Mixing
Orthogonal Projections and eigenvalues
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Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
(a) Give two real numbers that are eigenvalues of $A$.
I think that one of the eigenvalues will be $0$ due to the fact that if $V$ is a subspace the orthogonal complement will also be a subspace. And projection takes any vector to $0$ so I believe $0$ will be one? But I don’t know if this is the correct logic and stuck on what the other will be.
linear-algebra eigenvalues-eigenvectors
edited Jul 26 at 21:50
rogerl
16.5k22745
asked Jul 26 at 21:34
Molly
303
add a comment |Â
up vote
2
down vote
favorite
Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
(a) Give two real numbers that are eigenvalues of $A$.
I think that one of the eigenvalues will be $0$ due to the fact that if $V$ is a subspace the orthogonal complement will also be a subspace. And projection takes any vector to $0$ so I believe $0$ will be one? But I don’t know if this is the correct logic and stuck on what the other will be.
linear-algebra eigenvalues-eigenvectors
edited Jul 26 at 21:50
rogerl
16.5k22745
asked Jul 26 at 21:34
Molly
303
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
(a) Give two real numbers that are eigenvalues of $A$.
I think that one of the eigenvalues will be $0$ due to the fact that if $V$ is a subspace the orthogonal complement will also be a subspace. And projection takes any vector to $0$ so I believe $0$ will be one? But I don’t know if this is the correct logic and stuck on what the other will be.
linear-algebra eigenvalues-eigenvectors
edited Jul 26 at 21:50
rogerl
16.5k22745
asked Jul 26 at 21:34
Molly
303
Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
(a) Give two real numbers that are eigenvalues of $A$.
I think that one of the eigenvalues will be $0$ due to the fact that if $V$ is a subspace the orthogonal complement will also be a subspace. And projection takes any vector to $0$ so I believe $0$ will be one? But I don’t know if this is the correct logic and stuck on what the other will be.
linear-algebra eigenvalues-eigenvectors
edited Jul 26 at 21:50
rogerl
16.5k22745
asked Jul 26 at 21:34
Molly
303
edited Jul 26 at 21:50
rogerl
16.5k22745
edited Jul 26 at 21:50
rogerl
16.5k22745
edited Jul 26 at 21:50
rogerl
16.5k22745
16.5k22745
asked Jul 26 at 21:34
Molly
303
asked Jul 26 at 21:34
Molly
303
asked Jul 26 at 21:34
Molly
303
303
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday
add a comment |Â
2 Answers
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up vote
1
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You are right indeed
for any $vin V^perp$ we have $Pv=0 implies lambda =0$
for any $vin V$ we have $Pv=v implies lambda =1$
answered Jul 26 at 21:37
gimusi
65k73583
add a comment |Â
up vote
1
down vote
Hint:
$A$ satisfies $A^2 = A$ or $A(A-I) = 0$ so the eigenvalues are contained in $0,1$.
For $x in V, xne 0$ we have $Ax = x$, and for $x in V^perp, xne 0$ we have $Ax = 0$ so $0$ and $1$ are indeed eigenvalues.
answered Jul 26 at 21:37
mechanodroid
22.2k52041
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You are right indeed
for any $vin V^perp$ we have $Pv=0 implies lambda =0$
for any $vin V$ we have $Pv=v implies lambda =1$
answered Jul 26 at 21:37
gimusi
65k73583
add a comment |Â
up vote
1
down vote
You are right indeed
for any $vin V^perp$ we have $Pv=0 implies lambda =0$
for any $vin V$ we have $Pv=v implies lambda =1$
answered Jul 26 at 21:37
gimusi
65k73583
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are right indeed
for any $vin V^perp$ we have $Pv=0 implies lambda =0$
for any $vin V$ we have $Pv=v implies lambda =1$
answered Jul 26 at 21:37
gimusi
65k73583
You are right indeed
for any $vin V^perp$ we have $Pv=0 implies lambda =0$
for any $vin V$ we have $Pv=v implies lambda =1$
answered Jul 26 at 21:37
gimusi
65k73583
answered Jul 26 at 21:37
gimusi
65k73583
answered Jul 26 at 21:37
gimusi
65k73583
answered Jul 26 at 21:37
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint:
$A$ satisfies $A^2 = A$ or $A(A-I) = 0$ so the eigenvalues are contained in $0,1$.
For $x in V, xne 0$ we have $Ax = x$, and for $x in V^perp, xne 0$ we have $Ax = 0$ so $0$ and $1$ are indeed eigenvalues.
answered Jul 26 at 21:37
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
Hint:
$A$ satisfies $A^2 = A$ or $A(A-I) = 0$ so the eigenvalues are contained in $0,1$.
For $x in V, xne 0$ we have $Ax = x$, and for $x in V^perp, xne 0$ we have $Ax = 0$ so $0$ and $1$ are indeed eigenvalues.
answered Jul 26 at 21:37
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
$A$ satisfies $A^2 = A$ or $A(A-I) = 0$ so the eigenvalues are contained in $0,1$.
For $x in V, xne 0$ we have $Ax = x$, and for $x in V^perp, xne 0$ we have $Ax = 0$ so $0$ and $1$ are indeed eigenvalues.
answered Jul 26 at 21:37
mechanodroid
22.2k52041
Hint:
$A$ satisfies $A^2 = A$ or $A(A-I) = 0$ so the eigenvalues are contained in $0,1$.
For $x in V, xne 0$ we have $Ax = x$, and for $x in V^perp, xne 0$ we have $Ax = 0$ so $0$ and $1$ are indeed eigenvalues.
answered Jul 26 at 21:37
mechanodroid
22.2k52041
answered Jul 26 at 21:37
mechanodroid
22.2k52041
answered Jul 26 at 21:37
mechanodroid
22.2k52041
answered Jul 26 at 21:37
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
yesterday