Find basis for intersection of span of vectors and equation
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I have two subspaces $U=x_1+2x_2+x_3-3x_4+x_5=0$ and $V=text{span}((1, 2, 0, 1, -1)^t, (-1, 0, 3, 2, 0)^t, (1, 0, 0, 0, 1)^t, (0, 2, -3, -1, -3))$
I rewrote $U$ as
$
begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 & -1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{pmatrix}
begin{pmatrix} x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
$.
I also figured out that the last vector of $V$ is linearly dependent on the other three, so only the first three are necessary for the span, I put this into a matrix and performed row reduction:
$begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5 end{pmatrix}=
begin{pmatrix}
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & -2 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{pmatrix}
begin{pmatrix}
lambda_1 \ lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
Then I tried inserting the vectors of $V$ into the equation for $U$ like so: $(lambda_1 + lambda_4)+2(lambda_2-lambda_4)+(lambda_3-2lambda_4)-3(0)+0=0$. I then rewrote as
$begin{pmatrix}
lambda_1 \
lambda_2 \
lambda_3 \
lambda_4 \
end{pmatrix}
=
lambda_2
begin{pmatrix}
-2 \ 1 \ 0 \ 0
end{pmatrix}
+
lambda_3
begin{pmatrix}
-1 \ 0 \ 1\ 0
end{pmatrix}
+
lambda_4
begin{pmatrix}
3 \ 0 \ 0 \ 1
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 \
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
end{pmatrix}
begin{pmatrix}
lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
And by rewriting into row-echelon form, I found that all three columns together form a basis for $Ucap V$.
But using this suggestion, I found that the dimension of the intersection is 2, ( I put the first four (all of them) columns of the matrix in the rewritten $U$ form and the first three columns of the rewritten $V$ matrix next to each other and performed some row reduction) and found five pivot elements and two non-pivot variables, so I concluded that the dimension is two, in contradiction with all of the above.
So my question is, is my thinking incorrect in that I can put the vectors of one subspace into the equation of another, or am I making a mistake here, or did I not correctly apply the suggested answer link?
linear-algebra vector-spaces
asked Dec 1 at 12:41
The Coding Wombat
1879
add a comment |
up vote
1
down vote
favorite
I have two subspaces $U=x_1+2x_2+x_3-3x_4+x_5=0$ and $V=text{span}((1, 2, 0, 1, -1)^t, (-1, 0, 3, 2, 0)^t, (1, 0, 0, 0, 1)^t, (0, 2, -3, -1, -3))$
I rewrote $U$ as
$
begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 & -1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{pmatrix}
begin{pmatrix} x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
$.
I also figured out that the last vector of $V$ is linearly dependent on the other three, so only the first three are necessary for the span, I put this into a matrix and performed row reduction:
$begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5 end{pmatrix}=
begin{pmatrix}
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & -2 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{pmatrix}
begin{pmatrix}
lambda_1 \ lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
Then I tried inserting the vectors of $V$ into the equation for $U$ like so: $(lambda_1 + lambda_4)+2(lambda_2-lambda_4)+(lambda_3-2lambda_4)-3(0)+0=0$. I then rewrote as
$begin{pmatrix}
lambda_1 \
lambda_2 \
lambda_3 \
lambda_4 \
end{pmatrix}
=
lambda_2
begin{pmatrix}
-2 \ 1 \ 0 \ 0
end{pmatrix}
+
lambda_3
begin{pmatrix}
-1 \ 0 \ 1\ 0
end{pmatrix}
+
lambda_4
begin{pmatrix}
3 \ 0 \ 0 \ 1
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 \
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
end{pmatrix}
begin{pmatrix}
lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
And by rewriting into row-echelon form, I found that all three columns together form a basis for $Ucap V$.
But using this suggestion, I found that the dimension of the intersection is 2, ( I put the first four (all of them) columns of the matrix in the rewritten $U$ form and the first three columns of the rewritten $V$ matrix next to each other and performed some row reduction) and found five pivot elements and two non-pivot variables, so I concluded that the dimension is two, in contradiction with all of the above.
So my question is, is my thinking incorrect in that I can put the vectors of one subspace into the equation of another, or am I making a mistake here, or did I not correctly apply the suggested answer link?
linear-algebra vector-spaces
asked Dec 1 at 12:41
The Coding Wombat
1879
How are the columns of that last matrix a basis for any subspace of $mathbb R^5$ when they’re not even elements of $mathbb R^5$ in the first place?
– amd
Dec 2 at 1:59
@amd Hmm, Ucap V should indeed be a subspace of R^5, where is my analysis going wrong?
– The Coding Wombat
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have two subspaces $U=x_1+2x_2+x_3-3x_4+x_5=0$ and $V=text{span}((1, 2, 0, 1, -1)^t, (-1, 0, 3, 2, 0)^t, (1, 0, 0, 0, 1)^t, (0, 2, -3, -1, -3))$
I rewrote $U$ as
$
begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 & -1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{pmatrix}
begin{pmatrix} x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
$.
I also figured out that the last vector of $V$ is linearly dependent on the other three, so only the first three are necessary for the span, I put this into a matrix and performed row reduction:
$begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5 end{pmatrix}=
begin{pmatrix}
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & -2 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{pmatrix}
begin{pmatrix}
lambda_1 \ lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
Then I tried inserting the vectors of $V$ into the equation for $U$ like so: $(lambda_1 + lambda_4)+2(lambda_2-lambda_4)+(lambda_3-2lambda_4)-3(0)+0=0$. I then rewrote as
$begin{pmatrix}
lambda_1 \
lambda_2 \
lambda_3 \
lambda_4 \
end{pmatrix}
=
lambda_2
begin{pmatrix}
-2 \ 1 \ 0 \ 0
end{pmatrix}
+
lambda_3
begin{pmatrix}
-1 \ 0 \ 1\ 0
end{pmatrix}
+
lambda_4
begin{pmatrix}
3 \ 0 \ 0 \ 1
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 \
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
end{pmatrix}
begin{pmatrix}
lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
And by rewriting into row-echelon form, I found that all three columns together form a basis for $Ucap V$.
