1 + alpha cdot delta cdot u_2$ where $delta = -lambda$ so $Span(v) ⊆ Ucap V$
$endgroup$
– Nikolaj
Mar 27 at 22:56
$begingroup$
@Nikolaj That's right! More generally, every linear combination of vectors in a span is a vector in the span.
$endgroup$
– Michael Biro
Mar 28 at 0:38
add a comment |
$begingroup$
Hint: Can you show that $v in U cap V$?
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
$endgroup$
1
$begingroup$
@Javi It's not a critique or a clarification request. It's a hint meant to guide OP to the answer and is fairly standard practice on MSE. See math.meta.stackexchange.com/questions/28969/…
$endgroup$
– Michael Biro
Mar 27 at 21:01
$begingroup$
yes $v = -u_1 + u_2 = -v_1 + v_2$ (egreg also showed this. Not sure how to proceed from here though.
$endgroup$
– Nikolaj
Mar 27 at 22:24
1
$begingroup$
Ahh so since $Span(v)=alpha v$ then $alpha v = alpha cdot (-gamma) cdot v_1 + beta cdot v_2,$ where $beta = - gamma$ and $alpha v = alpha cdot (-lambda) cdot u_1 + alpha cdot delta cdot u_2$ where $delta = -lambda$ so $Span(v) ⊆ Ucap V$
$endgroup$
– Nikolaj
Mar 27 at 22:56
$begingroup$
@Nikolaj That's right! More generally, every linear combination of vectors in a span is a vector in the span.
$endgroup$
– Michael Biro
Mar 28 at 0:38
add a comment |
$begingroup$
Hint: Can you show that $v in U cap V$?
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
$endgroup$
Hint: Can you show that $v in U cap V$?
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
answered Mar 27 at 19:50
Michael BiroMichael Biro
11.4k21831
11.4k21831
1
$begingroup$
@Javi It's not a critique or a clarification request. It's a hint meant to guide OP to the answer and is fairly standard practice on MSE. See math.meta.stackexchange.com/questions/28969/…
$endgroup$
– Michael Biro
Mar 27 at 21:01
$begingroup$
yes $v = -u_1 + u_2 = -v_1 + v_2$ (egreg also showed this. Not sure how to proceed from here though.
$endgroup$
– Nikolaj
Mar 27 at 22:24
1
$begingroup$
Ahh so since $Span(v)=alpha v$ then $alpha v = alpha cdot (-gamma) cdot v_1 + beta cdot v_2,$ where $beta = - gamma$ and $alpha v = alpha cdot (-lambda) cdot u_1 + alpha cdot delta cdot u_2$ where $delta = -lambda$ so $Span(v) ⊆ Ucap V$
$endgroup$
– Nikolaj
Mar 27 at 22:56
$begingroup$
@Nikolaj That's right! More generally, every linear combination of vectors in a span is a vector in the span.
$endgroup$
– Michael Biro
Mar 28 at 0:38
add a comment |
1
$begingroup$
@Javi It's not a critique or a clarification request. It's a hint meant to guide OP to the answer and is fairly standard practice on MSE. See math.meta.stackexchange.com/questions/28969/…
$endgroup$
– Michael Biro
Mar 27 at 21:01
$begingroup$
yes $v = -u_1 + u_2 = -v_1 + v_2$ (egreg also showed this. Not sure how to proceed from here though.
$endgroup$
– Nikolaj
Mar 27 at 22:24
1
$begingroup$
Ahh so since $Span(v)=alpha v$ then $alpha v = alpha cdot (-gamma) cdot v_1 + beta cdot v_2,$ where $beta = - gamma$ and $alpha v = alpha cdot (-lambda) cdot u_1 + alpha cdot delta cdot u_2$ where $delta = -lambda$ so $Span(v) ⊆ Ucap V$
$endgroup$
– Nikolaj
Mar 27 at 22:56
$begingroup$
@Nikolaj That's right! More generally, every linear combination of vectors in a span is a vector in the span.
