Mixing
Linear system solutions
$begingroup$
Assume we have linear system in form
$$
Ax=b
$$
Then I know that the system has solution when $text{rank} A =text{rank} (A|b) $.
Can you tell me more properties like this one which helps us to know whether the system has any solutions or maybe 6 has only one solution using the determinant of matrix and other characteristics of it?
linear-algebra matrices
asked Jan 8 at 15:21
avan1235avan1235
3217
$endgroup$
add a comment |
$begingroup$
Assume we have linear system in form
$$
Ax=b
$$
Then I know that the system has solution when $text{rank} A =text{rank} (A|b) $.
Can you tell me more properties like this one which helps us to know whether the system has any solutions or maybe 6 has only one solution using the determinant of matrix and other characteristics of it?
linear-algebra matrices
asked Jan 8 at 15:21
avan1235avan1235
3217
$endgroup$
$begingroup$
Hm... something like Cramer's rule for example?
$endgroup$
– chickenNinja123
Jan 8 at 15:25
$begingroup$
Yes but more like this 😉
$endgroup$
– avan1235
Jan 8 at 15:26
add a comment |
$begingroup$
Assume we have linear system in form
$$
Ax=b
$$
Then I know that the system has solution when $text{rank} A =text{rank} (A|b) $.
Can you tell me more properties like this one which helps us to know whether the system has any solutions or maybe 6 has only one solution using the determinant of matrix and other characteristics of it?
linear-algebra matrices
asked Jan 8 at 15:21
avan1235avan1235
3217
$endgroup$
Assume we have linear system in form
$$
Ax=b
$$
Then I know that the system has solution when $text{rank} A =text{rank} (A|b) $.
Can you tell me more properties like this one which helps us to know whether the system has any solutions or maybe 6 has only one solution using the determinant of matrix and other characteristics of it?
linear-algebra matrices
linear-algebra matrices
asked Jan 8 at 15:21
avan1235avan1235
3217
asked Jan 8 at 15:21
avan1235avan1235
3217
asked Jan 8 at 15:21
avan1235avan1235
3217
asked Jan 8 at 15:21
avan1235avan1235
3217
asked Jan 8 at 15:21
avan1235avan1235
3217
3217
$begingroup$
Hm... something like Cramer's rule for example?
$endgroup$
– chickenNinja123
Jan 8 at 15:25
$begingroup$
Yes but more like this 😉
$endgroup$
– avan1235
Jan 8 at 15:26
add a comment |
$begingroup$
Hm... something like Cramer's rule for example?
$endgroup$
– chickenNinja123
Jan 8 at 15:25
$begingroup$
Yes but more like this 😉
$endgroup$
– avan1235
Jan 8 at 15:26
$begingroup$
Hm... something like Cramer's rule for example?
$endgroup$
– chickenNinja123
Jan 8 at 15:25
$begingroup$
Hm... something like Cramer's rule for example?
$endgroup$
– chickenNinja123
Jan 8 at 15:25
$begingroup$
Yes but more like this 😉
$endgroup$
– avan1235
Jan 8 at 15:26
$begingroup$
Yes but more like this 😉
$endgroup$
– avan1235
Jan 8 at 15:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $A$ be an $mtimes n$ matrix.
- If $operatorname{rank}(A)=m$, there always exists a solution.
- If $operatorname{rank}(A)=n$, there might be solutions or not, but if, then they are unique.
- Consequently, if $operatorname{rank}=m=n$, there always exists a unique solution. One may rephrase this to saying that $det(A)neq 0$.
answered Jan 8 at 15:29
JamesJames
801215
$endgroup$
$begingroup$
On what are they based?
$endgroup$
– avan1235
Jan 8 at 15:30
$begingroup$
This holds over any field. Most common examples would be $mathbb R$ and $mathbb C$. If this is what you mean.
