dimension and basis of a solution space based on the rank of a matrix and three vectors
$begingroup$
I am new to linear algebra and have been wrestling with the following question for hours. I have no teacher, and the answer book provides no key. Can anyone suggest a way to go about this?
Let $Ainmathbb{R}^{ 4times4}$, with rank $2$.
Suppose that the vectors
$u=(2,1,2,0),, v=(1,-1,2,4),, w=(1,0,2,-1)$
are solutions to the linear system $Ax=b$.
a. find the dimension of the solution space of the system $Ax=0$ and find its basis.
b. find the general solution to the system $Ax=b$.
Thank you very much
linear-algebra
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
asked Dec 22 '18 at 19:51
daltadalta
1238
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra and have been wrestling with the following question for hours. I have no teacher, and the answer book provides no key. Can anyone suggest a way to go about this?
Let $Ainmathbb{R}^{ 4times4}$, with rank $2$.
Suppose that the vectors
$u=(2,1,2,0),, v=(1,-1,2,4),, w=(1,0,2,-1)$
are solutions to the linear system $Ax=b$.
a. find the dimension of the solution space of the system $Ax=0$ and find its basis.
b. find the general solution to the system $Ax=b$.
Thank you very much
linear-algebra
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
asked Dec 22 '18 at 19:51
daltadalta
1238
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra and have been wrestling with the following question for hours. I have no teacher, and the answer book provides no key. Can anyone suggest a way to go about this?
Let $Ainmathbb{R}^{ 4times4}$, with rank $2$.
Suppose that the vectors
$u=(2,1,2,0),, v=(1,-1,2,4),, w=(1,0,2,-1)$
are solutions to the linear system $Ax=b$.
a. find the dimension of the solution space of the system $Ax=0$ and find its basis.
b. find the general solution to the system $Ax=b$.
Thank you very much
linear-algebra
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
asked Dec 22 '18 at 19:51
daltadalta
1238
$endgroup$
I am new to linear algebra and have been wrestling with the following question for hours. I have no teacher, and the answer book provides no key. Can anyone suggest a way to go about this?
Let $Ainmathbb{R}^{ 4times4}$, with rank $2$.
Suppose that the vectors
$u=(2,1,2,0),, v=(1,-1,2,4),, w=(1,0,2,-1)$
are solutions to the linear system $Ax=b$.
a. find the dimension of the solution space of the system $Ax=0$ and find its basis.
b. find the general solution to the system $Ax=b$.
Thank you very much
linear-algebra
linear-algebra
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
asked Dec 22 '18 at 19:51
daltadalta
1238
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
asked Dec 22 '18 at 19:51
daltadalta
1238
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
edited Dec 22 '18 at 19:55
Martín Vacas Vignolo
3,816623
3,816623
asked Dec 22 '18 at 19:51
daltadalta
1238
asked Dec 22 '18 at 19:51
daltadalta
1238
asked Dec 22 '18 at 19:51
daltadalta
1238
1238
add a comment |
add a comment |
1 Answer
1
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$begingroup$
$A$ is a matrix of rank $2$ which means that it has a $2$ dimensional kernel. Since $u,v,w$ have the same image, $u-v$ and $u-w$ are in the kernel. Since these two vectors are linearly independent, they span the kernel.
To get all solutions of $Ax=b$, you can take any solution and translate it with kernel elements, i.e every solution of the equation is of the form $u+c(u-v)+d(u-w)$ where $c,dinmathbb{R}$.
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
$endgroup$
$begingroup$
So the solution space is an affine linear space!
$endgroup$
– Mustang
Dec 22 '18 at 20:57
1
$begingroup$
Yes, it is. This is in general true. For a matrix $A$ and a vector $b$ in the image of $A$, the solutions of $Ax=b$ are always given by $u+N(A)$ where $u$ is a solution of the equation and $N(A)$ is the kernel of $A$.
$endgroup$
– Levent
Dec 22 '18 at 20:58
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
$A$ is a matrix of rank $2$ which means that it has a $2$ dimensional kernel. Since $u,v,w$ have the same image, $u-v$ and $u-w$ are in the kernel. Since these two vectors are linearly independent, they span the kernel.
To get all solutions of $Ax=b$, you can take any solution and translate it with kernel elements, i.e every solution of the equation is of the form $u+c(u-v)+d(u-w)$ where $c,dinmathbb{R}$.