$endgroup$
– Michael Biro
Mar 28 at 0:38
1
1
$begingroup$
@Javi It's not a critique or a clarification request. It's a hint meant to guide OP to the answer and is fairly standard practice on MSE. See math.meta.stackexchange.com/questions/28969/…
$endgroup$
– Michael Biro
Mar 27 at 21:01
$begingroup$
@Javi It's not a critique or a clarification request. It's a hint meant to guide OP to the answer and is fairly standard practice on MSE. See math.meta.stackexchange.com/questions/28969/…
$endgroup$
– Michael Biro
Mar 27 at 21:01
$begingroup$
yes $v = -u_1 + u_2 = -v_1 + v_2$ (egreg also showed this. Not sure how to proceed from here though.
$endgroup$
– Nikolaj
Mar 27 at 22:24
$begingroup$
yes $v = -u_1 + u_2 = -v_1 + v_2$ (egreg also showed this. Not sure how to proceed from here though.
$endgroup$
– Nikolaj
Mar 27 at 22:24
1
1
$begingroup$
Ahh so since $Span(v)=alpha v$ then $alpha v = alpha cdot (-gamma) cdot v_1 + beta cdot v_2,$ where $beta = - gamma$ and $alpha v = alpha cdot (-lambda) cdot u_1 + alpha cdot delta cdot u_2$ where $delta = -lambda$ so $Span(v) ⊆ Ucap V$
$endgroup$
– Nikolaj
Mar 27 at 22:56
$begingroup$
Ahh so since $Span(v)=alpha v$ then $alpha v = alpha cdot (-gamma) cdot v_1 + beta cdot v_2,$ where $beta = - gamma$ and $alpha v = alpha cdot (-lambda) cdot u_1 + alpha cdot delta cdot u_2$ where $delta = -lambda$ so $Span(v) ⊆ Ucap V$
$endgroup$
– Nikolaj
Mar 27 at 22:56
$begingroup$
@Nikolaj That's right! More generally, every linear combination of vectors in a span is a vector in the span.
$endgroup$
– Michael Biro
Mar 28 at 0:38
$begingroup$
@Nikolaj That's right! More generally, every linear combination of vectors in a span is a vector in the span.
$endgroup$
– Michael Biro
Mar 28 at 0:38
add a comment |
$begingroup$
Suppose a vector belongs to the intersection, so it can be written as
$$
alphabeginbmatrix 1 \ 2 \ 3 endbmatrix
+
betabeginbmatrix 2 \ 3 \ 4 endbmatrix
=
gammabeginbmatrix 1 \ 1 \ 2 endbmatrix
+
deltabeginbmatrix 2 \ 2 \ 3 end bmatrix
$$
Let's then look at the null space of the matrix
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
$$
A standard elimination yields
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & -1 & 1 & 2 \
0 & -2 & 1 & 3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & -1 & -1
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 2 & 0 & -1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
$$
Thus we get $alpha=-delta$, $beta=delta$, $gamma=-delta$.
Choosing $delta=1$ and $gamma=-1$, we get the vector
$$
v=-u_1+u_2=-v_1+v_2=beginbmatrix 1 \ 1 \ 1 endbmatrixin Ucap V
$$
The equations imply that the dimension of $Ucap V$ is $1$, so $Ucap V=operatornameSpanv$. If you want a confirm of this, the elimination implies that the rank of $[u_1 u_2 v_1 v_2]$, which equals the rank of $[u_1 u_2 -v_1 -v_2]$ is $3$ and Grassmann's formula yields $dim(Ucap V)=dim U+dim V-dim(U+V)=2+2-3=1$.
answered Mar 27 at 21:34
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Suppose a vector belongs to the intersection, so it can be written as
$$
alphabeginbmatrix 1 \ 2 \ 3 endbmatrix
+
betabeginbmatrix 2 \ 3 \ 4 endbmatrix
=
gammabeginbmatrix 1 \ 1 \ 2 endbmatrix
+
deltabeginbmatrix 2 \ 2 \ 3 end bmatrix
$$
Let's then look at the null space of the matrix
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
$$
A standard elimination yields
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & -1 & 1 & 2 \
0 & -2 & 1 & 3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & -1 & -1
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 2 & 0 & -1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
$$
Thus we get $alpha=-delta$, $beta=delta$, $gamma=-delta$.
