Mixing
identical solution spaces based on matrices of the same order and the same rank
$begingroup$
I came across a statement that:
if A and B are matrices of the same order, and the rank of B = the rank of AB, then the solution space of the homogenous system $AB underline{x}=0$ is identical to the solution space to the homogenous system $B underline{x}=0$.
No proof is provided, and I am a bit puzzled at why this would be the case? Can anyone provide some insight?
Thank you!
linear-algebra matrices
asked Jan 8 at 18:01
daltadalta
1098
$endgroup$
add a comment |
$begingroup$
I came across a statement that:
if A and B are matrices of the same order, and the rank of B = the rank of AB, then the solution space of the homogenous system $AB underline{x}=0$ is identical to the solution space to the homogenous system $B underline{x}=0$.
No proof is provided, and I am a bit puzzled at why this would be the case? Can anyone provide some insight?
Thank you!
linear-algebra matrices
asked Jan 8 at 18:01
daltadalta
1098
$endgroup$
add a comment |
$begingroup$
I came across a statement that:
if A and B are matrices of the same order, and the rank of B = the rank of AB, then the solution space of the homogenous system $AB underline{x}=0$ is identical to the solution space to the homogenous system $B underline{x}=0$.
No proof is provided, and I am a bit puzzled at why this would be the case? Can anyone provide some insight?
Thank you!
linear-algebra matrices
asked Jan 8 at 18:01
daltadalta
1098
$endgroup$
I came across a statement that:
if A and B are matrices of the same order, and the rank of B = the rank of AB, then the solution space of the homogenous system $AB underline{x}=0$ is identical to the solution space to the homogenous system $B underline{x}=0$.
No proof is provided, and I am a bit puzzled at why this would be the case? Can anyone provide some insight?
Thank you!
linear-algebra matrices
linear-algebra matrices
asked Jan 8 at 18:01
daltadalta
1098
asked Jan 8 at 18:01
daltadalta
1098
asked Jan 8 at 18:01
daltadalta
1098
asked Jan 8 at 18:01
daltadalta
1098
asked Jan 8 at 18:01
daltadalta
1098
1098
add a comment |
add a comment |
2 Answers
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The rows of $AB$ are linear combinations of the rows of $B$. If the ranks of $B,AB$ are identical, the rowspace of $B$ and $AB$ is the same, and the nullspace of $B,AB$ is also the same, being the orthogonal complement of the row space.
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
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$begingroup$
In general, $x$ will be a solution to $ABx = 0$ if either $Bx = 0$, or $Bx neq 0$ and $ABx = 0$. Consequently, every solution to $Bx = 0$ is also a solution to $ABx = 0$.
If we are given than $AB$ and $B$ have the same rank, however, then we can conclude (by the rank-nullity theorem) that the nullspaces of $AB$ and $B$ have the same dimension. So, the solution space to $Bx = 0$ is a subspace of $ABx$, but both subspaces have the same dimension. So, the subspaces (i.e. the solutions spaces) must be identical.
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
$endgroup$
$begingroup$
I'm not sure the first line is correct. Did you mean $ABx=0$?
$endgroup$
– Shubham Johri
Jan 8 at 19:31
$begingroup$
@ShubhamJohri yes, good catch
$endgroup$
– Omnomnomnom
Jan 8 at 19:35
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
The rows of $AB$ are linear combinations of the rows of $B$. If the ranks of $B,AB$ are identical, the rowspace of $B$ and $AB$ is the same, and the nullspace of $B,AB$ is also the same, being the orthogonal complement of the row space.
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
The rows of $AB$ are linear combinations of the rows of $B$. If the ranks of $B,AB$ are identical, the rowspace of $B$ and $AB$ is the same, and the nullspace of $B,AB$ is also the same, being the orthogonal complement of the row space.
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
$endgroup$
add a comment |
$begingroup$
The rows of $AB$ are linear combinations of the rows of $B$. If the ranks of $B,AB$ are identical, the rowspace of $B$ and $AB$ is the same, and the nullspace of $B,AB$ is also the same, being the orthogonal complement of the row space.
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
$endgroup$
The rows of $AB$ are linear combinations of the rows of $B$. If the ranks of $B,AB$ are identical, the rowspace of $B$ and $AB$ is the same, and the nullspace of $B,AB$ is also the same, being the orthogonal complement of the row space.
