Representing a linear combination of matrices as some kind of inner product.
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If I have a linear combination of two numbers $lambda_1y_1+lambda_2y_2$, I can represent it as $mathbf y cdotvec lambda$, i.e., an inner product of two vectors.
If I have a linear combination of two vectors $lambda_1mathbf y_1+lambda_2 mathbf y_2$, I can represent it as matrix-vector product $mathbf Yvec lambda$.
How can I do something similar for linear combination of two matrices $lambda_1 Y_1+lambda_2 Y_2$?
linear-algebra products
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
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add a comment |
$begingroup$
If I have a linear combination of two numbers $lambda_1y_1+lambda_2y_2$, I can represent it as $mathbf y cdotvec lambda$, i.e., an inner product of two vectors.
If I have a linear combination of two vectors $lambda_1mathbf y_1+lambda_2 mathbf y_2$, I can represent it as matrix-vector product $mathbf Yvec lambda$.
How can I do something similar for linear combination of two matrices $lambda_1 Y_1+lambda_2 Y_2$?
linear-algebra products
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
$endgroup$
$begingroup$
Nothing stops us in using a matrix $pmatrix{Y_1\Y_2}$ with matrix entries..
$endgroup$
– Berci
Jan 8 at 8:21
$begingroup$
... and if the definition of the linear function is independent of basis then $pmatrix{Y_1\Y_2}$ is a (1,2) tensor which maps the vector $pmatrix{lambda_1\ lambda_2}$ to the (1,1) tensor $lambda_1Y_2 + lambda_2Y_2$.
$endgroup$
– gandalf61
Jan 8 at 9:59
add a comment |
$begingroup$
If I have a linear combination of two numbers $lambda_1y_1+lambda_2y_2$, I can represent it as $mathbf y cdotvec lambda$, i.e., an inner product of two vectors.
If I have a linear combination of two vectors $lambda_1mathbf y_1+lambda_2 mathbf y_2$, I can represent it as matrix-vector product $mathbf Yvec lambda$.
How can I do something similar for linear combination of two matrices $lambda_1 Y_1+lambda_2 Y_2$?
linear-algebra products
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
$endgroup$
If I have a linear combination of two numbers $lambda_1y_1+lambda_2y_2$, I can represent it as $mathbf y cdotvec lambda$, i.e., an inner product of two vectors.
If I have a linear combination of two vectors $lambda_1mathbf y_1+lambda_2 mathbf y_2$, I can represent it as matrix-vector product $mathbf Yvec lambda$.
How can I do something similar for linear combination of two matrices $lambda_1 Y_1+lambda_2 Y_2$?
linear-algebra products
linear-algebra products
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
asked Jan 8 at 7:10
Тимофей ЛомоносовТимофей Ломоносов
508314
508314
$begingroup$
Nothing stops us in using a matrix $pmatrix{Y_1\Y_2}$ with matrix entries..
$endgroup$
– Berci
Jan 8 at 8:21
$begingroup$
... and if the definition of the linear function is independent of basis then $pmatrix{Y_1\Y_2}$ is a (1,2) tensor which maps the vector $pmatrix{lambda_1\ lambda_2}$ to the (1,1) tensor $lambda_1Y_2 + lambda_2Y_2$.
$endgroup$
– gandalf61
Jan 8 at 9:59
add a comment |
$begingroup$
Nothing stops us in using a matrix $pmatrix{Y_1\Y_2}$ with matrix entries..
$endgroup$
– Berci
Jan 8 at 8:21
$begingroup$
... and if the definition of the linear function is independent of basis then $pmatrix{Y_1\Y_2}$ is a (1,2) tensor which maps the vector $pmatrix{lambda_1\ lambda_2}$ to the (1,1) tensor $lambda_1Y_2 + lambda_2Y_2$.
$endgroup$
– gandalf61
Jan 8 at 9:59
$begingroup$
Nothing stops us in using a matrix $pmatrix{Y_1\Y_2}$ with matrix entries..
$endgroup$
– Berci
Jan 8 at 8:21
$begingroup$
Nothing stops us in using a matrix $pmatrix{Y_1\Y_2}$ with matrix entries..
$endgroup$
– Berci
Jan 8 at 8:21
$begingroup$
... and if the definition of the linear function is independent of basis then $pmatrix{Y_1\Y_2}$ is a (1,2) tensor which maps the vector $pmatrix{lambda_1\ lambda_2}$ to the (1,1) tensor $lambda_1Y_2 + lambda_2Y_2$.
$endgroup$
– gandalf61
Jan 8 at 9:59
$begingroup$
... and if the definition of the linear function is independent of basis then $pmatrix{Y_1\Y_2}$ is a (1,2) tensor which maps the vector $pmatrix{lambda_1\ lambda_2}$ to the (1,1) tensor $lambda_1Y_2 + lambda_2Y_2$.