$endgroup$
– James
Jan 8 at 15:31
$begingroup$
But I meant a concept of proof 😅
$endgroup$
– avan1235
Jan 8 at 15:35
$begingroup$
Well, this is rephrasing the notions of surjectivity, injectivity and bijectivity. For example, since the rank gives the dimension of the image, the first one says that $A$ is surjective. This is, for every point in the image set there is a point that is mapped to this. In other words a solution to the linear system.
$endgroup$
– James
Jan 8 at 15:38
$begingroup$
@avan1235 does this help you?
$endgroup$
– James
Jan 9 at 9:36
|
show 2 more comments
$begingroup$
The determinant is always defined by the square matrix. Consequently, if a square matrix $A$ has non-zero determinant, this will lead to the invertibility and thus $x=A^{-1}b$ is the unique solution.
However, you cannot judge the existence and uniqueness of the linear equations by using the determinant if $A$ is not a square matrix. Simply consider like follows. Assume $A$ is a $m$-row, $n$-column matrix. If $A$ is full row rank which means that $A$ has $m$ linearly independent columns which can span the entire $m$ space (refer as range $R$). So that for any vector $b$ belongs to $R$, the solution always exists.
On the other hand, if $rank(A)=n$, all the columns in $A$ are linearly independent and $x=0$ would be the only solution of $Ax=0$. Notice that the full column rank condition does not give any information on the range space. The only conclusion can be made is that, the solution is unique if it has solution.
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $A$ be an $mtimes n$ matrix.
- If $operatorname{rank}(A)=m$, there always exists a solution.
- If $operatorname{rank}(A)=n$, there might be solutions or not, but if, then they are unique.
- Consequently, if $operatorname{rank}=m=n$, there always exists a unique solution. One may rephrase this to saying that $det(A)neq 0$.
answered Jan 8 at 15:29
JamesJames
801215
$endgroup$
$begingroup$
On what are they based?
$endgroup$
– avan1235
Jan 8 at 15:30
$begingroup$
This holds over any field. Most common examples would be $mathbb R$ and $mathbb C$. If this is what you mean.
$endgroup$
– James
Jan 8 at 15:31
$begingroup$
But I meant a concept of proof 😅
$endgroup$
– avan1235
Jan 8 at 15:35
$begingroup$
Well, this is rephrasing the notions of surjectivity, injectivity and bijectivity. For example, since the rank gives the dimension of the image, the first one says that $A$ is surjective. This is, for every point in the image set there is a point that is mapped to this. In other words a solution to the linear system.
$endgroup$
– James
Jan 8 at 15:38
$begingroup$
@avan1235 does this help you?
$endgroup$
– James
Jan 9 at 9:36
|
show 2 more comments
$begingroup$
Let $A$ be an $mtimes n$ matrix.
- If $operatorname{rank}(A)=m$, there always exists a solution.
- If $operatorname{rank}(A)=n$, there might be solutions or not, but if, then they are unique.
- Consequently, if $operatorname{rank}=m=n$, there always exists a unique solution. One may rephrase this to saying that $det(A)neq 0$.
answered Jan 8 at 15:29
JamesJames
801215
$endgroup$
$begingroup$
On what are they based?
$endgroup$
– avan1235
Jan 8 at 15:30
$begingroup$
This holds over any field. Most common examples would be $mathbb R$ and $mathbb C$. If this is what you mean.
$endgroup$
– James
Jan 8 at 15:31
$begingroup$
But I meant a concept of proof 😅
$endgroup$
– avan1235
Jan 8 at 15:35
$begingroup$
Well, this is rephrasing the notions of surjectivity, injectivity and bijectivity. For example, since the rank gives the dimension of the image, the first one says that $A$ is surjective. This is, for every point in the image set there is a point that is mapped to this. In other words a solution to the linear system.
$endgroup$
– James
Jan 8 at 15:38
$begingroup$
@avan1235 does this help you?
$endgroup$
– James
Jan 9 at 9:36
|
show 2 more comments
$begingroup$
Let $A$ be an $mtimes n$ matrix.
- If $operatorname{rank}(A)=m$, there always exists a solution.
