Understanding the relation between Sweep operator and Graph rotation.
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
11.5k21634
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
add a comment |
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
11.5k21634
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
add a comment |
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
11.5k21634
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
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
11.5k21634
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
edited Dec 29 '18 at 14:16
Chris Godsil
11.5k21634
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
edited Dec 29 '18 at 14:16
Chris Godsil
11.5k21634
edited Dec 29 '18 at 14:16
Chris Godsil
11.5k21634
edited Dec 29 '18 at 14:16
Chris Godsil
11.5k21634
11.5k21634
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
asked Dec 29 '18 at 10:16
papabicepspapabiceps
322113
322113
add a comment |
add a comment |
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