Matrices restricted to a subspace
Let $Q$ be a $ntimes n$ stochastic matrix and $mathcal S$ be the following subspace of $mathbb R^n$:
$$mathcal S:=left{xinmathbb R^n: sum_{i=1}^nx_i=0 right}, .$$
In a paper that I'm reading, there is a concept that I do not know: the restriction of $Q$ to $mathcal S$, (denoted by $Q|_{mathcal S}$). What does it mean? For example, if I have a given matrix $Q$, how could I calculate $Q|_{mathcal S}$?
linear-algebra
asked Nov 22 '17 at 15:22
Mark
3,43751846
add a comment |
Let $Q$ be a $ntimes n$ stochastic matrix and $mathcal S$ be the following subspace of $mathbb R^n$:
$$mathcal S:=left{xinmathbb R^n: sum_{i=1}^nx_i=0 right}, .$$
In a paper that I'm reading, there is a concept that I do not know: the restriction of $Q$ to $mathcal S$, (denoted by $Q|_{mathcal S}$). What does it mean? For example, if I have a given matrix $Q$, how could I calculate $Q|_{mathcal S}$?
linear-algebra
asked Nov 22 '17 at 15:22
Mark
3,43751846
add a comment |
Let $Q$ be a $ntimes n$ stochastic matrix and $mathcal S$ be the following subspace of $mathbb R^n$:
$$mathcal S:=left{xinmathbb R^n: sum_{i=1}^nx_i=0 right}, .$$
In a paper that I'm reading, there is a concept that I do not know: the restriction of $Q$ to $mathcal S$, (denoted by $Q|_{mathcal S}$). What does it mean? For example, if I have a given matrix $Q$, how could I calculate $Q|_{mathcal S}$?
linear-algebra
asked Nov 22 '17 at 15:22
Mark
3,43751846
Let $Q$ be a $ntimes n$ stochastic matrix and $mathcal S$ be the following subspace of $mathbb R^n$:
$$mathcal S:=left{xinmathbb R^n: sum_{i=1}^nx_i=0 right}, .$$
In a paper that I'm reading, there is a concept that I do not know: the restriction of $Q$ to $mathcal S$, (denoted by $Q|_{mathcal S}$). What does it mean? For example, if I have a given matrix $Q$, how could I calculate $Q|_{mathcal S}$?
linear-algebra
linear-algebra
asked Nov 22 '17 at 15:22
Mark
3,43751846
asked Nov 22 '17 at 15:22
Mark
3,43751846
asked Nov 22 '17 at 15:22
Mark
3,43751846
asked Nov 22 '17 at 15:22
Mark
3,43751846
asked Nov 22 '17 at 15:22
Mark
3,43751846
3,43751846
add a comment |
add a comment |
1 Answer
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As a matrix (stochastic or not), $Q$ can be thought of as a mapping $$Bbb R^n to Bbb R^n : v mapsto Qv$$
Thus the domain and codomain of $Q$ are $Bbb R^n$. Since $S subset Bbb R^n$, we can consider the map $$S to Bbb R^n : v mapsto Qv$$
instead. The only difference from $Q$ is that the domain is just $S$. This map is the restriction $Q|_S$ of $Q$ to $S$.
Since $S$ is in fact a vector subspace of $Bbb R^n$, $Q|_S$ is also linear over $S$. If you wanted to represent is as a matrix, first you would need to pick a basis ${{v_i}}_{i=1}^{n-1}$ for $S$ (since $S$ is n-1 dimensional), then you could express the matrix elements as $Q_{sij} = langle v_i, Qv_jrangle$. What matrix you get depends on the basis elements you selected.
For example, if you simply drop any one of the standard basis elements of $Bbb R^n$, the remaining elements will form a basis for this particular space $S$. The resulting matrix will be original matrix $Q$ with one column removed. The column removed corresponds to the standard basis element you dropped.
edited Dec 26 '18 at 19:24
Community♦
1
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
add a comment |
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1 Answer
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As a matrix (stochastic or not), $Q$ can be thought of as a mapping $$Bbb R^n to Bbb R^n : v mapsto Qv$$
Thus the domain and codomain of $Q$ are $Bbb R^n$. Since $S subset Bbb R^n$, we can consider the map $$S to Bbb R^n : v mapsto Qv$$
instead. The only difference from $Q$ is that the domain is just $S$. This map is the restriction $Q|_S$ of $Q$ to $S$.
