Kernel and image of matrix: What are they? Why do they exist?What's an intuitive way to think about the...
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Kernel and image of matrix: What are they? Why do they exist?
What's an intuitive way to think about the determinant?LU Decomposition StepsSimplified method for symmetric matrix determinantsIf $A$ and $B$ are matrices where $BA$ is a defined matrix product, find a counterexample…Linear Transformation and Matricesa very basic definition cannot be found in most abstract algebra booksDetermine the values(s) of h and/or k such that the matrix is the augmented matrix of a consistent system.Intuitive understanding of Tensors of a higher rank (2 and above)Are all non-defective matrices normalIs the characteristic polynomial of a matrix $det(lambda I-A)$ or$ det(A-lambda I)$?
$begingroup$
I've been trying to get an understanding of the Kernel of image of matrices. I'm studying them in college right now, but the problem is, while I can find a ton of resources on how to find them given a matrix by following steps, I haven't been able to find anything that explains what they are intuitively. Also why have them in the first place?
I hope someone who understands these concepts better can help me.
linear-algebra matrices definition intuition motivation
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked 2 hours ago
EliorElior
153
$endgroup$
add a comment |
$begingroup$
I've been trying to get an understanding of the Kernel of image of matrices. I'm studying them in college right now, but the problem is, while I can find a ton of resources on how to find them given a matrix by following steps, I haven't been able to find anything that explains what they are intuitively. Also why have them in the first place?
I hope someone who understands these concepts better can help me.
linear-algebra matrices definition intuition motivation
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked 2 hours ago
EliorElior
153
$endgroup$
1
$begingroup$
Usually, we say the "Null space" of matrix, rather than the "Kernel" of matrix. Also the "Column space" of matrix rather than "Image" of matrix. Kernel and Image are using while talking linear transformation.
$endgroup$
– Jade Pang
2 hours ago
add a comment |
$begingroup$
I've been trying to get an understanding of the Kernel of image of matrices. I'm studying them in college right now, but the problem is, while I can find a ton of resources on how to find them given a matrix by following steps, I haven't been able to find anything that explains what they are intuitively. Also why have them in the first place?
I hope someone who understands these concepts better can help me.
linear-algebra matrices definition intuition motivation
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked 2 hours ago
EliorElior
153
$endgroup$
I've been trying to get an understanding of the Kernel of image of matrices. I'm studying them in college right now, but the problem is, while I can find a ton of resources on how to find them given a matrix by following steps, I haven't been able to find anything that explains what they are intuitively. Also why have them in the first place?
I hope someone who understands these concepts better can help me.
linear-algebra matrices definition intuition motivation
linear-algebra matrices definition intuition motivation
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked 2 hours ago
EliorElior
153
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
asked 2 hours ago
EliorElior
153
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
13.5k72550
13.5k72550
asked 2 hours ago
EliorElior
153
asked 2 hours ago
EliorElior
153
asked 2 hours ago
EliorElior
153
153
1
$begingroup$
Usually, we say the "Null space" of matrix, rather than the "Kernel" of matrix. Also the "Column space" of matrix rather than "Image" of matrix. Kernel and Image are using while talking linear transformation.
$endgroup$
– Jade Pang
2 hours ago
add a comment |
1
$begingroup$
Usually, we say the "Null space" of matrix, rather than the "Kernel" of matrix. Also the "Column space" of matrix rather than "Image" of matrix. Kernel and Image are using while talking linear transformation.
$endgroup$
– Jade Pang
2 hours ago
1
1
$begingroup$
Usually, we say the "Null space" of matrix, rather than the "Kernel" of matrix. Also the "Column space" of matrix rather than "Image" of matrix. Kernel and Image are using while talking linear transformation.
$endgroup$
– Jade Pang
2 hours ago
$begingroup$
Usually, we say the "Null space" of matrix, rather than the "Kernel" of matrix. Also the "Column space" of matrix rather than "Image" of matrix. Kernel and Image are using while talking linear transformation.
$endgroup$
– Jade Pang
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The matrix corresponds to a linear function. Generally speaking, for a function $f:Xto Y$, the kernel of $f$ is $f^{-1}(0)$ and the image is $f(X)$.
Here $Y$ is a group, ring, field, vector space etc so it has a zero element $0in Y$.
