Eigenvalues and Eigenvectors of special matrix $A=u u^T$ 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)Eigenvalues and eigenvectorsEigenvalues and eigenvectors of plane reflection operatorLinear Algebra Eigenvalues and EigenvectorsLinear combination of eigenvectors with positive eigenvalueseigenvalues and eigenvectors of a matrix - formulaDetermining a $4times4$ matrix knowing $3$ of its $4$ eigenvectors and eigenvaluesEigenvalues and eigenvectors - uniquenessEigenvectors and eigenvalues- finding the matrixEigenvectors of the repeated eigenvalues for a symmetric matrixEigenvectors and eigenvalues of the zero matrix
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Eigenvalues and Eigenvectors of special matrix $A=u u^T$
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)Eigenvalues and eigenvectorsEigenvalues and eigenvectors of plane reflection operatorLinear Algebra Eigenvalues and EigenvectorsLinear combination of eigenvectors with positive eigenvalueseigenvalues and eigenvectors of a matrix - formulaDetermining a $4times4$ matrix knowing $3$ of its $4$ eigenvectors and eigenvaluesEigenvalues and eigenvectors - uniquenessEigenvectors and eigenvalues- finding the matrixEigenvectors of the repeated eigenvalues for a symmetric matrixEigenvectors and eigenvalues of the zero matrix
$begingroup$
How to calculate the eigenvalues and eigenvectors of the special matrix $A=u u^T$, where $u in R^n$.
I wrote down the matrix, which is the linear combination of vector $u$ by itself, and clearly has zero as a determinant, so I don't know how to continue.
Am I missing something?
linear-algebra
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
$endgroup$
add a comment |
$begingroup$
How to calculate the eigenvalues and eigenvectors of the special matrix $A=u u^T$, where $u in R^n$.
I wrote down the matrix, which is the linear combination of vector $u$ by itself, and clearly has zero as a determinant, so I don't know how to continue.
Am I missing something?
linear-algebra
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
$endgroup$
1
$begingroup$
Try to compute $Au$ and note that $u^Tu$ is a number. Also argue that $A$ has an $(n-1)$-dimensional kernel.
$endgroup$
– amsmath
Mar 24 at 18:57
$begingroup$
$u$ is an eigenvector, and so is anything orthogonal to it.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:59
add a comment |
$begingroup$
How to calculate the eigenvalues and eigenvectors of the special matrix $A=u u^T$, where $u in R^n$.
I wrote down the matrix, which is the linear combination of vector $u$ by itself, and clearly has zero as a determinant, so I don't know how to continue.
Am I missing something?
linear-algebra
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
$endgroup$
How to calculate the eigenvalues and eigenvectors of the special matrix $A=u u^T$, where $u in R^n$.
I wrote down the matrix, which is the linear combination of vector $u$ by itself, and clearly has zero as a determinant, so I don't know how to continue.
Am I missing something?
linear-algebra
linear-algebra
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
asked Mar 24 at 18:56
EduardoGMEduardoGM
777
777
1
$begingroup$
Try to compute $Au$ and note that $u^Tu$ is a number. Also argue that $A$ has an $(n-1)$-dimensional kernel.
$endgroup$
– amsmath
Mar 24 at 18:57
$begingroup$
$u$ is an eigenvector, and so is anything orthogonal to it.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:59
add a comment |
1
$begingroup$
Try to compute $Au$ and note that $u^Tu$ is a number. Also argue that $A$ has an $(n-1)$-dimensional kernel.
$endgroup$
– amsmath
Mar 24 at 18:57
$begingroup$
$u$ is an eigenvector, and so is anything orthogonal to it.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:59
1
1
$begingroup$
Try to compute $Au$ and note that $u^Tu$ is a number. Also argue that $A$ has an $(n-1)$-dimensional kernel.
$endgroup$
– amsmath
Mar 24 at 18:57
$begingroup$
Try to compute $Au$ and note that $u^Tu$ is a number. Also argue that $A$ has an $(n-1)$-dimensional kernel.
$endgroup$
– amsmath
Mar 24 at 18:57
$begingroup$
$u$ is an eigenvector, and so is anything orthogonal to it.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:59
$begingroup$
$u$ is an eigenvector, and so is anything orthogonal to it.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
Observe that$$Au=uu^Tu=|u|^2u$$and $$forall w,w^Tu=0to Aw=uu^Tw=0$$
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Nice! So $||u||^2$ is an eigenvalue and $u$ is the eigenvector, but I don't understand the second line !
