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Linear algebra: Proof of number of solutions in a linear system. What is going on?
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In my linear algebra book I see this:
I am confused as to why the nonzero column matrix of $x_b = x_1 - x_2$ is a solution but not the zero one? I am also confused as to why $A(x_1 - x_2) = O$
For context, I just read the passage below and I see how the column matrices don't have to be zero matrices to be a solution to the equation $Ax = 0$:
For example, the matrix:
$$
left[ beginarrayc
1\
4\
7\
endarray right]
$$
is a non zero matrix but is a solution to the above example. But in the proof at the beginning, why does $A(x_1 - x_2) = 0$ and why is the nonzero column matrix a solution to the homogeneous system of linear equations Ax = 0? I might need an example to elucidate this concept. How do you know the column matrix is non zero?
Lastly, what is the proof doing?
BTW, I understand homogeneous to mean:
linear-algebra
asked Jul 17 at 18:42
Jwan622
1,61211224
add a comment |Â
up vote
1
down vote
favorite
In my linear algebra book I see this:
I am confused as to why the nonzero column matrix of $x_b = x_1 - x_2$ is a solution but not the zero one? I am also confused as to why $A(x_1 - x_2) = O$
For context, I just read the passage below and I see how the column matrices don't have to be zero matrices to be a solution to the equation $Ax = 0$:
For example, the matrix:
$$
left[ beginarrayc
1\
4\
7\
endarray right]
$$
is a non zero matrix but is a solution to the above example. But in the proof at the beginning, why does $A(x_1 - x_2) = 0$ and why is the nonzero column matrix a solution to the homogeneous system of linear equations Ax = 0? I might need an example to elucidate this concept. How do you know the column matrix is non zero?
Lastly, what is the proof doing?
BTW, I understand homogeneous to mean:
linear-algebra
asked Jul 17 at 18:42
Jwan622
1,61211224
1
It states that $x_1$ & $x_2$ are distinct solutions, which means $x_1 neq x_2$. Thus, $x_b = x_1-x_2 neq 0$
– gd1035
Jul 17 at 18:45
Use linearity to compute $A(x_1-x_2)$.
– Christian Sykes
Jul 17 at 18:45
Oh I see, it's a nonzero solution because we assume they are two different solutions. got it.
– Jwan622
Jul 19 at 14:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In my linear algebra book I see this:
I am confused as to why the nonzero column matrix of $x_b = x_1 - x_2$ is a solution but not the zero one? I am also confused as to why $A(x_1 - x_2) = O$
For context, I just read the passage below and I see how the column matrices don't have to be zero matrices to be a solution to the equation $Ax = 0$:
For example, the matrix:
$$
left[ beginarrayc
1\
4\
7\
endarray right]
$$
is a non zero matrix but is a solution to the above example. But in the proof at the beginning, why does $A(x_1 - x_2) = 0$ and why is the nonzero column matrix a solution to the homogeneous system of linear equations Ax = 0? I might need an example to elucidate this concept. How do you know the column matrix is non zero?
Lastly, what is the proof doing?
BTW, I understand homogeneous to mean:
linear-algebra
asked Jul 17 at 18:42
Jwan622
1,61211224
In my linear algebra book I see this:
I am confused as to why the nonzero column matrix of $x_b = x_1 - x_2$ is a solution but not the zero one? I am also confused as to why $A(x_1 - x_2) = O$
For context, I just read the passage below and I see how the column matrices don't have to be zero matrices to be a solution to the equation $Ax = 0$:
For example, the matrix:
$$
left[ beginarrayc
1\
4\
7\
endarray right]
$$
is a non zero matrix but is a solution to the above example. But in the proof at the beginning, why does $A(x_1 - x_2) = 0$ and why is the nonzero column matrix a solution to the homogeneous system of linear equations Ax = 0? I might need an example to elucidate this concept. How do you know the column matrix is non zero?
Lastly, what is the proof doing?
BTW, I understand homogeneous to mean:
linear-algebra
asked Jul 17 at 18:42
Jwan622
1,61211224
asked Jul 17 at 18:42
Jwan622
1,61211224
asked Jul 17 at 18:42
Jwan622
1,61211224
asked Jul 17 at 18:42
Jwan622
1,61211224
1,61211224
1
It states that $x_1$ & $x_2$ are distinct solutions, which means $x_1 neq x_2$. Thus, $x_b = x_1-x_2 neq 0$
– gd1035
Jul 17 at 18:45
Use linearity to compute $A(x_1-x_2)$.
