Understanding the relation between Sweep operator and Graph rotation.
$begingroup$
In this blog post, the sweep operator from numerical linear algebra was explained. It was all clear until, the post says that
Any ${n times n}$ matrix ${A}$ creates a graph ${hbox{Graph}[A] := { (X, AX): X in {bf R}^n }}$ (where we think of ${{bf R}^n}$ as the space of column vectors). This graph is an ${n}$-dimensional subspace of ${{bf R}^n times {bf R}^n}$. Conversely, most subspaces of ${{bf R}^n times {bf R}^n}$ arises as graphs; there are some that fail the vertical line test, but these are a positive codimension set of counterexamples.
We use ${e_1,dots,e_n,f_1,dots,f_n}$ to denote the standard basis of ${{bf R}^n times {bf R}^n}$, with ${e_1,dots,e_n}$ the standard basis for the first factor of ${{bf R}^n}$ and ${f_1,dots,f_n}$ the standard basis for the second factor. The operation of sweeping the ${k^{th}}$ entry then corresponds to a ninety degree rotation ${hbox{Rot}_k: {bf R}^n times {bf R}^n rightarrow {bf R}^n times {bf R}^n}$ in the ${e_k,f_k}$ plane, that sends ${f_k}$ to ${e_k}$ (and ${e_k}$ to ${-f_k}$), keeping all other basis vectors fixed: thus we have
$$displaystyle hbox{Graph}[ hbox{Sweep}_k[A] ] = hbox{Rot}_k hbox{Graph}[A] $$
I don't understand the definition of graph here, I don't know how to visualize the above transformation.
linear-algebra numerical-linear-algebra
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
$endgroup$
add a comment |
$begingroup$
In this blog post, the sweep operator from numerical linear algebra was explained. It was all clear until, the post says that
Any ${n times n}$ matrix ${A}$ creates a graph ${hbox{Graph}[A] := { (X, AX): X in {bf R}^n }}$ (where we think of ${{bf R}^n}$ as the space of column vectors). This graph is an ${n}$-dimensional subspace of ${{bf R}^n times {bf R}^n}$. Conversely, most subspaces of ${{bf R}^n times {bf R}^n}$ arises as graphs; there are some that fail the vertical line test, but these are a positive codimension set of counterexamples.
We use ${e_1,dots,e_n,f_1,dots,f_n}$ to denote the standard basis of ${{bf R}^n times {bf R}^n}$, with ${e_1,dots,e_n}$ the standard basis for the first factor of ${{bf R}^n}$ and ${f_1,dots,f_n}$ the standard basis for the second factor. The operation of sweeping the ${k^{th}}$ entry then corresponds to a ninety degree rotation ${hbox{Rot}_k: {bf R}^n times {bf R}^n rightarrow {bf R}^n times {bf R}^n}$ in the ${e_k,f_k}$ plane, that sends ${f_k}$ to ${e_k}$ (and ${e_k}$ to ${-f_k}$), keeping all other basis vectors fixed: thus we have
$$displaystyle hbox{Graph}[ hbox{Sweep}_k[A] ] = hbox{Rot}_k hbox{Graph}[A] $$
I don't understand the definition of graph here, I don't know how to visualize the above transformation.
linear-algebra numerical-linear-algebra
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
$endgroup$
add a comment |
$begingroup$
In this blog post, the sweep operator from numerical linear algebra was explained. It was all clear until, the post says that
Any ${n times n}$ matrix ${A}$ creates a graph ${hbox{Graph}[A] := { (X, AX): X in {bf R}^n }}$ (where we think of ${{bf R}^n}$ as the space of column vectors). This graph is an ${n}$-dimensional subspace of ${{bf R}^n times {bf R}^n}$. Conversely, most subspaces of ${{bf R}^n times {bf R}^n}$ arises as graphs; there are some that fail the vertical line test, but these are a positive codimension set of counterexamples.
We use ${e_1,dots,e_n,f_1,dots,f_n}$ to denote the standard basis of ${{bf R}^n times {bf R}^n}$, with ${e_1,dots,e_n}$ the standard basis for the first factor of ${{bf R}^n}$ and ${f_1,dots,f_n}$ the standard basis for the second factor. The operation of sweeping the ${k^{th}}$ entry then corresponds to a ninety degree rotation ${hbox{Rot}_k: {bf R}^n times {bf R}^n rightarrow {bf R}^n times {bf R}^n}$ in the ${e_k,f_k}$ plane, that sends ${f_k}$ to ${e_k}$ (and ${e_k}$ to ${-f_k}$), keeping all other basis vectors fixed: thus we have
$$displaystyle hbox{Graph}[ hbox{Sweep}_k[A] ] = hbox{Rot}_k hbox{Graph}[A] $$
I don't understand the definition of graph here, I don't know how to visualize the above transformation.
linear-algebra numerical-linear-algebra
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
$endgroup$
In this blog post, the sweep operator from numerical linear algebra was explained. It was all clear until, the post says that
Any ${n times n}$ matrix ${A}$ creates a graph ${hbox{Graph}[A] := { (X, AX): X in {bf R}^n }}$ (where we think of ${{bf R}^n}$ as the space of column vectors). This graph is an ${n}$-dimensional subspace of ${{bf R}^n times {bf R}^n}$. Conversely, most subspaces of ${{bf R}^n times {bf R}^n}$ arises as graphs; there are some that fail the vertical line test, but these are a positive codimension set of counterexamples.
We use ${e_1,dots,e_n,f_1,dots,f_n}$ to denote the standard basis of ${{bf R}^n times {bf R}^n}$, with ${e_1,dots,e_n}$ the standard basis for the first factor of ${{bf R}^n}$ and ${f_1,dots,f_n}$ the standard basis for the second factor. The operation of sweeping the ${k^{th}}$ entry then corresponds to a ninety degree rotation ${hbox{Rot}_k: {bf R}^n times {bf R}^n rightarrow {bf R}^n times {bf R}^n}$ in the ${e_k,f_k}$ plane, that sends ${f_k}$ to ${e_k}$ (and ${e_k}$ to ${-f_k}$), keeping all other basis vectors fixed: thus we have
$$displaystyle hbox{Graph}[ hbox{Sweep}_k[A] ] = hbox{Rot}_k hbox{Graph}[A] $$
I don't understand the definition of graph here, I don't know how to visualize the above transformation.
linear-algebra numerical-linear-algebra
linear-algebra numerical-linear-algebra
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
edited Dec 29 '18 at 14:16
Chris Godsil
11.8k21635
11.8k21635
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
asked Dec 29 '18 at 10:16
papabicepspapabiceps
324114
324114
add a comment |
add a comment |
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