Matrices with constant column vectors in linear algebra [closed]
I'm doing an assignment in linear algebra that involves a notation that I don't understand it is simply two ones ($mathbf{11}$). I understand that a $mathbf{1}$ in linear algebra means a sum vector but does that mean that $mathbf{11}$ is simply the matrix counterpart "sum matrix"?.
For example,
$$mathbf{1}_{2} = begin{bmatrix} 1 \ 1end{bmatrix}.$$
Would it be correct to assume that
$$begin{bmatrix}mathbf{1}_{2}mathbf{1}_{2}end{bmatrix} = begin{bmatrix} 1 & 1 \ 1 & 1end{bmatrix}?$$
I'm asking because there is an equation that involves the transpose of $mathbf{11}$ but since it is all ones and it is square that doesn't do anything.
linear-algebra matrices
edited Dec 27 '18 at 6:39
dantopa
6,42932042
asked Dec 27 '18 at 1:49
Mads
1
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as unclear what you're asking by amWhy, Lord Shark the Unknown, Eevee Trainer, KReiser, user91500 Dec 27 '18 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 7 more comments
I'm doing an assignment in linear algebra that involves a notation that I don't understand it is simply two ones ($mathbf{11}$). I understand that a $mathbf{1}$ in linear algebra means a sum vector but does that mean that $mathbf{11}$ is simply the matrix counterpart "sum matrix"?.
For example,
$$mathbf{1}_{2} = begin{bmatrix} 1 \ 1end{bmatrix}.$$
Would it be correct to assume that
$$begin{bmatrix}mathbf{1}_{2}mathbf{1}_{2}end{bmatrix} = begin{bmatrix} 1 & 1 \ 1 & 1end{bmatrix}?$$
I'm asking because there is an equation that involves the transpose of $mathbf{11}$ but since it is all ones and it is square that doesn't do anything.
linear-algebra matrices
edited Dec 27 '18 at 6:39
dantopa
6,42932042
asked Dec 27 '18 at 1:49
Mads
1
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as unclear what you're asking by amWhy, Lord Shark the Unknown, Eevee Trainer, KReiser, user91500 Dec 27 '18 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Could you share the exact problem/text you're looking at? It may be helpful
– Alex
Dec 27 '18 at 1:51
I'm calculating the pagerank of adjacency matrices. On a slide there has been given an equation that involves the notation 11**^T and the only information ive been given is that **1 is a vector of n ones where n of course is the dimensios of the adjacency matrix.
– Mads
Dec 27 '18 at 2:02
So nothing about what the "contraction" of two 1 's is
– Mads
Dec 27 '18 at 2:03
1
I have never seen this notation before and I've had a two-semester linear algebra course. If I were you I'd ask for clarification to whoever assigned this to you.
– Victor S.
Dec 27 '18 at 3:08
That's the problem, we have holliday so they won't be able to answer. Im pretty sure it means what I think though, I just have to be sure.
– Mads
Dec 27 '18 at 3:25
|
show 7 more comments
I'm doing an assignment in linear algebra that involves a notation that I don't understand it is simply two ones ($mathbf{11}$). I understand that a $mathbf{1}$ in linear algebra means a sum vector but does that mean that $mathbf{11}$ is simply the matrix counterpart "sum matrix"?.
For example,
$$mathbf{1}_{2} = begin{bmatrix} 1 \ 1end{bmatrix}.$$
Would it be correct to assume that
$$begin{bmatrix}mathbf{1}_{2}mathbf{1}_{2}end{bmatrix} = begin{bmatrix} 1 & 1 \ 1 & 1end{bmatrix}?$$
I'm asking because there is an equation that involves the transpose of $mathbf{11}$ but since it is all ones and it is square that doesn't do anything.
linear-algebra matrices
edited Dec 27 '18 at 6:39
dantopa
6,42932042
asked Dec 27 '18 at 1:49
Mads
1
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm doing an assignment in linear algebra that involves a notation that I don't understand it is simply two ones ($mathbf{11}$). I understand that a $mathbf{1}$ in linear algebra means a sum vector but does that mean that $mathbf{11}$ is simply the matrix counterpart "sum matrix"?.