– Christian Sykes
Jul 17 at 18:45
Oh I see, it's a nonzero solution because we assume they are two different solutions. got it.
– Jwan622
Jul 19 at 14:24
add a comment |Â
1
It states that $x_1$ & $x_2$ are distinct solutions, which means $x_1 neq x_2$. Thus, $x_b = x_1-x_2 neq 0$
– gd1035
Jul 17 at 18:45
Use linearity to compute $A(x_1-x_2)$.
– Christian Sykes
Jul 17 at 18:45
Oh I see, it's a nonzero solution because we assume they are two different solutions. got it.
– Jwan622
Jul 19 at 14:24
1
1
It states that $x_1$ & $x_2$ are distinct solutions, which means $x_1 neq x_2$. Thus, $x_b = x_1-x_2 neq 0$
– gd1035
Jul 17 at 18:45
It states that $x_1$ & $x_2$ are distinct solutions, which means $x_1 neq x_2$. Thus, $x_b = x_1-x_2 neq 0$
– gd1035
Jul 17 at 18:45
Use linearity to compute $A(x_1-x_2)$.
– Christian Sykes
Jul 17 at 18:45
Use linearity to compute $A(x_1-x_2)$.
– Christian Sykes
Jul 17 at 18:45
Oh I see, it's a nonzero solution because we assume they are two different solutions. got it.
– Jwan622
Jul 19 at 14:24
Oh I see, it's a nonzero solution because we assume they are two different solutions. got it.
– Jwan622
Jul 19 at 14:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Since $x_1$ and $x_2$ are both solutions to $Ax=b$ we get $$A(x_1 -x_2) = Ax_1 -Ax_2 = b-b =0$$ So $x_h=x_1-x_2$ is a solution to $Ax=0.$
Since $x_1ne x_2$ we get $x_h=x_1-x_2 ne 0$
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
2
down vote
Solutions of nonhomogeneous equations have the form
$$
beginbmatrix rm any ; solution ;of \ Ax=b endbmatrix = beginbmatrix rm another ; solution ;of \ Ax=b endbmatrix + beginbmatrix rm some ; solution ;of \ Ax=0 endbmatrix.
$$
It means that you can desribe all of them by knowing only one of them + how the solutions of the corresponding homogeneous equation look like.
Example.
Let us look at the following equation:
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 10 \ 0 endbmatrix
$$
It has many solutions. For example,
$$
beginbmatrix x_1 \ y_1 endbmatrix = beginbmatrix 5 \ 0 endbmatrix, quad beginbmatrix x_2 \ y_2 endbmatrix = beginbmatrix 5 \ 10 endbmatrix, quad beginbmatrix x_3 \ y_3 endbmatrix = beginbmatrix 5 \ -200 endbmatrix.
$$
But what is the structure of these solutions? One can see that they all have the form
$$
beginbmatrix x_n \ y_n endbmatrix = beginbmatrix 5 \ 0 endbmatrix + beginbmatrix 0 \ d endbmatrix.
$$
It turns out (and your book gives you a proof!) that the second term $beginbmatrix 0 \ d endbmatrix$ is the common form of all the solutions of the corresponding homogeneous equation
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 0 \ 0 endbmatrix.
$$
So you can obtain any solution of the nonhomogeneous equation by taking any particular one of its solutions ($beginbmatrix 5 \ 0 endbmatrix$) and adding some solution of the homogeneous one ($beginbmatrix 0 \ 0 endbmatrix$,$beginbmatrix 0 \ 10 endbmatrix$,$beginbmatrix 0 \ -200 endbmatrix$).
If this answer is not enough, please ask more precisely.
answered Jul 17 at 19:58
Zeekless
31110
I am confused as to why the nonzero column matrix of xb=x1−x2 is a solution but not the zero one? I am also confused as to why A(x1−x2)=O
– Jwan622
Jul 19 at 14:16
@Jwan622, equation $0 cdot x = 0$ has a zero solution and a non-zero solution. Equation $1 cdot x = 0$ has only zero solution. Equation $0 cdot x = 1$ has no solution at all. Of course, $x_b$ may be zero (if $x_1=x_2$), but if $x_1 neq x_2$, it is not.
