Intuitions on understanding system of linear equations as linear transformations
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I am looking for an explanation of (in)homogeneity of linear systems of equations from a perspective of linear transformations.
For example, the following connections between linear transformations and linear systems of equations I understand:
1) If the system is homogenous it has at least $mathbf{0}$ as a solution.
Transformation perspective:
$mathbf{0}$ is a solution, as $Tmathbf{0} = mathbf{0}$ for all linear transformations $T$; trivial from definition of matrix multiplication.
There might be other vectors that $T$ sends to $0$.
2) If the system is invertible, is has exactly one solution.
Transformation perspective:
If it is invertible, it is a bijective mapping, thus there is exactly one vector that satisfies the system. If there were more, it would not be bijective.
So, specifically I would like to understand from a transformations perspective the following:
From a linear transformations perspective, what does it mean for a system of equatiosn to be homogenous (apart from $mathbf{0}$ being a trivial member of the nullspace of the transformation.)
From a linear transformations perspective, what does it mean for a system of equatiosn to be inhomogeneous
what are free variables from the perspective of linear transformations
from the perspective of linear transformations why are there infinitely many solutions if there are more variables than equations ( = basis vectors from the point of view of transformation?)
linear-algebra linear-transformations systems-of-equations
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
$endgroup$
|
show 4 more comments
$begingroup$
I am looking for an explanation of (in)homogeneity of linear systems of equations from a perspective of linear transformations.
For example, the following connections between linear transformations and linear systems of equations I understand:
1) If the system is homogenous it has at least $mathbf{0}$ as a solution.
Transformation perspective:
$mathbf{0}$ is a solution, as $Tmathbf{0} = mathbf{0}$ for all linear transformations $T$; trivial from definition of matrix multiplication.
There might be other vectors that $T$ sends to $0$.
2) If the system is invertible, is has exactly one solution.
Transformation perspective:
If it is invertible, it is a bijective mapping, thus there is exactly one vector that satisfies the system. If there were more, it would not be bijective.
So, specifically I would like to understand from a transformations perspective the following:
From a linear transformations perspective, what does it mean for a system of equatiosn to be homogenous (apart from $mathbf{0}$ being a trivial member of the nullspace of the transformation.)
From a linear transformations perspective, what does it mean for a system of equatiosn to be inhomogeneous
what are free variables from the perspective of linear transformations
from the perspective of linear transformations why are there infinitely many solutions if there are more variables than equations ( = basis vectors from the point of view of transformation?)
linear-algebra linear-transformations systems-of-equations
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
$endgroup$
1
$begingroup$
You understand the notion of rank?
$endgroup$
– Randall
Dec 29 '18 at 2:36
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Yes, it is the number on linearly independent vectors in a matrix, which is the same as the dimensionality of the transformation.
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– user3578468
Dec 29 '18 at 2:42
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What exactly do you mean by your phrase “dimensionality of the transformation?”
$endgroup$
– amd
Dec 29 '18 at 3:04
$begingroup$
As Randall is suggesting, any linear transformation can be represented as a matrix (which we will designated as A). If the matrix has full rank then f(x) = Ax is a bijective mapping. If A does not have full rank, then the null space is non-trivial and for any output, there are infinitely many inputs.
$endgroup$
– NicNic8
Dec 29 '18 at 3:10
$begingroup$
@amd the number of independent columns/ basis vectors.
$endgroup$
– user3578468
Dec 29 '18 at 3:13
|
show 4 more comments
$begingroup$
I am looking for an explanation of (in)homogeneity of linear systems of equations from a perspective of linear transformations.
For example, the following connections between linear transformations and linear systems of equations I understand:
1) If the system is homogenous it has at least $mathbf{0}$ as a solution.
Transformation perspective:
$mathbf{0}$ is a solution, as $Tmathbf{0} = mathbf{0}$ for all linear transformations $T$; trivial from definition of matrix multiplication.
There might be other vectors that $T$ sends to $0$.
2) If the system is invertible, is has exactly one solution.
Transformation perspective:
If it is invertible, it is a bijective mapping, thus there is exactly one vector that satisfies the system. If there were more, it would not be bijective.
So, specifically I would like to understand from a transformations perspective the following:
From a linear transformations perspective, what does it mean for a system of equatiosn to be homogenous (apart from $mathbf{0}$ being a trivial member of the nullspace of the transformation.)
From a linear transformations perspective, what does it mean for a system of equatiosn to be inhomogeneous
what are free variables from the perspective of linear transformations
from the perspective of linear transformations why are there infinitely many solutions if there are more variables than equations ( = basis vectors from the point of view of transformation?)
linear-algebra linear-transformations systems-of-equations
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
$endgroup$
I am looking for an explanation of (in)homogeneity of linear systems of equations from a perspective of linear transformations.
For example, the following connections between linear transformations and linear systems of equations I understand:
1) If the system is homogenous it has at least $mathbf{0}$ as a solution.
Transformation perspective:
$mathbf{0}$ is a solution, as $Tmathbf{0} = mathbf{0}$ for all linear transformations $T$; trivial from definition of matrix multiplication.
There might be other vectors that $T$ sends to $0$.
2) If the system is invertible, is has exactly one solution.
Transformation perspective:
If it is invertible, it is a bijective mapping, thus there is exactly one vector that satisfies the system. If there were more, it would not be bijective.