$endgroup$
– gandalf61
Jan 8 at 9:59
add a comment |
1 Answer
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$begingroup$
In the language of Linear Algebra, the $y$'s are looked upon as vectors in some vector space, and what you do is writing those vectors' coordinate representation as columns of a matrix.
Hence this requires fixing a coordinate system (i.e. a basis) for the space first.
In your first example, the numbers can in fact be looked upon as vectors in the $1$-dimensional space, which is why each column's height in $mathbf{y}$ is $1$.
In your third example, matrices (say $M_{n times m}(mathbb{R})$) can also be viewed as a $nm$-dimensional space, but writing linear combinations in a similar way requires writing each matrix as a column vector. For example,
$begin{bmatrix}
1 \
2 \
3 \
4 \
end{bmatrix}$ for
$begin{bmatrix}
1 & 2 \
3 & 4 \
end{bmatrix}$.
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
In the language of Linear Algebra, the $y$'s are looked upon as vectors in some vector space, and what you do is writing those vectors' coordinate representation as columns of a matrix.
Hence this requires fixing a coordinate system (i.e. a basis) for the space first.
In your first example, the numbers can in fact be looked upon as vectors in the $1$-dimensional space, which is why each column's height in $mathbf{y}$ is $1$.
In your third example, matrices (say $M_{n times m}(mathbb{R})$) can also be viewed as a $nm$-dimensional space, but writing linear combinations in a similar way requires writing each matrix as a column vector. For example,
$begin{bmatrix}
1 \
2 \
3 \
4 \
end{bmatrix}$ for
$begin{bmatrix}
1 & 2 \
3 & 4 \
end{bmatrix}$.
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In the language of Linear Algebra, the $y$'s are looked upon as vectors in some vector space, and what you do is writing those vectors' coordinate representation as columns of a matrix.
Hence this requires fixing a coordinate system (i.e. a basis) for the space first.
In your first example, the numbers can in fact be looked upon as vectors in the $1$-dimensional space, which is why each column's height in $mathbf{y}$ is $1$.
In your third example, matrices (say $M_{n times m}(mathbb{R})$) can also be viewed as a $nm$-dimensional space, but writing linear combinations in a similar way requires writing each matrix as a column vector. For example,
$begin{bmatrix}
1 \
2 \
3 \
4 \
end{bmatrix}$ for
$begin{bmatrix}
1 & 2 \
3 & 4 \
end{bmatrix}$.
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In the language of Linear Algebra, the $y$'s are looked upon as vectors in some vector space, and what you do is writing those vectors' coordinate representation as columns of a matrix.
Hence this requires fixing a coordinate system (i.e. a basis) for the space first.
In your first example, the numbers can in fact be looked upon as vectors in the $1$-dimensional space, which is why each column's height in $mathbf{y}$ is $1$.
In your third example, matrices (say $M_{n times m}(mathbb{R})$) can also be viewed as a $nm$-dimensional space, but writing linear combinations in a similar way requires writing each matrix as a column vector. For example,
$begin{bmatrix}
1 \
2 \
3 \
4 \
end{bmatrix}$ for
$begin{bmatrix}
1 & 2 \
3 & 4 \
end{bmatrix}$.
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In the language of Linear Algebra, the $y$'s are looked upon as vectors in some vector space, and what you do is writing those vectors' coordinate representation as columns of a matrix.
Hence this requires fixing a coordinate system (i.e. a basis) for the space first.
In your first example, the numbers can in fact be looked upon as vectors in the $1$-dimensional space, which is why each column's height in $mathbf{y}$ is $1$.
In your third example, matrices (say $M_{n times m}(mathbb{R})$) can also be viewed as a $nm$-dimensional space, but writing linear combinations in a similar way requires writing each matrix as a column vector. For example,
$begin{bmatrix}
1 \
2 \
3 \
4 \
end{bmatrix}$ for
$begin{bmatrix}
1 & 2 \
3 & 4 \
end{bmatrix}$.
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
answered Jan 8 at 8:28
Yanger MaYanger Ma
1014
1014
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yanger Ma is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
Nothing stops us in using a matrix $pmatrix{Y_1\Y_2}$ with matrix entries..
$endgroup$
– Berci
Jan 8 at 8:21
$begingroup$
... and if the definition of the linear function is independent of basis then $pmatrix{Y_1\Y_2}$ is a (1,2) tensor which maps the vector $pmatrix{lambda_1\ lambda_2}$ to the (1,1) tensor $lambda_1Y_2 + lambda_2Y_2$.
$endgroup$
– gandalf61
Jan 8 at 9:59