– Zeekless
Jul 19 at 14:39
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since $x_1$ and $x_2$ are both solutions to $Ax=b$ we get $$A(x_1 -x_2) = Ax_1 -Ax_2 = b-b =0$$ So $x_h=x_1-x_2$ is a solution to $Ax=0.$
Since $x_1ne x_2$ we get $x_h=x_1-x_2 ne 0$
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
2
down vote
accepted
Since $x_1$ and $x_2$ are both solutions to $Ax=b$ we get $$A(x_1 -x_2) = Ax_1 -Ax_2 = b-b =0$$ So $x_h=x_1-x_2$ is a solution to $Ax=0.$
Since $x_1ne x_2$ we get $x_h=x_1-x_2 ne 0$
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since $x_1$ and $x_2$ are both solutions to $Ax=b$ we get $$A(x_1 -x_2) = Ax_1 -Ax_2 = b-b =0$$ So $x_h=x_1-x_2$ is a solution to $Ax=0.$
Since $x_1ne x_2$ we get $x_h=x_1-x_2 ne 0$
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
Since $x_1$ and $x_2$ are both solutions to $Ax=b$ we get $$A(x_1 -x_2) = Ax_1 -Ax_2 = b-b =0$$ So $x_h=x_1-x_2$ is a solution to $Ax=0.$
Since $x_1ne x_2$ we get $x_h=x_1-x_2 ne 0$
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
answered Jul 17 at 19:31
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
up vote
2
down vote
Solutions of nonhomogeneous equations have the form
$$
beginbmatrix rm any ; solution ;of \ Ax=b endbmatrix = beginbmatrix rm another ; solution ;of \ Ax=b endbmatrix + beginbmatrix rm some ; solution ;of \ Ax=0 endbmatrix.
$$
It means that you can desribe all of them by knowing only one of them + how the solutions of the corresponding homogeneous equation look like.
Example.
Let us look at the following equation:
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 10 \ 0 endbmatrix
$$
It has many solutions. For example,
$$
beginbmatrix x_1 \ y_1 endbmatrix = beginbmatrix 5 \ 0 endbmatrix, quad beginbmatrix x_2 \ y_2 endbmatrix = beginbmatrix 5 \ 10 endbmatrix, quad beginbmatrix x_3 \ y_3 endbmatrix = beginbmatrix 5 \ -200 endbmatrix.
$$
But what is the structure of these solutions? One can see that they all have the form
$$
beginbmatrix x_n \ y_n endbmatrix = beginbmatrix 5 \ 0 endbmatrix + beginbmatrix 0 \ d endbmatrix.
$$
It turns out (and your book gives you a proof!) that the second term $beginbmatrix 0 \ d endbmatrix$ is the common form of all the solutions of the corresponding homogeneous equation
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 0 \ 0 endbmatrix.
$$
So you can obtain any solution of the nonhomogeneous equation by taking any particular one of its solutions ($beginbmatrix 5 \ 0 endbmatrix$) and adding some solution of the homogeneous one ($beginbmatrix 0 \ 0 endbmatrix$,$beginbmatrix 0 \ 10 endbmatrix$,$beginbmatrix 0 \ -200 endbmatrix$).
If this answer is not enough, please ask more precisely.
answered Jul 17 at 19:58
Zeekless
31110
I am confused as to why the nonzero column matrix of xb=x1−x2 is a solution but not the zero one? I am also confused as to why A(x1−x2)=O
– Jwan622
Jul 19 at 14:16
@Jwan622, equation $0 cdot x = 0$ has a zero solution and a non-zero solution. Equation $1 cdot x = 0$ has only zero solution. Equation $0 cdot x = 1$ has no solution at all. Of course, $x_b$ may be zero (if $x_1=x_2$), but if $x_1 neq x_2$, it is not.
– Zeekless
Jul 19 at 14:39
add a comment |Â
up vote
2
down vote
Solutions of nonhomogeneous equations have the form
$$
beginbmatrix rm any ; solution ;of \ Ax=b endbmatrix = beginbmatrix rm another ; solution ;of \ Ax=b endbmatrix + beginbmatrix rm some ; solution ;of \ Ax=0 endbmatrix.
$$
It means that you can desribe all of them by knowing only one of them + how the solutions of the corresponding homogeneous equation look like.