- If $operatorname{rank}(A)=n$, there might be solutions or not, but if, then they are unique.
- Consequently, if $operatorname{rank}=m=n$, there always exists a unique solution. One may rephrase this to saying that $det(A)neq 0$.
answered Jan 8 at 15:29
JamesJames
801215
$endgroup$
Let $A$ be an $mtimes n$ matrix.
- If $operatorname{rank}(A)=m$, there always exists a solution.
- If $operatorname{rank}(A)=n$, there might be solutions or not, but if, then they are unique.
- Consequently, if $operatorname{rank}=m=n$, there always exists a unique solution. One may rephrase this to saying that $det(A)neq 0$.
answered Jan 8 at 15:29
JamesJames
801215
edited Jan 8 at 15:30
answered Jan 8 at 15:29
JamesJames
801215
answered Jan 8 at 15:29
JamesJames
801215
answered Jan 8 at 15:29
JamesJames
801215
801215
$begingroup$
On what are they based?
$endgroup$
– avan1235
Jan 8 at 15:30
$begingroup$
This holds over any field. Most common examples would be $mathbb R$ and $mathbb C$. If this is what you mean.
$endgroup$
– James
Jan 8 at 15:31
$begingroup$
But I meant a concept of proof 😅
$endgroup$
– avan1235
Jan 8 at 15:35
$begingroup$
Well, this is rephrasing the notions of surjectivity, injectivity and bijectivity. For example, since the rank gives the dimension of the image, the first one says that $A$ is surjective. This is, for every point in the image set there is a point that is mapped to this. In other words a solution to the linear system.
$endgroup$
– James
Jan 8 at 15:38
$begingroup$
@avan1235 does this help you?
$endgroup$
– James
Jan 9 at 9:36
|
show 2 more comments
$begingroup$
On what are they based?
$endgroup$
– avan1235
Jan 8 at 15:30
$begingroup$
This holds over any field. Most common examples would be $mathbb R$ and $mathbb C$. If this is what you mean.
$endgroup$
– James
Jan 8 at 15:31
$begingroup$
But I meant a concept of proof 😅
$endgroup$
– avan1235
Jan 8 at 15:35
$begingroup$
Well, this is rephrasing the notions of surjectivity, injectivity and bijectivity. For example, since the rank gives the dimension of the image, the first one says that $A$ is surjective. This is, for every point in the image set there is a point that is mapped to this. In other words a solution to the linear system.
$endgroup$
– James
Jan 8 at 15:38
$begingroup$
@avan1235 does this help you?
$endgroup$
– James
Jan 9 at 9:36
$begingroup$
On what are they based?
$endgroup$
– avan1235
Jan 8 at 15:30
$begingroup$
On what are they based?
$endgroup$
– avan1235
Jan 8 at 15:30
$begingroup$
This holds over any field. Most common examples would be $mathbb R$ and $mathbb C$. If this is what you mean.
$endgroup$
– James
Jan 8 at 15:31
$begingroup$
This holds over any field. Most common examples would be $mathbb R$ and $mathbb C$. If this is what you mean.
$endgroup$
– James
Jan 8 at 15:31
$begingroup$
But I meant a concept of proof 😅
$endgroup$
– avan1235
Jan 8 at 15:35
$begingroup$
But I meant a concept of proof 😅
$endgroup$
– avan1235
Jan 8 at 15:35
$begingroup$
Well, this is rephrasing the notions of surjectivity, injectivity and bijectivity. For example, since the rank gives the dimension of the image, the first one says that $A$ is surjective. This is, for every point in the image set there is a point that is mapped to this. In other words a solution to the linear system.
$endgroup$
– James
Jan 8 at 15:38
$begingroup$
Well, this is rephrasing the notions of surjectivity, injectivity and bijectivity. For example, since the rank gives the dimension of the image, the first one says that $A$ is surjective. This is, for every point in the image set there is a point that is mapped to this. In other words a solution to the linear system.
$endgroup$
– James
Jan 8 at 15:38
$begingroup$
@avan1235 does this help you?
