What are the eigenvalues of matrix $I-uu^T$ where where $u$ is a unit vector and $I$ is the identity matrix in $mathbb R^n$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)about diagonal matrix and eigenvaluesReconstruct the original symmetric matrix given Eigen values and the longest Eigen vectorHow to solve the eigenvalues of a complex matrix of very high condition number?Eigenvalues of block matrix where blocks are relatedFind the eigenvalues and corresponding eigen vectors of the matrixWhat are the eigenvalues of the following symmetric matrix?Question about eigenvalues of a matrix and powers and trace of a matrixAre the generalized $lambda$ eigenvalues of a matrix T and its transpose the same?Eigenvalues and Eigenvectors without matrix $A$Find the matrix $Ain mathbbR^3times 3$
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What are the eigenvalues of matrix $I-uu^T$ where where $u$ is a unit vector and $I$ is the identity matrix in $mathbb R^n$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)about diagonal matrix and eigenvaluesReconstruct the original symmetric matrix given Eigen values and the longest Eigen vectorHow to solve the eigenvalues of a complex matrix of very high condition number?Eigenvalues of block matrix where blocks are relatedFind the eigenvalues and corresponding eigen vectors of the matrixWhat are the eigenvalues of the following symmetric matrix?Question about eigenvalues of a matrix and powers and trace of a matrixAre the generalized $lambda$ eigenvalues of a matrix T and its transpose the same?Eigenvalues and Eigenvectors without matrix $A$Find the matrix $Ain mathbbR^3times 3$
$begingroup$
I have tried to find an eigenvalue $0$, by $(I-uu^T)u=u-1u=0u$. I have felt that others are $1$,but that's no really reliable.
eigenvalues-eigenvectors
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
$endgroup$
add a comment |
$begingroup$
I have tried to find an eigenvalue $0$, by $(I-uu^T)u=u-1u=0u$. I have felt that others are $1$,but that's no really reliable.
eigenvalues-eigenvectors
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
$endgroup$
$begingroup$
This is a projection matrix
$endgroup$
– Soumik Ghosh
Mar 31 at 13:44
add a comment |
$begingroup$
I have tried to find an eigenvalue $0$, by $(I-uu^T)u=u-1u=0u$. I have felt that others are $1$,but that's no really reliable.
eigenvalues-eigenvectors
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
$endgroup$
I have tried to find an eigenvalue $0$, by $(I-uu^T)u=u-1u=0u$. I have felt that others are $1$,but that's no really reliable.
eigenvalues-eigenvectors
eigenvalues-eigenvectors
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
edited Mar 31 at 13:45
Siddharth Bhat
3,1671918
3,1671918
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
asked Mar 31 at 13:40
Maxius XuMaxius Xu
113
113
$begingroup$
This is a projection matrix
$endgroup$
– Soumik Ghosh
Mar 31 at 13:44
add a comment |
$begingroup$
This is a projection matrix
$endgroup$
– Soumik Ghosh
Mar 31 at 13:44
$begingroup$
This is a projection matrix
$endgroup$
– Soumik Ghosh
Mar 31 at 13:44
$begingroup$
This is a projection matrix
$endgroup$
– Soumik Ghosh
Mar 31 at 13:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Put $;U:=Spanu;$ and denote by $;langle .rangle;$ the usual inner product, then for any $;win U^perp;$ we get:
$$(I-uu^t)w=w-u(u^tw)=w-langle w,urangle u=wimplies w$$
is an eigenvector with eigenvalue $;1;$ . Complete the answer by using what you already did.
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Put $;U:=Spanu;$ and denote by $;langle .rangle;$ the usual inner product, then for any $;win U^perp;$ we get:
$$(I-uu^t)w=w-u(u^tw)=w-langle w,urangle u=wimplies w$$
is an eigenvector with eigenvalue $;1;$ . Complete the answer by using what you already did.
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
Put $;U:=Spanu;$ and denote by $;langle .rangle;$ the usual inner product, then for any $;win U^perp;$ we get:
$$(I-uu^t)w=w-u(u^tw)=w-langle w,urangle u=wimplies w$$
is an eigenvector with eigenvalue $;1;$ . Complete the answer by using what you already did.
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
$endgroup$
add a comment |
$begingroup$
Put $;U:=Spanu;$ and denote by $;langle .rangle;$ the usual inner product, then for any $;win U^perp;$ we get:
$$(I-uu^t)w=w-u(u^tw)=w-langle w,urangle u=wimplies w$$
is an eigenvector with eigenvalue $;1;$ . Complete the answer by using what you already did.
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
$endgroup$
Put $;U:=Spanu;$ and denote by $;langle .rangle;$ the usual inner product, then for any $;win U^perp;$ we get:
$$(I-uu^t)w=w-u(u^tw)=w-langle w,urangle u=wimplies w$$
is an eigenvector with eigenvalue $;1;$ . Complete the answer by using what you already did.
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
answered Mar 31 at 13:48
DonAntonioDonAntonio
180k1495233
180k1495233
add a comment |
add a comment |
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$begingroup$
This is a projection matrix
$endgroup$
– Soumik Ghosh
Mar 31 at 13:44