Example.
Let us look at the following equation:
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 10 \ 0 endbmatrix
$$
It has many solutions. For example,
$$
beginbmatrix x_1 \ y_1 endbmatrix = beginbmatrix 5 \ 0 endbmatrix, quad beginbmatrix x_2 \ y_2 endbmatrix = beginbmatrix 5 \ 10 endbmatrix, quad beginbmatrix x_3 \ y_3 endbmatrix = beginbmatrix 5 \ -200 endbmatrix.
$$
But what is the structure of these solutions? One can see that they all have the form
$$
beginbmatrix x_n \ y_n endbmatrix = beginbmatrix 5 \ 0 endbmatrix + beginbmatrix 0 \ d endbmatrix.
$$
It turns out (and your book gives you a proof!) that the second term $beginbmatrix 0 \ d endbmatrix$ is the common form of all the solutions of the corresponding homogeneous equation
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 0 \ 0 endbmatrix.
$$
So you can obtain any solution of the nonhomogeneous equation by taking any particular one of its solutions ($beginbmatrix 5 \ 0 endbmatrix$) and adding some solution of the homogeneous one ($beginbmatrix 0 \ 0 endbmatrix$,$beginbmatrix 0 \ 10 endbmatrix$,$beginbmatrix 0 \ -200 endbmatrix$).
If this answer is not enough, please ask more precisely.
answered Jul 17 at 19:58
Zeekless
31110
I am confused as to why the nonzero column matrix of xb=x1−x2 is a solution but not the zero one? I am also confused as to why A(x1−x2)=O
– Jwan622
Jul 19 at 14:16
@Jwan622, equation $0 cdot x = 0$ has a zero solution and a non-zero solution. Equation $1 cdot x = 0$ has only zero solution. Equation $0 cdot x = 1$ has no solution at all. Of course, $x_b$ may be zero (if $x_1=x_2$), but if $x_1 neq x_2$, it is not.
– Zeekless
Jul 19 at 14:39
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Solutions of nonhomogeneous equations have the form
$$
beginbmatrix rm any ; solution ;of \ Ax=b endbmatrix = beginbmatrix rm another ; solution ;of \ Ax=b endbmatrix + beginbmatrix rm some ; solution ;of \ Ax=0 endbmatrix.
$$
It means that you can desribe all of them by knowing only one of them + how the solutions of the corresponding homogeneous equation look like.
Example.
Let us look at the following equation:
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 10 \ 0 endbmatrix
$$
It has many solutions. For example,
$$
beginbmatrix x_1 \ y_1 endbmatrix = beginbmatrix 5 \ 0 endbmatrix, quad beginbmatrix x_2 \ y_2 endbmatrix = beginbmatrix 5 \ 10 endbmatrix, quad beginbmatrix x_3 \ y_3 endbmatrix = beginbmatrix 5 \ -200 endbmatrix.
$$
But what is the structure of these solutions? One can see that they all have the form
$$
beginbmatrix x_n \ y_n endbmatrix = beginbmatrix 5 \ 0 endbmatrix + beginbmatrix 0 \ d endbmatrix.
$$
It turns out (and your book gives you a proof!) that the second term $beginbmatrix 0 \ d endbmatrix$ is the common form of all the solutions of the corresponding homogeneous equation
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 0 \ 0 endbmatrix.
$$
So you can obtain any solution of the nonhomogeneous equation by taking any particular one of its solutions ($beginbmatrix 5 \ 0 endbmatrix$) and adding some solution of the homogeneous one ($beginbmatrix 0 \ 0 endbmatrix$,$beginbmatrix 0 \ 10 endbmatrix$,$beginbmatrix 0 \ -200 endbmatrix$).
If this answer is not enough, please ask more precisely.
answered Jul 17 at 19:58
Zeekless
31110
Solutions of nonhomogeneous equations have the form
$$
beginbmatrix rm any ; solution ;of \ Ax=b endbmatrix = beginbmatrix rm another ; solution ;of \ Ax=b endbmatrix + beginbmatrix rm some ; solution ;of \ Ax=0 endbmatrix.
$$
It means that you can desribe all of them by knowing only one of them + how the solutions of the corresponding homogeneous equation look like.
Example.