$endgroup$
– EduardoGM
Mar 24 at 19:14
1
$begingroup$
That's right. The second line implies that another eigenvalue of the matrix $A$ is zero with its eigenvectors being all orthogonal to $u$ (which also states that the multiplicity of its $0$-value eigenvalue is $n-1$)
$endgroup$
– Mostafa Ayaz
Mar 24 at 19:17
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Hint
Observe that$$Au=uu^Tu=|u|^2u$$and $$forall w,w^Tu=0to Aw=uu^Tw=0$$
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Nice! So $||u||^2$ is an eigenvalue and $u$ is the eigenvector, but I don't understand the second line !
$endgroup$
– EduardoGM
Mar 24 at 19:14
1
$begingroup$
That's right. The second line implies that another eigenvalue of the matrix $A$ is zero with its eigenvectors being all orthogonal to $u$ (which also states that the multiplicity of its $0$-value eigenvalue is $n-1$)
$endgroup$
– Mostafa Ayaz
Mar 24 at 19:17
add a comment |
$begingroup$
Hint
Observe that$$Au=uu^Tu=|u|^2u$$and $$forall w,w^Tu=0to Aw=uu^Tw=0$$
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Nice! So $||u||^2$ is an eigenvalue and $u$ is the eigenvector, but I don't understand the second line !
$endgroup$
– EduardoGM
Mar 24 at 19:14
1
$begingroup$
That's right. The second line implies that another eigenvalue of the matrix $A$ is zero with its eigenvectors being all orthogonal to $u$ (which also states that the multiplicity of its $0$-value eigenvalue is $n-1$)
$endgroup$
– Mostafa Ayaz
Mar 24 at 19:17
add a comment |
$begingroup$
Hint
Observe that$$Au=uu^Tu=|u|^2u$$and $$forall w,w^Tu=0to Aw=uu^Tw=0$$
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
Observe that$$Au=uu^Tu=|u|^2u$$and $$forall w,w^Tu=0to Aw=uu^Tw=0$$
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 24 at 19:02
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
$begingroup$
Nice! So $||u||^2$ is an eigenvalue and $u$ is the eigenvector, but I don't understand the second line !
$endgroup$
– EduardoGM
Mar 24 at 19:14
1
$begingroup$
That's right. The second line implies that another eigenvalue of the matrix $A$ is zero with its eigenvectors being all orthogonal to $u$ (which also states that the multiplicity of its $0$-value eigenvalue is $n-1$)
$endgroup$
– Mostafa Ayaz
Mar 24 at 19:17
add a comment |
$begingroup$
Nice! So $||u||^2$ is an eigenvalue and $u$ is the eigenvector, but I don't understand the second line !
$endgroup$
– EduardoGM
Mar 24 at 19:14
1
$begingroup$
That's right. The second line implies that another eigenvalue of the matrix $A$ is zero with its eigenvectors being all orthogonal to $u$ (which also states that the multiplicity of its $0$-value eigenvalue is $n-1$)
$endgroup$
– Mostafa Ayaz
Mar 24 at 19:17
$begingroup$
Nice! So $||u||^2$ is an eigenvalue and $u$ is the eigenvector, but I don't understand the second line !
$endgroup$
– EduardoGM
Mar 24 at 19:14
$begingroup$
Nice! So $||u||^2$ is an eigenvalue and $u$ is the eigenvector, but I don't understand the second line !
$endgroup$
– EduardoGM
Mar 24 at 19:14
1
1
$begingroup$
That's right. The second line implies that another eigenvalue of the matrix $A$ is zero with its eigenvectors being all orthogonal to $u$ (which also states that the multiplicity of its $0$-value eigenvalue is $n-1$)
$endgroup$
– Mostafa Ayaz
Mar 24 at 19:17
$begingroup$
That's right. The second line implies that another eigenvalue of the matrix $A$ is zero with its eigenvectors being all orthogonal to $u$ (which also states that the multiplicity of its $0$-value eigenvalue is $n-1$)
$endgroup$
– Mostafa Ayaz
Mar 24 at 19:17
add a comment |
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$begingroup$
Try to compute $Au$ and note that $u^Tu$ is a number. Also argue that $A$ has an $(n-1)$-dimensional kernel.
$endgroup$
– amsmath
Mar 24 at 18:57
$begingroup$
$u$ is an eigenvector, and so is anything orthogonal to it.
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:59