Find solution space of $ Ax = b $
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Let $ A $ be a $ 4 times 4 $ matrix over $ mathbf R $ with a rank of 3. $ (1,2,0,-1) , (0,2,1,1) $ are solutions to $ Ax = b $ and I need to find the solution space of the $ Ax = 0$ and general solution of $ Ax = b $
If the rank of the $ A $ is $ 3$ then the dimension of solution space of $ Ax= 0 $ is $ 1$ and I guess any vector that is a solution for $ Ax = 0 $ can't be spanned by $ (1,2,0,-1),(0,2,1,1)$ but how do I proceed from here?
linear-algebra matrices matrix-equations
asked Aug 27 at 17:17
bm1125
38016
add a comment |Â
up vote
1
down vote
favorite
Let $ A $ be a $ 4 times 4 $ matrix over $ mathbf R $ with a rank of 3. $ (1,2,0,-1) , (0,2,1,1) $ are solutions to $ Ax = b $ and I need to find the solution space of the $ Ax = 0$ and general solution of $ Ax = b $
If the rank of the $ A $ is $ 3$ then the dimension of solution space of $ Ax= 0 $ is $ 1$ and I guess any vector that is a solution for $ Ax = 0 $ can't be spanned by $ (1,2,0,-1),(0,2,1,1)$ but how do I proceed from here?
linear-algebra matrices matrix-equations
asked Aug 27 at 17:17
bm1125
38016
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $ A $ be a $ 4 times 4 $ matrix over $ mathbf R $ with a rank of 3. $ (1,2,0,-1) , (0,2,1,1) $ are solutions to $ Ax = b $ and I need to find the solution space of the $ Ax = 0$ and general solution of $ Ax = b $
If the rank of the $ A $ is $ 3$ then the dimension of solution space of $ Ax= 0 $ is $ 1$ and I guess any vector that is a solution for $ Ax = 0 $ can't be spanned by $ (1,2,0,-1),(0,2,1,1)$ but how do I proceed from here?
linear-algebra matrices matrix-equations
asked Aug 27 at 17:17
bm1125
38016
Let $ A $ be a $ 4 times 4 $ matrix over $ mathbf R $ with a rank of 3. $ (1,2,0,-1) , (0,2,1,1) $ are solutions to $ Ax = b $ and I need to find the solution space of the $ Ax = 0$ and general solution of $ Ax = b $
If the rank of the $ A $ is $ 3$ then the dimension of solution space of $ Ax= 0 $ is $ 1$ and I guess any vector that is a solution for $ Ax = 0 $ can't be spanned by $ (1,2,0,-1),(0,2,1,1)$ but how do I proceed from here?
linear-algebra matrices matrix-equations
asked Aug 27 at 17:17
bm1125
38016
asked Aug 27 at 17:17
bm1125
38016
asked Aug 27 at 17:17
bm1125
38016
asked Aug 27 at 17:17
bm1125
38016
38016
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Think about it geometrically. You’ve determined that the solution space is one-dimensional, i.e., that it’s some line in $mathbb R^4$. You’ve also got two known points on this line. I suspect that you’ll be able to take it from here.
More generally, if $mathbf v$ and $mathbf w$ are distinct solutions to your inhomogeneous equation, then $Amathbf v-Amathbf w = A(mathbf v-mathbf w)=0$, so $mathbf v-mathbf w$ is a solution to the homogeneous equation. In this case, the null space is one-dimensional, so you now have a vector that spans the space. Thus all solutions to the inhomogeneous equation are of the form $mathbf v + lambda(mathbf v-mathbf w)$, $lambdainmathbb R$, or, more symmetrically, $(1-lambda)mathbf v+lambdamathbf w$. This is just the line through $mathbf v$ and $mathbf w$, as above.
answered Aug 27 at 18:35
amd
26.6k21046
so I understand that all solutions of $ Ax = 0 $ should be $ lambda(v-w) $ but why did you write $ v+ lambda(v-w) $ ??
– bm1125
Aug 27 at 18:58
@bm1125 All solutions to the inhomogeneous system are of that form—it’s just the line through $mathbf v$ and $mathbf w$.
– amd
Aug 27 at 18:59
add a comment |Â
up vote
3
down vote
Take $A(1,2,0,-1) - A (0,2,1,1) = b-b =0.$ Since the rank is 3, then the nullity is 1, and so the kernel is spanned by $(1,2,0,-1) - (0,2,1,1)=(1,0,-1,-2).$
answered Aug 27 at 17:22
Chickenmancer
3,057622
wow! thanks! and now to find general solution to $ Ax = b$ I should just find 4th vector that is linearly idependent of those three??
– bm1125
Aug 27 at 17:56
This is a little complicated, since $b$ is unknown. However, assuming $x$ satisfies $Ax=b,$ and $vintextspan(1,0,−1,−2),$ then $x+v$ describes the most general solution.