Note the kernel $f^{-1}(0)$ is a subspace (subgroup, etc) of $X$ and the image $f(X)$ is a subspace of $Y$.
There is a nice sequence of maps
$$0toker fto Xto operatorname{im} fto Y.$$
answered 2 hours ago
MPWMPW
30.4k12157
$endgroup$
$begingroup$
So the kernel would be the whatever you need to input to 'f' in order to get 0 (or 0 vector)?
$endgroup$
– Elior
2 hours ago
$begingroup$
@Elior : Precisely. It’s everything that gets mapped to zero.
$endgroup$
– MPW
2 hours ago
$begingroup$
That makes sense. I'm a little confused by the sequence of maps at the end of you question (I may be interpreting it wrong). Is that showing the relationship between 0 and the ker(f), and then the ker(f) being part of the domain X, then same for im(f) being part of codomain. Why is there an arrow from X to im(f)? How does the domain relate to the image?
$endgroup$
– Elior
2 hours ago
$begingroup$
The arrow you mention is the action of the map $f$ of $X$ onto its image. The arrows before and after that can be thought of as simple inclusion maps. Try googling “exact sequence” for more info.
$endgroup$
– MPW
2 hours ago
add a comment |
$begingroup$
Well they exist because they kind of define the output given the input or if there is a unique solution of an inverse problem etc.
They both defines sets. Kernel defines a set of vectors such that take one of them and multiply it with your vector, you end up with 0 vector. Image on the other hand defines a "target set". Take any vector that can be multiplied with your matrix and multiply them. You cannot end up with a vector that does not belong to the image of your matrix.
answered 2 hours ago
starhdstarhd
11
$endgroup$
$begingroup$
What is an example of a use for the kernel (in mathematics or any other field)?
$endgroup$
– Elior
2 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The matrix corresponds to a linear function. Generally speaking, for a function $f:Xto Y$, the kernel of $f$ is $f^{-1}(0)$ and the image is $f(X)$.
Here $Y$ is a group, ring, field, vector space etc so it has a zero element $0in Y$.
Note the kernel $f^{-1}(0)$ is a subspace (subgroup, etc) of $X$ and the image $f(X)$ is a subspace of $Y$.
There is a nice sequence of maps
$$0toker fto Xto operatorname{im} fto Y.$$
answered 2 hours ago
MPWMPW
30.4k12157
$endgroup$
$begingroup$
So the kernel would be the whatever you need to input to 'f' in order to get 0 (or 0 vector)?
$endgroup$
– Elior
2 hours ago
$begingroup$
@Elior : Precisely. It’s everything that gets mapped to zero.
$endgroup$
– MPW
2 hours ago
$begingroup$
That makes sense. I'm a little confused by the sequence of maps at the end of you question (I may be interpreting it wrong). Is that showing the relationship between 0 and the ker(f), and then the ker(f) being part of the domain X, then same for im(f) being part of codomain. Why is there an arrow from X to im(f)? How does the domain relate to the image?
$endgroup$
– Elior
2 hours ago
$begingroup$
The arrow you mention is the action of the map $f$ of $X$ onto its image. The arrows before and after that can be thought of as simple inclusion maps. Try googling “exact sequence” for more info.
$endgroup$
– MPW
2 hours ago
add a comment |
$begingroup$
The matrix corresponds to a linear function. Generally speaking, for a function $f:Xto Y$, the kernel of $f$ is $f^{-1}(0)$ and the image is $f(X)$.
Here $Y$ is a group, ring, field, vector space etc so it has a zero element $0in Y$.
Note the kernel $f^{-1}(0)$ is a subspace (subgroup, etc) of $X$ and the image $f(X)$ is a subspace of $Y$.
There is a nice sequence of maps
$$0toker fto Xto operatorname{im} fto Y.$$
answered 2 hours ago
MPWMPW
30.4k12157
$endgroup$
$begingroup$
So the kernel would be the whatever you need to input to 'f' in order to get 0 (or 0 vector)?
$endgroup$
– Elior
2 hours ago
$begingroup$
@Elior : Precisely. It’s everything that gets mapped to zero.
