Show that R (reflection operator) is diagonalizable.
$begingroup$
Let be $V$ a real vector space (finite dimension) with standard scalar product, $S$ an $V-$subspace and $R: Vrightarrow V$ the reflection operator on $S$. Show that $R$ is a diagonalizable operator.
I can think of how to do it on the plane - using eigenvalues and eigenvectors. But this question, I honestly do not know how to begin.
linear-algebra linear-transformations euclidean-geometry diagonalization
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
$endgroup$
add a comment |
$begingroup$
Let be $V$ a real vector space (finite dimension) with standard scalar product, $S$ an $V-$subspace and $R: Vrightarrow V$ the reflection operator on $S$. Show that $R$ is a diagonalizable operator.
I can think of how to do it on the plane - using eigenvalues and eigenvectors. But this question, I honestly do not know how to begin.
linear-algebra linear-transformations euclidean-geometry diagonalization
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
$endgroup$
add a comment |
$begingroup$
Let be $V$ a real vector space (finite dimension) with standard scalar product, $S$ an $V-$subspace and $R: Vrightarrow V$ the reflection operator on $S$. Show that $R$ is a diagonalizable operator.
I can think of how to do it on the plane - using eigenvalues and eigenvectors. But this question, I honestly do not know how to begin.
linear-algebra linear-transformations euclidean-geometry diagonalization
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
$endgroup$
Let be $V$ a real vector space (finite dimension) with standard scalar product, $S$ an $V-$subspace and $R: Vrightarrow V$ the reflection operator on $S$. Show that $R$ is a diagonalizable operator.
I can think of how to do it on the plane - using eigenvalues and eigenvectors. But this question, I honestly do not know how to begin.
linear-algebra linear-transformations euclidean-geometry diagonalization
linear-algebra linear-transformations euclidean-geometry diagonalization
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
edited Dec 2 '18 at 18:34
Servaes
22.5k33793
22.5k33793
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
asked Dec 2 '18 at 18:26
Juliana de SouzaJuliana de Souza
656
656
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT: Decompose $V$ as $Soplus S^{perp}$. What do $Rvert_S$ and $Rvert_{S^{perp}}$ look like?
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
$endgroup$
1
$begingroup$
Very good hint.
$endgroup$
– rschwieb
Dec 2 '18 at 18:33
add a comment |
$begingroup$
Since A is reflection operator so it will satisfy the equation $A^2=I$. So it can be easily seen that the minimal polynomial of A will annihilate the polynomial $(x^2-1)$ so minimal polynomial of A is either (x-1) or (x+1) or (x-1)(x+1). In each case minimal polynomial have only linear factors. Therefore the operator A is diagnolisable.
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
HINT: Decompose $V$ as $Soplus S^{perp}$. What do $Rvert_S$ and $Rvert_{S^{perp}}$ look like?
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
$endgroup$
1
$begingroup$
Very good hint.
$endgroup$
– rschwieb
Dec 2 '18 at 18:33
add a comment |
$begingroup$
HINT: Decompose $V$ as $Soplus S^{perp}$. What do $Rvert_S$ and $Rvert_{S^{perp}}$ look like?
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
$endgroup$
1
$begingroup$
Very good hint.
$endgroup$
– rschwieb
Dec 2 '18 at 18:33
add a comment |
$begingroup$
HINT: Decompose $V$ as $Soplus S^{perp}$. What do $Rvert_S$ and $Rvert_{S^{perp}}$ look like?
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
$endgroup$
HINT: Decompose $V$ as $Soplus S^{perp}$. What do $Rvert_S$ and $Rvert_{S^{perp}}$ look like?
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
answered Dec 2 '18 at 18:30
ServaesServaes
22.5k33793
22.5k33793
1
$begingroup$
Very good hint.
$endgroup$
– rschwieb
Dec 2 '18 at 18:33
add a comment |
1
$begingroup$
Very good hint.
$endgroup$
– rschwieb
Dec 2 '18 at 18:33
1
1
$begingroup$
Very good hint.
$endgroup$
– rschwieb
Dec 2 '18 at 18:33
$begingroup$
Very good hint.
$endgroup$
– rschwieb
Dec 2 '18 at 18:33
add a comment |
$begingroup$
Since A is reflection operator so it will satisfy the equation $A^2=I$. So it can be easily seen that the minimal polynomial of A will annihilate the polynomial $(x^2-1)$ so minimal polynomial of A is either (x-1) or (x+1) or (x-1)(x+1). In each case minimal polynomial have only linear factors. Therefore the operator A is diagnolisable.
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
$endgroup$
add a comment |
$begingroup$
Since A is reflection operator so it will satisfy the equation $A^2=I$. So it can be easily seen that the minimal polynomial of A will annihilate the polynomial $(x^2-1)$ so minimal polynomial of A is either (x-1) or (x+1) or (x-1)(x+1). In each case minimal polynomial have only linear factors. Therefore the operator A is diagnolisable.
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
$endgroup$
add a comment |
$begingroup$
Since A is reflection operator so it will satisfy the equation $A^2=I$. So it can be easily seen that the minimal polynomial of A will annihilate the polynomial $(x^2-1)$ so minimal polynomial of A is either (x-1) or (x+1) or (x-1)(x+1). In each case minimal polynomial have only linear factors. Therefore the operator A is diagnolisable.
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
$endgroup$
Since A is reflection operator so it will satisfy the equation $A^2=I$. So it can be easily seen that the minimal polynomial of A will annihilate the polynomial $(x^2-1)$ so minimal polynomial of A is either (x-1) or (x+1) or (x-1)(x+1). In each case minimal polynomial have only linear factors. Therefore the operator A is diagnolisable.
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
answered Dec 2 '18 at 18:47
MANI SHANKAR PANDEYMANI SHANKAR PANDEY
477
477
add a comment |
add a comment |
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