For example,
$$mathbf{1}_{2} = begin{bmatrix} 1 \ 1end{bmatrix}.$$
Would it be correct to assume that
$$begin{bmatrix}mathbf{1}_{2}mathbf{1}_{2}end{bmatrix} = begin{bmatrix} 1 & 1 \ 1 & 1end{bmatrix}?$$
I'm asking because there is an equation that involves the transpose of $mathbf{11}$ but since it is all ones and it is square that doesn't do anything.
linear-algebra matrices
linear-algebra matrices
edited Dec 27 '18 at 6:39
dantopa
6,42932042
asked Dec 27 '18 at 1:49
Mads
1
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 27 '18 at 6:39
dantopa
6,42932042
asked Dec 27 '18 at 1:49
Mads
1
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 27 '18 at 6:39
dantopa
6,42932042
edited Dec 27 '18 at 6:39
dantopa
6,42932042
edited Dec 27 '18 at 6:39
dantopa
6,42932042
6,42932042
asked Dec 27 '18 at 1:49
Mads
1
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 27 '18 at 1:49
Mads
1
asked Dec 27 '18 at 1:49
Mads
1
1
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mads is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as unclear what you're asking by amWhy, Lord Shark the Unknown, Eevee Trainer, KReiser, user91500 Dec 27 '18 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by amWhy, Lord Shark the Unknown, Eevee Trainer, KReiser, user91500 Dec 27 '18 at 9:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Could you share the exact problem/text you're looking at? It may be helpful
– Alex
Dec 27 '18 at 1:51
I'm calculating the pagerank of adjacency matrices. On a slide there has been given an equation that involves the notation 11**^T and the only information ive been given is that **1 is a vector of n ones where n of course is the dimensios of the adjacency matrix.
– Mads
Dec 27 '18 at 2:02
So nothing about what the "contraction" of two 1 's is
– Mads
Dec 27 '18 at 2:03
1
I have never seen this notation before and I've had a two-semester linear algebra course. If I were you I'd ask for clarification to whoever assigned this to you.
– Victor S.
Dec 27 '18 at 3:08
That's the problem, we have holliday so they won't be able to answer. Im pretty sure it means what I think though, I just have to be sure.
– Mads
Dec 27 '18 at 3:25
|
show 7 more comments
2
Could you share the exact problem/text you're looking at? It may be helpful
– Alex
Dec 27 '18 at 1:51
I'm calculating the pagerank of adjacency matrices. On a slide there has been given an equation that involves the notation 11**^T and the only information ive been given is that **1 is a vector of n ones where n of course is the dimensios of the adjacency matrix.
– Mads
Dec 27 '18 at 2:02
So nothing about what the "contraction" of two 1 's is
– Mads
Dec 27 '18 at 2:03
1
I have never seen this notation before and I've had a two-semester linear algebra course. If I were you I'd ask for clarification to whoever assigned this to you.
– Victor S.
Dec 27 '18 at 3:08
That's the problem, we have holliday so they won't be able to answer. Im pretty sure it means what I think though, I just have to be sure.
– Mads
Dec 27 '18 at 3:25
2
2
Could you share the exact problem/text you're looking at? It may be helpful
– Alex
Dec 27 '18 at 1:51
Could you share the exact problem/text you're looking at? It may be helpful
– Alex
Dec 27 '18 at 1:51
I'm calculating the pagerank of adjacency matrices. On a slide there has been given an equation that involves the notation 11**^T and the only information ive been given is that **1 is a vector of n ones where n of course is the dimensios of the adjacency matrix.
– Mads
Dec 27 '18 at 2:02
I'm calculating the pagerank of adjacency matrices. On a slide there has been given an equation that involves the notation 11**^T and the only information ive been given is that **1 is a vector of n ones where n of course is the dimensios of the adjacency matrix.
– Mads
Dec 27 '18 at 2:02
So nothing about what the "contraction" of two 1 's is
– Mads
Dec 27 '18 at 2:03
So nothing about what the "contraction" of two 1 's is
– Mads
Dec 27 '18 at 2:03
1
1
I have never seen this notation before and I've had a two-semester linear algebra course. If I were you I'd ask for clarification to whoever assigned this to you.
– Victor S.
Dec 27 '18 at 3:08
I have never seen this notation before and I've had a two-semester linear algebra course. If I were you I'd ask for clarification to whoever assigned this to you.
– Victor S.
Dec 27 '18 at 3:08
That's the problem, we have holliday so they won't be able to answer. Im pretty sure it means what I think though, I just have to be sure.
– Mads
Dec 27 '18 at 3:25
That's the problem, we have holliday so they won't be able to answer. Im pretty sure it means what I think though, I just have to be sure.
– Mads
Dec 27 '18 at 3:25
|
show 7 more comments
1 Answer
1
active
oldest
votes
The question is a bit vague; best guess follows.