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
$endgroup$
$begingroup$
So the solution space is an affine linear space!
$endgroup$
– Mustang
Dec 22 '18 at 20:57
1
$begingroup$
Yes, it is. This is in general true. For a matrix $A$ and a vector $b$ in the image of $A$, the solutions of $Ax=b$ are always given by $u+N(A)$ where $u$ is a solution of the equation and $N(A)$ is the kernel of $A$.
$endgroup$
– Levent
Dec 22 '18 at 20:58
add a comment |
$begingroup$
$A$ is a matrix of rank $2$ which means that it has a $2$ dimensional kernel. Since $u,v,w$ have the same image, $u-v$ and $u-w$ are in the kernel. Since these two vectors are linearly independent, they span the kernel.
To get all solutions of $Ax=b$, you can take any solution and translate it with kernel elements, i.e every solution of the equation is of the form $u+c(u-v)+d(u-w)$ where $c,dinmathbb{R}$.
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
$endgroup$
$begingroup$
So the solution space is an affine linear space!
$endgroup$
– Mustang
Dec 22 '18 at 20:57
1
$begingroup$
Yes, it is. This is in general true. For a matrix $A$ and a vector $b$ in the image of $A$, the solutions of $Ax=b$ are always given by $u+N(A)$ where $u$ is a solution of the equation and $N(A)$ is the kernel of $A$.
$endgroup$
– Levent
Dec 22 '18 at 20:58
add a comment |
$begingroup$
$A$ is a matrix of rank $2$ which means that it has a $2$ dimensional kernel. Since $u,v,w$ have the same image, $u-v$ and $u-w$ are in the kernel. Since these two vectors are linearly independent, they span the kernel.
To get all solutions of $Ax=b$, you can take any solution and translate it with kernel elements, i.e every solution of the equation is of the form $u+c(u-v)+d(u-w)$ where $c,dinmathbb{R}$.
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
$endgroup$
$A$ is a matrix of rank $2$ which means that it has a $2$ dimensional kernel. Since $u,v,w$ have the same image, $u-v$ and $u-w$ are in the kernel. Since these two vectors are linearly independent, they span the kernel.
To get all solutions of $Ax=b$, you can take any solution and translate it with kernel elements, i.e every solution of the equation is of the form $u+c(u-v)+d(u-w)$ where $c,dinmathbb{R}$.
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
answered Dec 22 '18 at 20:37
LeventLevent
2,729925
2,729925
$begingroup$
So the solution space is an affine linear space!
$endgroup$
– Mustang
Dec 22 '18 at 20:57
1
$begingroup$
Yes, it is. This is in general true. For a matrix $A$ and a vector $b$ in the image of $A$, the solutions of $Ax=b$ are always given by $u+N(A)$ where $u$ is a solution of the equation and $N(A)$ is the kernel of $A$.
$endgroup$
– Levent
Dec 22 '18 at 20:58
add a comment |
$begingroup$
So the solution space is an affine linear space!
$endgroup$
– Mustang
Dec 22 '18 at 20:57
1
$begingroup$
Yes, it is. This is in general true. For a matrix $A$ and a vector $b$ in the image of $A$, the solutions of $Ax=b$ are always given by $u+N(A)$ where $u$ is a solution of the equation and $N(A)$ is the kernel of $A$.
$endgroup$
– Levent
Dec 22 '18 at 20:58
$begingroup$
So the solution space is an affine linear space!
$endgroup$
– Mustang
Dec 22 '18 at 20:57
$begingroup$
So the solution space is an affine linear space!
$endgroup$
– Mustang
Dec 22 '18 at 20:57
1
1
$begingroup$
Yes, it is. This is in general true. For a matrix $A$ and a vector $b$ in the image of $A$, the solutions of $Ax=b$ are always given by $u+N(A)$ where $u$ is a solution of the equation and $N(A)$ is the kernel of $A$.
$endgroup$
– Levent
Dec 22 '18 at 20:58
$begingroup$
Yes, it is. This is in general true. For a matrix $A$ and a vector $b$ in the image of $A$, the solutions of $Ax=b$ are always given by $u+N(A)$ where $u$ is a solution of the equation and $N(A)$ is the kernel of $A$.
$endgroup$
– Levent
Dec 22 '18 at 20:58
add a comment |
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