$endgroup$
– James
Jan 9 at 9:36
$begingroup$
@avan1235 does this help you?
$endgroup$
– James
Jan 9 at 9:36
|
show 2 more comments
$begingroup$
The determinant is always defined by the square matrix. Consequently, if a square matrix $A$ has non-zero determinant, this will lead to the invertibility and thus $x=A^{-1}b$ is the unique solution.
However, you cannot judge the existence and uniqueness of the linear equations by using the determinant if $A$ is not a square matrix. Simply consider like follows. Assume $A$ is a $m$-row, $n$-column matrix. If $A$ is full row rank which means that $A$ has $m$ linearly independent columns which can span the entire $m$ space (refer as range $R$). So that for any vector $b$ belongs to $R$, the solution always exists.
On the other hand, if $rank(A)=n$, all the columns in $A$ are linearly independent and $x=0$ would be the only solution of $Ax=0$. Notice that the full column rank condition does not give any information on the range space. The only conclusion can be made is that, the solution is unique if it has solution.
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
$endgroup$
add a comment |
$begingroup$
The determinant is always defined by the square matrix. Consequently, if a square matrix $A$ has non-zero determinant, this will lead to the invertibility and thus $x=A^{-1}b$ is the unique solution.
However, you cannot judge the existence and uniqueness of the linear equations by using the determinant if $A$ is not a square matrix. Simply consider like follows. Assume $A$ is a $m$-row, $n$-column matrix. If $A$ is full row rank which means that $A$ has $m$ linearly independent columns which can span the entire $m$ space (refer as range $R$). So that for any vector $b$ belongs to $R$, the solution always exists.
On the other hand, if $rank(A)=n$, all the columns in $A$ are linearly independent and $x=0$ would be the only solution of $Ax=0$. Notice that the full column rank condition does not give any information on the range space. The only conclusion can be made is that, the solution is unique if it has solution.
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
$endgroup$
add a comment |
$begingroup$
The determinant is always defined by the square matrix. Consequently, if a square matrix $A$ has non-zero determinant, this will lead to the invertibility and thus $x=A^{-1}b$ is the unique solution.
However, you cannot judge the existence and uniqueness of the linear equations by using the determinant if $A$ is not a square matrix. Simply consider like follows. Assume $A$ is a $m$-row, $n$-column matrix. If $A$ is full row rank which means that $A$ has $m$ linearly independent columns which can span the entire $m$ space (refer as range $R$). So that for any vector $b$ belongs to $R$, the solution always exists.
On the other hand, if $rank(A)=n$, all the columns in $A$ are linearly independent and $x=0$ would be the only solution of $Ax=0$. Notice that the full column rank condition does not give any information on the range space. The only conclusion can be made is that, the solution is unique if it has solution.
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
$endgroup$
The determinant is always defined by the square matrix. Consequently, if a square matrix $A$ has non-zero determinant, this will lead to the invertibility and thus $x=A^{-1}b$ is the unique solution.
However, you cannot judge the existence and uniqueness of the linear equations by using the determinant if $A$ is not a square matrix. Simply consider like follows. Assume $A$ is a $m$-row, $n$-column matrix. If $A$ is full row rank which means that $A$ has $m$ linearly independent columns which can span the entire $m$ space (refer as range $R$). So that for any vector $b$ belongs to $R$, the solution always exists.
On the other hand, if $rank(A)=n$, all the columns in $A$ are linearly independent and $x=0$ would be the only solution of $Ax=0$. Notice that the full column rank condition does not give any information on the range space. The only conclusion can be made is that, the solution is unique if it has solution.
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
answered Jan 8 at 17:29
Wangkun XuWangkun Xu
485
485
add a comment |
add a comment |
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$begingroup$
Hm... something like Cramer's rule for example?
$endgroup$
– chickenNinja123
Jan 8 at 15:25
$begingroup$
Yes but more like this 😉
$endgroup$
– avan1235
Jan 8 at 15:26