$endgroup$
– MPW
2 hours ago
$begingroup$
That makes sense. I'm a little confused by the sequence of maps at the end of you question (I may be interpreting it wrong). Is that showing the relationship between 0 and the ker(f), and then the ker(f) being part of the domain X, then same for im(f) being part of codomain. Why is there an arrow from X to im(f)? How does the domain relate to the image?
$endgroup$
– Elior
2 hours ago
$begingroup$
The arrow you mention is the action of the map $f$ of $X$ onto its image. The arrows before and after that can be thought of as simple inclusion maps. Try googling “exact sequence” for more info.
$endgroup$
– MPW
2 hours ago
add a comment |
$begingroup$
The matrix corresponds to a linear function. Generally speaking, for a function $f:Xto Y$, the kernel of $f$ is $f^{-1}(0)$ and the image is $f(X)$.
Here $Y$ is a group, ring, field, vector space etc so it has a zero element $0in Y$.
Note the kernel $f^{-1}(0)$ is a subspace (subgroup, etc) of $X$ and the image $f(X)$ is a subspace of $Y$.
There is a nice sequence of maps
$$0toker fto Xto operatorname{im} fto Y.$$
answered 2 hours ago
MPWMPW
30.4k12157
$endgroup$
The matrix corresponds to a linear function. Generally speaking, for a function $f:Xto Y$, the kernel of $f$ is $f^{-1}(0)$ and the image is $f(X)$.
Here $Y$ is a group, ring, field, vector space etc so it has a zero element $0in Y$.
Note the kernel $f^{-1}(0)$ is a subspace (subgroup, etc) of $X$ and the image $f(X)$ is a subspace of $Y$.
There is a nice sequence of maps
$$0toker fto Xto operatorname{im} fto Y.$$
answered 2 hours ago
MPWMPW
30.4k12157
edited 2 hours ago
answered 2 hours ago
MPWMPW
30.4k12157
answered 2 hours ago
MPWMPW
30.4k12157
answered 2 hours ago
MPWMPW
30.4k12157
30.4k12157
$begingroup$
So the kernel would be the whatever you need to input to 'f' in order to get 0 (or 0 vector)?
$endgroup$
– Elior
2 hours ago
$begingroup$
@Elior : Precisely. It’s everything that gets mapped to zero.
$endgroup$
– MPW
2 hours ago
$begingroup$
That makes sense. I'm a little confused by the sequence of maps at the end of you question (I may be interpreting it wrong). Is that showing the relationship between 0 and the ker(f), and then the ker(f) being part of the domain X, then same for im(f) being part of codomain. Why is there an arrow from X to im(f)? How does the domain relate to the image?
$endgroup$
– Elior
2 hours ago
$begingroup$
The arrow you mention is the action of the map $f$ of $X$ onto its image. The arrows before and after that can be thought of as simple inclusion maps. Try googling “exact sequence” for more info.
$endgroup$
– MPW
2 hours ago
add a comment |
$begingroup$
So the kernel would be the whatever you need to input to 'f' in order to get 0 (or 0 vector)?
$endgroup$
– Elior
2 hours ago
$begingroup$
@Elior : Precisely. It’s everything that gets mapped to zero.
$endgroup$
– MPW
2 hours ago
$begingroup$
That makes sense. I'm a little confused by the sequence of maps at the end of you question (I may be interpreting it wrong). Is that showing the relationship between 0 and the ker(f), and then the ker(f) being part of the domain X, then same for im(f) being part of codomain. Why is there an arrow from X to im(f)? How does the domain relate to the image?
$endgroup$
– Elior
2 hours ago
$begingroup$
The arrow you mention is the action of the map $f$ of $X$ onto its image. The arrows before and after that can be thought of as simple inclusion maps. Try googling “exact sequence” for more info.
$endgroup$
– MPW
2 hours ago
$begingroup$
So the kernel would be the whatever you need to input to 'f' in order to get 0 (or 0 vector)?
$endgroup$
– Elior
2 hours ago
$begingroup$
So the kernel would be the whatever you need to input to 'f' in order to get 0 (or 0 vector)?
$endgroup$
– Elior
2 hours ago
$begingroup$
@Elior : Precisely. It’s everything that gets mapped to zero.
$endgroup$
– MPW
2 hours ago
$begingroup$
@Elior : Precisely. It’s everything that gets mapped to zero.