Let us look at the following equation:
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 10 \ 0 endbmatrix
$$
It has many solutions. For example,
$$
beginbmatrix x_1 \ y_1 endbmatrix = beginbmatrix 5 \ 0 endbmatrix, quad beginbmatrix x_2 \ y_2 endbmatrix = beginbmatrix 5 \ 10 endbmatrix, quad beginbmatrix x_3 \ y_3 endbmatrix = beginbmatrix 5 \ -200 endbmatrix.
$$
But what is the structure of these solutions? One can see that they all have the form
$$
beginbmatrix x_n \ y_n endbmatrix = beginbmatrix 5 \ 0 endbmatrix + beginbmatrix 0 \ d endbmatrix.
$$
It turns out (and your book gives you a proof!) that the second term $beginbmatrix 0 \ d endbmatrix$ is the common form of all the solutions of the corresponding homogeneous equation
$$
beginbmatrix 2 & 0 \ 0 & 0 endbmatrixbeginbmatrix x \ y endbmatrix = beginbmatrix 0 \ 0 endbmatrix.
$$
So you can obtain any solution of the nonhomogeneous equation by taking any particular one of its solutions ($beginbmatrix 5 \ 0 endbmatrix$) and adding some solution of the homogeneous one ($beginbmatrix 0 \ 0 endbmatrix$,$beginbmatrix 0 \ 10 endbmatrix$,$beginbmatrix 0 \ -200 endbmatrix$).
If this answer is not enough, please ask more precisely.
answered Jul 17 at 19:58
Zeekless
31110
answered Jul 17 at 19:58
Zeekless
31110
answered Jul 17 at 19:58
Zeekless
31110
answered Jul 17 at 19:58
Zeekless
31110
31110
I am confused as to why the nonzero column matrix of xb=x1−x2 is a solution but not the zero one? I am also confused as to why A(x1−x2)=O
– Jwan622
Jul 19 at 14:16
@Jwan622, equation $0 cdot x = 0$ has a zero solution and a non-zero solution. Equation $1 cdot x = 0$ has only zero solution. Equation $0 cdot x = 1$ has no solution at all. Of course, $x_b$ may be zero (if $x_1=x_2$), but if $x_1 neq x_2$, it is not.
– Zeekless
Jul 19 at 14:39
add a comment |Â
I am confused as to why the nonzero column matrix of xb=x1−x2 is a solution but not the zero one? I am also confused as to why A(x1−x2)=O
– Jwan622
Jul 19 at 14:16
@Jwan622, equation $0 cdot x = 0$ has a zero solution and a non-zero solution. Equation $1 cdot x = 0$ has only zero solution. Equation $0 cdot x = 1$ has no solution at all. Of course, $x_b$ may be zero (if $x_1=x_2$), but if $x_1 neq x_2$, it is not.
– Zeekless
Jul 19 at 14:39
I am confused as to why the nonzero column matrix of xb=x1−x2 is a solution but not the zero one? I am also confused as to why A(x1−x2)=O
– Jwan622
Jul 19 at 14:16
I am confused as to why the nonzero column matrix of xb=x1−x2 is a solution but not the zero one? I am also confused as to why A(x1−x2)=O
– Jwan622
Jul 19 at 14:16
@Jwan622, equation $0 cdot x = 0$ has a zero solution and a non-zero solution. Equation $1 cdot x = 0$ has only zero solution. Equation $0 cdot x = 1$ has no solution at all. Of course, $x_b$ may be zero (if $x_1=x_2$), but if $x_1 neq x_2$, it is not.
– Zeekless
Jul 19 at 14:39
@Jwan622, equation $0 cdot x = 0$ has a zero solution and a non-zero solution. Equation $1 cdot x = 0$ has only zero solution. Equation $0 cdot x = 1$ has no solution at all. Of course, $x_b$ may be zero (if $x_1=x_2$), but if $x_1 neq x_2$, it is not.
– Zeekless
Jul 19 at 14:39
add a comment |Â
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1
It states that $x_1$ & $x_2$ are distinct solutions, which means $x_1 neq x_2$. Thus, $x_b = x_1-x_2 neq 0$
– gd1035
Jul 17 at 18:45
Use linearity to compute $A(x_1-x_2)$.
– Christian Sykes
Jul 17 at 18:45
Oh I see, it's a nonzero solution because we assume they are two different solutions. got it.
– Jwan622
Jul 19 at 14:24