Finding matrices which P is null space
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Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$
a) Find one base of subspace $P^bot$
b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.
a) I found base $(1,1,1,1)$ that is easy
b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are
A=$beginbmatrix
1& 1& 1& 1\
1& 1& 1& 1\
1 & 1& 1& 1\
1 & 1& 1& 1
endbmatrix$.
linear-algebra matrices vector-spaces
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up vote
1
down vote
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Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$
a) Find one base of subspace $P^bot$
b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.
a) I found base $(1,1,1,1)$ that is easy
b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are
A=$beginbmatrix
1& 1& 1& 1\
1& 1& 1& 1\
1 & 1& 1& 1\
1 & 1& 1& 1
endbmatrix$.
linear-algebra matrices vector-spaces
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$
a) Find one base of subspace $P^bot$
b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.
a) I found base $(1,1,1,1)$ that is easy
b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are
A=$beginbmatrix
1& 1& 1& 1\
1& 1& 1& 1\
1 & 1& 1& 1\
1 & 1& 1& 1
endbmatrix$.
linear-algebra matrices vector-spaces
Let P is set all vectors $(x_1,x_2,x_3,x_4) in R^4$ for which $x_1 + x_2 + x_3 + x_4=0$
a) Find one base of subspace $P^bot$
b) Prove that P is subspace of $R^4$ then construct matrices which P is null space.
a) I found base $(1,1,1,1)$ that is easy
b) I know how to prove that $P$ is subspace, but I do not know if my matrix is good, because what means that P is null space, does that mean that for every vector $xin P$ $Ax=0$. So matrices $A$ are
A=$beginbmatrix
1& 1& 1& 1\
1& 1& 1& 1\
1 & 1& 1& 1\
1 & 1& 1& 1
endbmatrix$.
linear-algebra matrices vector-spaces
edited Aug 20 at 10:17
Bernard
111k635103
111k635103
asked Aug 20 at 10:15
Marko à  koriÃÂ
876
876
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3 Answers
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Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.
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a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.
b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.
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Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices
$$beginbmatrix
a& a& a& a\
a& a& a& a\
vdots& vdots& vdots& vdots\
a & a& a& a
endbmatrix$$
with $a neq 0$.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.
add a comment |Â
up vote
0
down vote
accepted
Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.
Your answers are fine, but I don't know if you are expected to find just one matrix whose null space is $P$. The question says 'matrices'.
edited Aug 20 at 11:39
Martin Sleziak
43.5k6113260
43.5k6113260
answered Aug 20 at 10:23
Kavi Rama Murthy
23.3k2933
23.3k2933
add a comment |Â
add a comment |Â
up vote
0
down vote
a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.
b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.
add a comment |Â
up vote
0
down vote
a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.
b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.
b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.
a) The vector $(1,1,1,1)$ is indeed the basis vector for $P^bot$. Just a warning that the question may implicitly also require you to prove your claim, which is not difficult in this particular case, but may cause loss of points.
b) Your matrix is ok, but you could also just have the matrix $[1,1,1,1]$ which has the same nullspace.
edited Aug 20 at 10:23
answered Aug 20 at 10:17
5xum
82.4k383147
82.4k383147
add a comment |Â
add a comment |Â
up vote
0
down vote
Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices
$$beginbmatrix
a& a& a& a\
a& a& a& a\
vdots& vdots& vdots& vdots\
a & a& a& a
endbmatrix$$
with $a neq 0$.
add a comment |Â
up vote
0
down vote
Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices
$$beginbmatrix
a& a& a& a\
a& a& a& a\
vdots& vdots& vdots& vdots\
a & a& a& a
endbmatrix$$
with $a neq 0$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices
$$beginbmatrix
a& a& a& a\
a& a& a& a\
vdots& vdots& vdots& vdots\
a & a& a& a
endbmatrix$$
with $a neq 0$.
Yes that's correct, a basis for $P$ is $(1,-1,0,0),(1,0,-1,0),(1,0,0,-1)$ and $(1,1,1,1)$ is a basis for $P^perp$, for point "b" more in general matrices $A$ are n-by-4 matrices
$$beginbmatrix
a& a& a& a\
a& a& a& a\
vdots& vdots& vdots& vdots\
a & a& a& a
endbmatrix$$
with $a neq 0$.
answered Aug 20 at 10:24
gimusi
68.7k73685
68.7k73685
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