So, specifically I would like to understand from a transformations perspective the following:
From a linear transformations perspective, what does it mean for a system of equatiosn to be homogenous (apart from $mathbf{0}$ being a trivial member of the nullspace of the transformation.)
From a linear transformations perspective, what does it mean for a system of equatiosn to be inhomogeneous
what are free variables from the perspective of linear transformations
from the perspective of linear transformations why are there infinitely many solutions if there are more variables than equations ( = basis vectors from the point of view of transformation?)
linear-algebra linear-transformations systems-of-equations
linear-algebra linear-transformations systems-of-equations
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
asked Dec 29 '18 at 2:35
user3578468user3578468
445312
445312
1
$begingroup$
You understand the notion of rank?
$endgroup$
– Randall
Dec 29 '18 at 2:36
$begingroup$
Yes, it is the number on linearly independent vectors in a matrix, which is the same as the dimensionality of the transformation.
$endgroup$
– user3578468
Dec 29 '18 at 2:42
$begingroup$
What exactly do you mean by your phrase “dimensionality of the transformation?”
$endgroup$
– amd
Dec 29 '18 at 3:04
$begingroup$
As Randall is suggesting, any linear transformation can be represented as a matrix (which we will designated as A). If the matrix has full rank then f(x) = Ax is a bijective mapping. If A does not have full rank, then the null space is non-trivial and for any output, there are infinitely many inputs.
$endgroup$
– NicNic8
Dec 29 '18 at 3:10
$begingroup$
@amd the number of independent columns/ basis vectors.
$endgroup$
– user3578468
Dec 29 '18 at 3:13
|
show 4 more comments
1
$begingroup$
You understand the notion of rank?
$endgroup$
– Randall
Dec 29 '18 at 2:36
$begingroup$
Yes, it is the number on linearly independent vectors in a matrix, which is the same as the dimensionality of the transformation.
$endgroup$
– user3578468
Dec 29 '18 at 2:42
$begingroup$
What exactly do you mean by your phrase “dimensionality of the transformation?”
$endgroup$
– amd
Dec 29 '18 at 3:04
$begingroup$
As Randall is suggesting, any linear transformation can be represented as a matrix (which we will designated as A). If the matrix has full rank then f(x) = Ax is a bijective mapping. If A does not have full rank, then the null space is non-trivial and for any output, there are infinitely many inputs.
$endgroup$
– NicNic8
Dec 29 '18 at 3:10
$begingroup$
@amd the number of independent columns/ basis vectors.
$endgroup$
– user3578468
Dec 29 '18 at 3:13
1
1
$begingroup$
You understand the notion of rank?
$endgroup$
– Randall
Dec 29 '18 at 2:36
$begingroup$
You understand the notion of rank?
$endgroup$
– Randall
Dec 29 '18 at 2:36
$begingroup$
Yes, it is the number on linearly independent vectors in a matrix, which is the same as the dimensionality of the transformation.
$endgroup$
– user3578468
Dec 29 '18 at 2:42
$begingroup$
Yes, it is the number on linearly independent vectors in a matrix, which is the same as the dimensionality of the transformation.
$endgroup$
– user3578468
Dec 29 '18 at 2:42
$begingroup$
What exactly do you mean by your phrase “dimensionality of the transformation?”
$endgroup$
– amd
Dec 29 '18 at 3:04
$begingroup$
What exactly do you mean by your phrase “dimensionality of the transformation?”
$endgroup$
– amd
Dec 29 '18 at 3:04
$begingroup$
As Randall is suggesting, any linear transformation can be represented as a matrix (which we will designated as A). If the matrix has full rank then f(x) = Ax is a bijective mapping. If A does not have full rank, then the null space is non-trivial and for any output, there are infinitely many inputs.
$endgroup$
– NicNic8
Dec 29 '18 at 3:10
$begingroup$
As Randall is suggesting, any linear transformation can be represented as a matrix (which we will designated as A). If the matrix has full rank then f(x) = Ax is a bijective mapping. If A does not have full rank, then the null space is non-trivial and for any output, there are infinitely many inputs.
$endgroup$
– NicNic8
Dec 29 '18 at 3:10
$begingroup$
@amd the number of independent columns/ basis vectors.
$endgroup$
– user3578468
Dec 29 '18 at 3:13
$begingroup$
@amd the number of independent columns/ basis vectors.
$endgroup$
– user3578468
Dec 29 '18 at 3:13
|
show 4 more comments
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1
$begingroup$
You understand the notion of rank?
$endgroup$
– Randall
Dec 29 '18 at 2:36
$begingroup$
Yes, it is the number on linearly independent vectors in a matrix, which is the same as the dimensionality of the transformation.
$endgroup$
– user3578468
Dec 29 '18 at 2:42
$begingroup$
What exactly do you mean by your phrase “dimensionality of the transformation?”
$endgroup$
– amd
Dec 29 '18 at 3:04
$begingroup$
As Randall is suggesting, any linear transformation can be represented as a matrix (which we will designated as A). If the matrix has full rank then f(x) = Ax is a bijective mapping. If A does not have full rank, then the null space is non-trivial and for any output, there are infinitely many inputs.
$endgroup$
– NicNic8
Dec 29 '18 at 3:10
$begingroup$
@amd the number of independent columns/ basis vectors.
$endgroup$
– user3578468
Dec 29 '18 at 3:13