The vector $mathbf{1}_{n}$ is a column vector of length $n > 0$ with identical unit entries:
$$
mathbf{1}_{3} = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}
$$
This is useful for creating constant vectors with the scalar product $cmathbf{1}$. And, as alluded, when used in a dot product, it produces a summation. For example, given a (real) vector $x$ of dimension 3,
$$
xcdotmathbf{1}_{3} = mathbf{1}_{3} cdot x = sum_{k=1}^{3}x_{k}.
$$
To address your specific question,
$$
mathbf{1}_{n} cdot mathbf{1}_{n} = sum_{k=1}^{n}1 = n.
$$
Many instructors encourage students to think of matrices as a combination of column vectors. For example, using two copies of $mathbf{1}_{3}$ yields
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 \ 1 & 1 \ 1 & 1 \end{bmatrix}
$$
In this instance, the transpose matrix is
$$
mathbf{A}^{T} = begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 end{bmatrix}
$$
As you suspected, the transpose of $n$ copies of the constant vector of length $n$ is equivalent to the original matrix:
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
$$
The transpose matrix is a composition of row vectors
$$
mathbf{A}^{T}
= begin{bmatrix} mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
=mathbf{A}
$$
An important class, symmetric matrices, are defined by the property
$$mathbf{A}=mathbf{A}^{T}$$
answered Dec 27 '18 at 6:22
dantopa
6,42932042
I'm sorry first time here, i could't figure out how to use math mode
– Mads
Dec 27 '18 at 6:26
@Mads: We are here to help.
– dantopa
Dec 27 '18 at 6:30
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The question is a bit vague; best guess follows.
The vector $mathbf{1}_{n}$ is a column vector of length $n > 0$ with identical unit entries:
$$
mathbf{1}_{3} = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}
$$
This is useful for creating constant vectors with the scalar product $cmathbf{1}$. And, as alluded, when used in a dot product, it produces a summation. For example, given a (real) vector $x$ of dimension 3,
$$
xcdotmathbf{1}_{3} = mathbf{1}_{3} cdot x = sum_{k=1}^{3}x_{k}.
$$
To address your specific question,
$$
mathbf{1}_{n} cdot mathbf{1}_{n} = sum_{k=1}^{n}1 = n.
$$
Many instructors encourage students to think of matrices as a combination of column vectors. For example, using two copies of $mathbf{1}_{3}$ yields
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 \ 1 & 1 \ 1 & 1 \end{bmatrix}
$$
In this instance, the transpose matrix is
$$
mathbf{A}^{T} = begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 end{bmatrix}
$$
As you suspected, the transpose of $n$ copies of the constant vector of length $n$ is equivalent to the original matrix:
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
$$
The transpose matrix is a composition of row vectors
$$
mathbf{A}^{T}
= begin{bmatrix} mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
=mathbf{A}
$$
An important class, symmetric matrices, are defined by the property
$$mathbf{A}=mathbf{A}^{T}$$
answered Dec 27 '18 at 6:22
dantopa
6,42932042
I'm sorry first time here, i could't figure out how to use math mode
– Mads
Dec 27 '18 at 6:26
@Mads: We are here to help.
– dantopa
Dec 27 '18 at 6:30
add a comment |
The question is a bit vague; best guess follows.
The vector $mathbf{1}_{n}$ is a column vector of length $n > 0$ with identical unit entries:
$$
mathbf{1}_{3} = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}
$$
This is useful for creating constant vectors with the scalar product $cmathbf{1}$. And, as alluded, when used in a dot product, it produces a summation. For example, given a (real) vector $x$ of dimension 3,
$$
xcdotmathbf{1}_{3} = mathbf{1}_{3} cdot x = sum_{k=1}^{3}x_{k}.
$$
To address your specific question,
$$
mathbf{1}_{n} cdot mathbf{1}_{n} = sum_{k=1}^{n}1 = n.
$$
Many instructors encourage students to think of matrices as a combination of column vectors. For example, using two copies of $mathbf{1}_{3}$ yields
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 \ 1 & 1 \ 1 & 1 \end{bmatrix}
$$
In this instance, the transpose matrix is
$$
mathbf{A}^{T} = begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 end{bmatrix}
$$
As you suspected, the transpose of $n$ copies of the constant vector of length $n$ is equivalent to the original matrix:
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
$$
The transpose matrix is a composition of row vectors
$$
mathbf{A}^{T}
= begin{bmatrix} mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
=mathbf{A}
$$
An important class, symmetric matrices, are defined by the property
$$mathbf{A}=mathbf{A}^{T}$$
answered Dec 27 '18 at 6:22
dantopa
6,42932042
I'm sorry first time here, i could't figure out how to use math mode
– Mads
Dec 27 '18 at 6:26
@Mads: We are here to help.
– dantopa
Dec 27 '18 at 6:30
add a comment |
The question is a bit vague; best guess follows.