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
answered Jan 8 at 18:29
Shubham JohriShubham Johri
5,082717
5,082717
add a comment |
add a comment |
$begingroup$
In general, $x$ will be a solution to $ABx = 0$ if either $Bx = 0$, or $Bx neq 0$ and $ABx = 0$. Consequently, every solution to $Bx = 0$ is also a solution to $ABx = 0$.
If we are given than $AB$ and $B$ have the same rank, however, then we can conclude (by the rank-nullity theorem) that the nullspaces of $AB$ and $B$ have the same dimension. So, the solution space to $Bx = 0$ is a subspace of $ABx$, but both subspaces have the same dimension. So, the subspaces (i.e. the solutions spaces) must be identical.
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
$endgroup$
$begingroup$
I'm not sure the first line is correct. Did you mean $ABx=0$?
$endgroup$
– Shubham Johri
Jan 8 at 19:31
$begingroup$
@ShubhamJohri yes, good catch
$endgroup$
– Omnomnomnom
Jan 8 at 19:35
add a comment |
$begingroup$
In general, $x$ will be a solution to $ABx = 0$ if either $Bx = 0$, or $Bx neq 0$ and $ABx = 0$. Consequently, every solution to $Bx = 0$ is also a solution to $ABx = 0$.
If we are given than $AB$ and $B$ have the same rank, however, then we can conclude (by the rank-nullity theorem) that the nullspaces of $AB$ and $B$ have the same dimension. So, the solution space to $Bx = 0$ is a subspace of $ABx$, but both subspaces have the same dimension. So, the subspaces (i.e. the solutions spaces) must be identical.
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
$endgroup$
$begingroup$
I'm not sure the first line is correct. Did you mean $ABx=0$?
$endgroup$
– Shubham Johri
Jan 8 at 19:31
$begingroup$
@ShubhamJohri yes, good catch
$endgroup$
– Omnomnomnom
Jan 8 at 19:35
add a comment |
$begingroup$
In general, $x$ will be a solution to $ABx = 0$ if either $Bx = 0$, or $Bx neq 0$ and $ABx = 0$. Consequently, every solution to $Bx = 0$ is also a solution to $ABx = 0$.
If we are given than $AB$ and $B$ have the same rank, however, then we can conclude (by the rank-nullity theorem) that the nullspaces of $AB$ and $B$ have the same dimension. So, the solution space to $Bx = 0$ is a subspace of $ABx$, but both subspaces have the same dimension. So, the subspaces (i.e. the solutions spaces) must be identical.
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
$endgroup$
In general, $x$ will be a solution to $ABx = 0$ if either $Bx = 0$, or $Bx neq 0$ and $ABx = 0$. Consequently, every solution to $Bx = 0$ is also a solution to $ABx = 0$.
If we are given than $AB$ and $B$ have the same rank, however, then we can conclude (by the rank-nullity theorem) that the nullspaces of $AB$ and $B$ have the same dimension. So, the solution space to $Bx = 0$ is a subspace of $ABx$, but both subspaces have the same dimension. So, the subspaces (i.e. the solutions spaces) must be identical.
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
edited Jan 8 at 19:35
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
answered Jan 8 at 18:31
OmnomnomnomOmnomnomnom
128k790179
128k790179
$begingroup$
I'm not sure the first line is correct. Did you mean $ABx=0$?
$endgroup$
– Shubham Johri
Jan 8 at 19:31
$begingroup$
@ShubhamJohri yes, good catch
$endgroup$
– Omnomnomnom
Jan 8 at 19:35
add a comment |
$begingroup$
I'm not sure the first line is correct. Did you mean $ABx=0$?
$endgroup$
– Shubham Johri
Jan 8 at 19:31
$begingroup$
@ShubhamJohri yes, good catch
$endgroup$
– Omnomnomnom
Jan 8 at 19:35
$begingroup$
I'm not sure the first line is correct. Did you mean $ABx=0$?
$endgroup$
– Shubham Johri
Jan 8 at 19:31
$begingroup$
I'm not sure the first line is correct. Did you mean $ABx=0$?
$endgroup$
– Shubham Johri
Jan 8 at 19:31
$begingroup$
@ShubhamJohri yes, good catch
$endgroup$
– Omnomnomnom
Jan 8 at 19:35
$begingroup$
@ShubhamJohri yes, good catch
$endgroup$
– Omnomnomnom
Jan 8 at 19:35
add a comment |
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