– Chickenmancer
Aug 27 at 19:37
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Think about it geometrically. You’ve determined that the solution space is one-dimensional, i.e., that it’s some line in $mathbb R^4$. You’ve also got two known points on this line. I suspect that you’ll be able to take it from here.
More generally, if $mathbf v$ and $mathbf w$ are distinct solutions to your inhomogeneous equation, then $Amathbf v-Amathbf w = A(mathbf v-mathbf w)=0$, so $mathbf v-mathbf w$ is a solution to the homogeneous equation. In this case, the null space is one-dimensional, so you now have a vector that spans the space. Thus all solutions to the inhomogeneous equation are of the form $mathbf v + lambda(mathbf v-mathbf w)$, $lambdainmathbb R$, or, more symmetrically, $(1-lambda)mathbf v+lambdamathbf w$. This is just the line through $mathbf v$ and $mathbf w$, as above.
answered Aug 27 at 18:35
amd
26.6k21046
so I understand that all solutions of $ Ax = 0 $ should be $ lambda(v-w) $ but why did you write $ v+ lambda(v-w) $ ??
– bm1125
Aug 27 at 18:58
@bm1125 All solutions to the inhomogeneous system are of that form—it’s just the line through $mathbf v$ and $mathbf w$.
– amd
Aug 27 at 18:59
add a comment |Â
up vote
1
down vote
accepted
Think about it geometrically. You’ve determined that the solution space is one-dimensional, i.e., that it’s some line in $mathbb R^4$. You’ve also got two known points on this line. I suspect that you’ll be able to take it from here.
More generally, if $mathbf v$ and $mathbf w$ are distinct solutions to your inhomogeneous equation, then $Amathbf v-Amathbf w = A(mathbf v-mathbf w)=0$, so $mathbf v-mathbf w$ is a solution to the homogeneous equation. In this case, the null space is one-dimensional, so you now have a vector that spans the space. Thus all solutions to the inhomogeneous equation are of the form $mathbf v + lambda(mathbf v-mathbf w)$, $lambdainmathbb R$, or, more symmetrically, $(1-lambda)mathbf v+lambdamathbf w$. This is just the line through $mathbf v$ and $mathbf w$, as above.
answered Aug 27 at 18:35
amd
26.6k21046
so I understand that all solutions of $ Ax = 0 $ should be $ lambda(v-w) $ but why did you write $ v+ lambda(v-w) $ ??
– bm1125
Aug 27 at 18:58
@bm1125 All solutions to the inhomogeneous system are of that form—it’s just the line through $mathbf v$ and $mathbf w$.
– amd
Aug 27 at 18:59
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Think about it geometrically. You’ve determined that the solution space is one-dimensional, i.e., that it’s some line in $mathbb R^4$. You’ve also got two known points on this line. I suspect that you’ll be able to take it from here.
More generally, if $mathbf v$ and $mathbf w$ are distinct solutions to your inhomogeneous equation, then $Amathbf v-Amathbf w = A(mathbf v-mathbf w)=0$, so $mathbf v-mathbf w$ is a solution to the homogeneous equation. In this case, the null space is one-dimensional, so you now have a vector that spans the space. Thus all solutions to the inhomogeneous equation are of the form $mathbf v + lambda(mathbf v-mathbf w)$, $lambdainmathbb R$, or, more symmetrically, $(1-lambda)mathbf v+lambdamathbf w$. This is just the line through $mathbf v$ and $mathbf w$, as above.
answered Aug 27 at 18:35
amd
26.6k21046
Think about it geometrically. You’ve determined that the solution space is one-dimensional, i.e., that it’s some line in $mathbb R^4$. You’ve also got two known points on this line. I suspect that you’ll be able to take it from here.
More generally, if $mathbf v$ and $mathbf w$ are distinct solutions to your inhomogeneous equation, then $Amathbf v-Amathbf w = A(mathbf v-mathbf w)=0$, so $mathbf v-mathbf w$ is a solution to the homogeneous equation. In this case, the null space is one-dimensional, so you now have a vector that spans the space. Thus all solutions to the inhomogeneous equation are of the form $mathbf v + lambda(mathbf v-mathbf w)$, $lambdainmathbb R$, or, more symmetrically, $(1-lambda)mathbf v+lambdamathbf w$. This is just the line through $mathbf v$ and $mathbf w$, as above.
answered Aug 27 at 18:35
amd
26.6k21046
edited Aug 27 at 19:00
answered Aug 27 at 18:35
amd
26.6k21046
answered Aug 27 at 18:35
amd
26.6k21046
answered Aug 27 at 18:35
amd
26.6k21046
26.6k21046
so I understand that all solutions of $ Ax = 0 $ should be $ lambda(v-w) $ but why did you write $ v+ lambda(v-w) $ ??
– bm1125
Aug 27 at 18:58
@bm1125 All solutions to the inhomogeneous system are of that form—it’s just the line through $mathbf v$ and $mathbf w$.
– amd
Aug 27 at 18:59
add a comment |Â
so I understand that all solutions of $ Ax = 0 $ should be $ lambda(v-w) $ but why did you write $ v+ lambda(v-w) $ ??