But using this suggestion, I found that the dimension of the intersection is 2, ( I put the first four (all of them) columns of the matrix in the rewritten $U$ form and the first three columns of the rewritten $V$ matrix next to each other and performed some row reduction) and found five pivot elements and two non-pivot variables, so I concluded that the dimension is two, in contradiction with all of the above.
So my question is, is my thinking incorrect in that I can put the vectors of one subspace into the equation of another, or am I making a mistake here, or did I not correctly apply the suggested answer link?
linear-algebra vector-spaces
asked Dec 1 at 12:41
The Coding Wombat
1879
I have two subspaces $U=x_1+2x_2+x_3-3x_4+x_5=0$ and $V=text{span}((1, 2, 0, 1, -1)^t, (-1, 0, 3, 2, 0)^t, (1, 0, 0, 0, 1)^t, (0, 2, -3, -1, -3))$
I rewrote $U$ as
$
begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 & -1 \
1 & 0 & 0 & 0 \
0 & 1 & 0 & 0 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{pmatrix}
begin{pmatrix} x_2 \ x_3 \ x_4 \ x_5
end{pmatrix}
$.
I also figured out that the last vector of $V$ is linearly dependent on the other three, so only the first three are necessary for the span, I put this into a matrix and performed row reduction:
$begin{pmatrix}x_1 \ x_2 \ x_3 \ x_4 \ x_5 end{pmatrix}=
begin{pmatrix}
1 & 0 & 0 & 1 \
0 & 1 & 0 & -1 \
0 & 0 & 1 & -2 \
0 & 0 & 0 & 0 \
0 & 0 & 0 & 0
end{pmatrix}
begin{pmatrix}
lambda_1 \ lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
Then I tried inserting the vectors of $V$ into the equation for $U$ like so: $(lambda_1 + lambda_4)+2(lambda_2-lambda_4)+(lambda_3-2lambda_4)-3(0)+0=0$. I then rewrote as
$begin{pmatrix}
lambda_1 \
lambda_2 \
lambda_3 \
lambda_4 \
end{pmatrix}
=
lambda_2
begin{pmatrix}
-2 \ 1 \ 0 \ 0
end{pmatrix}
+
lambda_3
begin{pmatrix}
-1 \ 0 \ 1\ 0
end{pmatrix}
+
lambda_4
begin{pmatrix}
3 \ 0 \ 0 \ 1
end{pmatrix}
=
begin{pmatrix}
-2 & -1 & 3 \
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 1
end{pmatrix}
begin{pmatrix}
lambda_2 \ lambda_3 \ lambda_4
end{pmatrix}$
And by rewriting into row-echelon form, I found that all three columns together form a basis for $Ucap V$.
But using this suggestion, I found that the dimension of the intersection is 2, ( I put the first four (all of them) columns of the matrix in the rewritten $U$ form and the first three columns of the rewritten $V$ matrix next to each other and performed some row reduction) and found five pivot elements and two non-pivot variables, so I concluded that the dimension is two, in contradiction with all of the above.
So my question is, is my thinking incorrect in that I can put the vectors of one subspace into the equation of another, or am I making a mistake here, or did I not correctly apply the suggested answer link?
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Dec 1 at 12:41
The Coding Wombat
1879
asked Dec 1 at 12:41
The Coding Wombat
1879
edited 2 days ago
asked Dec 1 at 12:41
The Coding Wombat
1879
asked Dec 1 at 12:41
The Coding Wombat
1879
asked Dec 1 at 12:41
The Coding Wombat
1879
1879
How are the columns of that last matrix a basis for any subspace of $mathbb R^5$ when they’re not even elements of $mathbb R^5$ in the first place?
– amd
Dec 2 at 1:59
@amd Hmm, Ucap V should indeed be a subspace of R^5, where is my analysis going wrong?
– The Coding Wombat
2 days ago
add a comment |
How are the columns of that last matrix a basis for any subspace of $mathbb R^5$ when they’re not even elements of $mathbb R^5$ in the first place?
– amd
Dec 2 at 1:59
@amd Hmm, Ucap V should indeed be a subspace of R^5, where is my analysis going wrong?
– The Coding Wombat
2 days ago
How are the columns of that last matrix a basis for any subspace of $mathbb R^5$ when they’re not even elements of $mathbb R^5$ in the first place?
– amd
Dec 2 at 1:59
How are the columns of that last matrix a basis for any subspace of $mathbb R^5$ when they’re not even elements of $mathbb R^5$ in the first place?
– amd
Dec 2 at 1:59
@amd Hmm, Ucap V should indeed be a subspace of R^5, where is my analysis going wrong?
– The Coding Wombat
2 days ago
@amd Hmm, Ucap V should indeed be a subspace of R^5, where is my analysis going wrong?
– The Coding Wombat
2 days ago
add a comment |
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How are the columns of that last matrix a basis for any subspace of $mathbb R^5$ when they’re not even elements of $mathbb R^5$ in the first place?
– amd
Dec 2 at 1:59
@amd Hmm, Ucap V should indeed be a subspace of R^5, where is my analysis going wrong?
– The Coding Wombat
2 days ago