$endgroup$
– MPW
2 hours ago
$begingroup$
That makes sense. I'm a little confused by the sequence of maps at the end of you question (I may be interpreting it wrong). Is that showing the relationship between 0 and the ker(f), and then the ker(f) being part of the domain X, then same for im(f) being part of codomain. Why is there an arrow from X to im(f)? How does the domain relate to the image?
$endgroup$
– Elior
2 hours ago
$begingroup$
That makes sense. I'm a little confused by the sequence of maps at the end of you question (I may be interpreting it wrong). Is that showing the relationship between 0 and the ker(f), and then the ker(f) being part of the domain X, then same for im(f) being part of codomain. Why is there an arrow from X to im(f)? How does the domain relate to the image?
$endgroup$
– Elior
2 hours ago
$begingroup$
The arrow you mention is the action of the map $f$ of $X$ onto its image. The arrows before and after that can be thought of as simple inclusion maps. Try googling “exact sequence” for more info.
$endgroup$
– MPW
2 hours ago
$begingroup$
The arrow you mention is the action of the map $f$ of $X$ onto its image. The arrows before and after that can be thought of as simple inclusion maps. Try googling “exact sequence” for more info.
$endgroup$
– MPW
2 hours ago
add a comment |
$begingroup$
Well they exist because they kind of define the output given the input or if there is a unique solution of an inverse problem etc.
They both defines sets. Kernel defines a set of vectors such that take one of them and multiply it with your vector, you end up with 0 vector. Image on the other hand defines a "target set". Take any vector that can be multiplied with your matrix and multiply them. You cannot end up with a vector that does not belong to the image of your matrix.
answered 2 hours ago
starhdstarhd
11
$endgroup$
$begingroup$
What is an example of a use for the kernel (in mathematics or any other field)?
$endgroup$
– Elior
2 hours ago
add a comment |
$begingroup$
Well they exist because they kind of define the output given the input or if there is a unique solution of an inverse problem etc.
They both defines sets. Kernel defines a set of vectors such that take one of them and multiply it with your vector, you end up with 0 vector. Image on the other hand defines a "target set". Take any vector that can be multiplied with your matrix and multiply them. You cannot end up with a vector that does not belong to the image of your matrix.
answered 2 hours ago
starhdstarhd
11
$endgroup$
$begingroup$
What is an example of a use for the kernel (in mathematics or any other field)?
$endgroup$
– Elior
2 hours ago
add a comment |
$begingroup$
Well they exist because they kind of define the output given the input or if there is a unique solution of an inverse problem etc.
They both defines sets. Kernel defines a set of vectors such that take one of them and multiply it with your vector, you end up with 0 vector. Image on the other hand defines a "target set". Take any vector that can be multiplied with your matrix and multiply them. You cannot end up with a vector that does not belong to the image of your matrix.
answered 2 hours ago
starhdstarhd
11
$endgroup$
Well they exist because they kind of define the output given the input or if there is a unique solution of an inverse problem etc.
They both defines sets. Kernel defines a set of vectors such that take one of them and multiply it with your vector, you end up with 0 vector. Image on the other hand defines a "target set". Take any vector that can be multiplied with your matrix and multiply them. You cannot end up with a vector that does not belong to the image of your matrix.
answered 2 hours ago
starhdstarhd
11
answered 2 hours ago
starhdstarhd
11
answered 2 hours ago
starhdstarhd
11
answered 2 hours ago
starhdstarhd
11
11
$begingroup$
What is an example of a use for the kernel (in mathematics or any other field)?
$endgroup$
– Elior
2 hours ago
add a comment |
$begingroup$
What is an example of a use for the kernel (in mathematics or any other field)?
$endgroup$
– Elior
2 hours ago
$begingroup$
What is an example of a use for the kernel (in mathematics or any other field)?
$endgroup$
– Elior
2 hours ago
$begingroup$
What is an example of a use for the kernel (in mathematics or any other field)?
$endgroup$
– Elior
2 hours ago
add a comment |
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$begingroup$
Usually, we say the "Null space" of matrix, rather than the "Kernel" of matrix. Also the "Column space" of matrix rather than "Image" of matrix. Kernel and Image are using while talking linear transformation.
$endgroup$
– Jade Pang
2 hours ago