– bm1125
Aug 27 at 18:58
@bm1125 All solutions to the inhomogeneous system are of that form—it’s just the line through $mathbf v$ and $mathbf w$.
– amd
Aug 27 at 18:59
so I understand that all solutions of $ Ax = 0 $ should be $ lambda(v-w) $ but why did you write $ v+ lambda(v-w) $ ??
– bm1125
Aug 27 at 18:58
so I understand that all solutions of $ Ax = 0 $ should be $ lambda(v-w) $ but why did you write $ v+ lambda(v-w) $ ??
– bm1125
Aug 27 at 18:58
@bm1125 All solutions to the inhomogeneous system are of that form—it’s just the line through $mathbf v$ and $mathbf w$.
– amd
Aug 27 at 18:59
@bm1125 All solutions to the inhomogeneous system are of that form—it’s just the line through $mathbf v$ and $mathbf w$.
– amd
Aug 27 at 18:59
add a comment |Â
up vote
3
down vote
Take $A(1,2,0,-1) - A (0,2,1,1) = b-b =0.$ Since the rank is 3, then the nullity is 1, and so the kernel is spanned by $(1,2,0,-1) - (0,2,1,1)=(1,0,-1,-2).$
answered Aug 27 at 17:22
Chickenmancer
3,057622
wow! thanks! and now to find general solution to $ Ax = b$ I should just find 4th vector that is linearly idependent of those three??
– bm1125
Aug 27 at 17:56
This is a little complicated, since $b$ is unknown. However, assuming $x$ satisfies $Ax=b,$ and $vintextspan(1,0,−1,−2),$ then $x+v$ describes the most general solution.
– Chickenmancer
Aug 27 at 19:37
add a comment |Â
up vote
3
down vote
Take $A(1,2,0,-1) - A (0,2,1,1) = b-b =0.$ Since the rank is 3, then the nullity is 1, and so the kernel is spanned by $(1,2,0,-1) - (0,2,1,1)=(1,0,-1,-2).$
answered Aug 27 at 17:22
Chickenmancer
3,057622
wow! thanks! and now to find general solution to $ Ax = b$ I should just find 4th vector that is linearly idependent of those three??
– bm1125
Aug 27 at 17:56
This is a little complicated, since $b$ is unknown. However, assuming $x$ satisfies $Ax=b,$ and $vintextspan(1,0,−1,−2),$ then $x+v$ describes the most general solution.
– Chickenmancer
Aug 27 at 19:37
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Take $A(1,2,0,-1) - A (0,2,1,1) = b-b =0.$ Since the rank is 3, then the nullity is 1, and so the kernel is spanned by $(1,2,0,-1) - (0,2,1,1)=(1,0,-1,-2).$
answered Aug 27 at 17:22
Chickenmancer
3,057622
Take $A(1,2,0,-1) - A (0,2,1,1) = b-b =0.$ Since the rank is 3, then the nullity is 1, and so the kernel is spanned by $(1,2,0,-1) - (0,2,1,1)=(1,0,-1,-2).$
answered Aug 27 at 17:22
Chickenmancer
3,057622
answered Aug 27 at 17:22
Chickenmancer
3,057622
answered Aug 27 at 17:22
Chickenmancer
3,057622
answered Aug 27 at 17:22
Chickenmancer
3,057622
3,057622
wow! thanks! and now to find general solution to $ Ax = b$ I should just find 4th vector that is linearly idependent of those three??
– bm1125
Aug 27 at 17:56
This is a little complicated, since $b$ is unknown. However, assuming $x$ satisfies $Ax=b,$ and $vintextspan(1,0,−1,−2),$ then $x+v$ describes the most general solution.
– Chickenmancer
Aug 27 at 19:37
add a comment |Â
wow! thanks! and now to find general solution to $ Ax = b$ I should just find 4th vector that is linearly idependent of those three??
– bm1125
Aug 27 at 17:56
This is a little complicated, since $b$ is unknown. However, assuming $x$ satisfies $Ax=b,$ and $vintextspan(1,0,−1,−2),$ then $x+v$ describes the most general solution.
– Chickenmancer
Aug 27 at 19:37
wow! thanks! and now to find general solution to $ Ax = b$ I should just find 4th vector that is linearly idependent of those three??
– bm1125
Aug 27 at 17:56
wow! thanks! and now to find general solution to $ Ax = b$ I should just find 4th vector that is linearly idependent of those three??
– bm1125
Aug 27 at 17:56
This is a little complicated, since $b$ is unknown. However, assuming $x$ satisfies $Ax=b,$ and $vintextspan(1,0,−1,−2),$ then $x+v$ describes the most general solution.
– Chickenmancer
Aug 27 at 19:37
This is a little complicated, since $b$ is unknown. However, assuming $x$ satisfies $Ax=b,$ and $vintextspan(1,0,−1,−2),$ then $x+v$ describes the most general solution.
– Chickenmancer
Aug 27 at 19:37
add a comment |Â
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