Since $S$ is in fact a vector subspace of $Bbb R^n$, $Q|_S$ is also linear over $S$. If you wanted to represent is as a matrix, first you would need to pick a basis ${{v_i}}_{i=1}^{n-1}$ for $S$ (since $S$ is n-1 dimensional), then you could express the matrix elements as $Q_{sij} = langle v_i, Qv_jrangle$. What matrix you get depends on the basis elements you selected.
For example, if you simply drop any one of the standard basis elements of $Bbb R^n$, the remaining elements will form a basis for this particular space $S$. The resulting matrix will be original matrix $Q$ with one column removed. The column removed corresponds to the standard basis element you dropped.
edited Dec 26 '18 at 19:24
Community♦
1
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
add a comment |
As a matrix (stochastic or not), $Q$ can be thought of as a mapping $$Bbb R^n to Bbb R^n : v mapsto Qv$$
Thus the domain and codomain of $Q$ are $Bbb R^n$. Since $S subset Bbb R^n$, we can consider the map $$S to Bbb R^n : v mapsto Qv$$
instead. The only difference from $Q$ is that the domain is just $S$. This map is the restriction $Q|_S$ of $Q$ to $S$.
Since $S$ is in fact a vector subspace of $Bbb R^n$, $Q|_S$ is also linear over $S$. If you wanted to represent is as a matrix, first you would need to pick a basis ${{v_i}}_{i=1}^{n-1}$ for $S$ (since $S$ is n-1 dimensional), then you could express the matrix elements as $Q_{sij} = langle v_i, Qv_jrangle$. What matrix you get depends on the basis elements you selected.
For example, if you simply drop any one of the standard basis elements of $Bbb R^n$, the remaining elements will form a basis for this particular space $S$. The resulting matrix will be original matrix $Q$ with one column removed. The column removed corresponds to the standard basis element you dropped.
edited Dec 26 '18 at 19:24
Community♦
1
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
add a comment |
As a matrix (stochastic or not), $Q$ can be thought of as a mapping $$Bbb R^n to Bbb R^n : v mapsto Qv$$
Thus the domain and codomain of $Q$ are $Bbb R^n$. Since $S subset Bbb R^n$, we can consider the map $$S to Bbb R^n : v mapsto Qv$$
instead. The only difference from $Q$ is that the domain is just $S$. This map is the restriction $Q|_S$ of $Q$ to $S$.
Since $S$ is in fact a vector subspace of $Bbb R^n$, $Q|_S$ is also linear over $S$. If you wanted to represent is as a matrix, first you would need to pick a basis ${{v_i}}_{i=1}^{n-1}$ for $S$ (since $S$ is n-1 dimensional), then you could express the matrix elements as $Q_{sij} = langle v_i, Qv_jrangle$. What matrix you get depends on the basis elements you selected.
For example, if you simply drop any one of the standard basis elements of $Bbb R^n$, the remaining elements will form a basis for this particular space $S$. The resulting matrix will be original matrix $Q$ with one column removed. The column removed corresponds to the standard basis element you dropped.
edited Dec 26 '18 at 19:24
Community♦
1
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
As a matrix (stochastic or not), $Q$ can be thought of as a mapping $$Bbb R^n to Bbb R^n : v mapsto Qv$$
Thus the domain and codomain of $Q$ are $Bbb R^n$. Since $S subset Bbb R^n$, we can consider the map $$S to Bbb R^n : v mapsto Qv$$
instead. The only difference from $Q$ is that the domain is just $S$. This map is the restriction $Q|_S$ of $Q$ to $S$.
Since $S$ is in fact a vector subspace of $Bbb R^n$, $Q|_S$ is also linear over $S$. If you wanted to represent is as a matrix, first you would need to pick a basis ${{v_i}}_{i=1}^{n-1}$ for $S$ (since $S$ is n-1 dimensional), then you could express the matrix elements as $Q_{sij} = langle v_i, Qv_jrangle$. What matrix you get depends on the basis elements you selected.
For example, if you simply drop any one of the standard basis elements of $Bbb R^n$, the remaining elements will form a basis for this particular space $S$. The resulting matrix will be original matrix $Q$ with one column removed. The column removed corresponds to the standard basis element you dropped.
edited Dec 26 '18 at 19:24
Community♦
1
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
edited Dec 26 '18 at 19:24
Community♦
1
edited Dec 26 '18 at 19:24
Community♦
1
edited Dec 26 '18 at 19:24
Community♦
1
1
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
answered Nov 22 '17 at 20:08
Paul Sinclair
19.3k21441
19.3k21441
add a comment |
add a comment |
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