The vector $mathbf{1}_{n}$ is a column vector of length $n > 0$ with identical unit entries:
$$
mathbf{1}_{3} = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}
$$
This is useful for creating constant vectors with the scalar product $cmathbf{1}$. And, as alluded, when used in a dot product, it produces a summation. For example, given a (real) vector $x$ of dimension 3,
$$
xcdotmathbf{1}_{3} = mathbf{1}_{3} cdot x = sum_{k=1}^{3}x_{k}.
$$
To address your specific question,
$$
mathbf{1}_{n} cdot mathbf{1}_{n} = sum_{k=1}^{n}1 = n.
$$
Many instructors encourage students to think of matrices as a combination of column vectors. For example, using two copies of $mathbf{1}_{3}$ yields
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 \ 1 & 1 \ 1 & 1 \end{bmatrix}
$$
In this instance, the transpose matrix is
$$
mathbf{A}^{T} = begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 end{bmatrix}
$$
As you suspected, the transpose of $n$ copies of the constant vector of length $n$ is equivalent to the original matrix:
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
$$
The transpose matrix is a composition of row vectors
$$
mathbf{A}^{T}
= begin{bmatrix} mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
=mathbf{A}
$$
An important class, symmetric matrices, are defined by the property
$$mathbf{A}=mathbf{A}^{T}$$
answered Dec 27 '18 at 6:22
dantopa
6,42932042
The question is a bit vague; best guess follows.
The vector $mathbf{1}_{n}$ is a column vector of length $n > 0$ with identical unit entries:
$$
mathbf{1}_{3} = begin{bmatrix} 1 \ 1 \ 1 end{bmatrix}
$$
This is useful for creating constant vectors with the scalar product $cmathbf{1}$. And, as alluded, when used in a dot product, it produces a summation. For example, given a (real) vector $x$ of dimension 3,
$$
xcdotmathbf{1}_{3} = mathbf{1}_{3} cdot x = sum_{k=1}^{3}x_{k}.
$$
To address your specific question,
$$
mathbf{1}_{n} cdot mathbf{1}_{n} = sum_{k=1}^{n}1 = n.
$$
Many instructors encourage students to think of matrices as a combination of column vectors. For example, using two copies of $mathbf{1}_{3}$ yields
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 \ 1 & 1 \ 1 & 1 \end{bmatrix}
$$
In this instance, the transpose matrix is
$$
mathbf{A}^{T} = begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 end{bmatrix}
$$
As you suspected, the transpose of $n$ copies of the constant vector of length $n$ is equivalent to the original matrix:
$$
mathbf{A}
= begin{bmatrix} mathbf{1}_{3} & mathbf{1}_{3} & mathbf{1}_{3} \end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
$$
The transpose matrix is a composition of row vectors
$$
mathbf{A}^{T}
= begin{bmatrix} mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} \ mathbf{1}_{3}^{T} end{bmatrix}
= begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}
=mathbf{A}
$$
An important class, symmetric matrices, are defined by the property
$$mathbf{A}=mathbf{A}^{T}$$
answered Dec 27 '18 at 6:22
dantopa
6,42932042
edited Dec 27 '18 at 6:42
answered Dec 27 '18 at 6:22
dantopa
6,42932042
answered Dec 27 '18 at 6:22
dantopa
6,42932042
answered Dec 27 '18 at 6:22
dantopa
6,42932042
6,42932042
I'm sorry first time here, i could't figure out how to use math mode
– Mads
Dec 27 '18 at 6:26
@Mads: We are here to help.
– dantopa
Dec 27 '18 at 6:30
add a comment |
I'm sorry first time here, i could't figure out how to use math mode
– Mads
Dec 27 '18 at 6:26
@Mads: We are here to help.
– dantopa
Dec 27 '18 at 6:30
I'm sorry first time here, i could't figure out how to use math mode
– Mads
Dec 27 '18 at 6:26
I'm sorry first time here, i could't figure out how to use math mode
– Mads
Dec 27 '18 at 6:26
@Mads: We are here to help.
– dantopa
Dec 27 '18 at 6:30
@Mads: We are here to help.
– dantopa
Dec 27 '18 at 6:30
add a comment |
2
Could you share the exact problem/text you're looking at? It may be helpful
– Alex
Dec 27 '18 at 1:51
I'm calculating the pagerank of adjacency matrices. On a slide there has been given an equation that involves the notation 11**^T and the only information ive been given is that **1 is a vector of n ones where n of course is the dimensios of the adjacency matrix.
– Mads
Dec 27 '18 at 2:02
So nothing about what the "contraction" of two 1 's is
– Mads
Dec 27 '18 at 2:03
1
I have never seen this notation before and I've had a two-semester linear algebra course. If I were you I'd ask for clarification to whoever assigned this to you.
– Victor S.
Dec 27 '18 at 3:08
That's the problem, we have holliday so they won't be able to answer. Im pretty sure it means what I think though, I just have to be sure.
– Mads
Dec 27 '18 at 3:25