Choosing $delta=1$ and $gamma=-1$, we get the vector
$$
v=-u_1+u_2=-v_1+v_2=beginbmatrix 1 \ 1 \ 1 endbmatrixin Ucap V
$$
The equations imply that the dimension of $Ucap V$ is $1$, so $Ucap V=operatornameSpanv$. If you want a confirm of this, the elimination implies that the rank of $[u_1 u_2 v_1 v_2]$, which equals the rank of $[u_1 u_2 -v_1 -v_2]$ is $3$ and Grassmann's formula yields $dim(Ucap V)=dim U+dim V-dim(U+V)=2+2-3=1$.
answered Mar 27 at 21:34
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
Suppose a vector belongs to the intersection, so it can be written as
$$
alphabeginbmatrix 1 \ 2 \ 3 endbmatrix
+
betabeginbmatrix 2 \ 3 \ 4 endbmatrix
=
gammabeginbmatrix 1 \ 1 \ 2 endbmatrix
+
deltabeginbmatrix 2 \ 2 \ 3 end bmatrix
$$
Let's then look at the null space of the matrix
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
$$
A standard elimination yields
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & -1 & 1 & 2 \
0 & -2 & 1 & 3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & -1 & -1
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 2 & 0 & -1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
$$
Thus we get $alpha=-delta$, $beta=delta$, $gamma=-delta$.
Choosing $delta=1$ and $gamma=-1$, we get the vector
$$
v=-u_1+u_2=-v_1+v_2=beginbmatrix 1 \ 1 \ 1 endbmatrixin Ucap V
$$
The equations imply that the dimension of $Ucap V$ is $1$, so $Ucap V=operatornameSpanv$. If you want a confirm of this, the elimination implies that the rank of $[u_1 u_2 v_1 v_2]$, which equals the rank of $[u_1 u_2 -v_1 -v_2]$ is $3$ and Grassmann's formula yields $dim(Ucap V)=dim U+dim V-dim(U+V)=2+2-3=1$.
answered Mar 27 at 21:34
egregegreg
185k1486206
$endgroup$
Suppose a vector belongs to the intersection, so it can be written as
$$
alphabeginbmatrix 1 \ 2 \ 3 endbmatrix
+
betabeginbmatrix 2 \ 3 \ 4 endbmatrix
=
gammabeginbmatrix 1 \ 1 \ 2 endbmatrix
+
deltabeginbmatrix 2 \ 2 \ 3 end bmatrix
$$
Let's then look at the null space of the matrix
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
$$
A standard elimination yields
$$
beginbmatrix
1 & 2 & -1 & -2 \
2 & 3 & -1 & -2 \
3 & 4 & -2 & -3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & -1 & 1 & 2 \
0 & -2 & 1 & 3
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & -1 & -1
endbmatrix
to
beginbmatrix
1 & 2 & -1 & -2 \
0 & 1 & -1 & -2 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 2 & 0 & -1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
to
beginbmatrix
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & 1
endbmatrix
$$
Thus we get $alpha=-delta$, $beta=delta$, $gamma=-delta$.
Choosing $delta=1$ and $gamma=-1$, we get the vector
$$
v=-u_1+u_2=-v_1+v_2=beginbmatrix 1 \ 1 \ 1 endbmatrixin Ucap V
$$
The equations imply that the dimension of $Ucap V$ is $1$, so $Ucap V=operatornameSpanv$. If you want a confirm of this, the elimination implies that the rank of $[u_1 u_2 v_1 v_2]$, which equals the rank of $[u_1 u_2 -v_1 -v_2]$ is $3$ and Grassmann's formula yields $dim(Ucap V)=dim U+dim V-dim(U+V)=2+2-3=1$.
answered Mar 27 at 21:34
egregegreg
185k1486206
answered Mar 27 at 21:34
egregegreg
185k1486206
answered Mar 27 at 21:34
egregegreg
185k1486206
answered Mar 27 at 21:34
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
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$begingroup$
It could be helpful to know the field underlying the vector space. Is it a vector space over real numbers?
$endgroup$
– NazimJ
Mar 27 at 20:02
$begingroup$
It is in the real vectorspace $mathbbR^3$. I have added it.
$endgroup$
– Nikolaj
Mar 27 at 20:17