Using eigenvectors as a basis for space 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)Find an orthogonal basis consisting of eigenvectorsLooking for orthogonal basis of eigenvectors using Gram Schmidt processChange of basis help turning a sub-matrix from complex to realConflicting answers for eigenvectorsBasis of the null spaceEigenvectors for prime numbers matricesSymmetric Matrix , Eigenvectors are not orthogonal to the same eigenvalue.diagonalizable matrices using eigenvectorsBasis consisting of eigenvectorsFind basis of fundamental subspaces with given eigenvalues and eigenvectors
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Using eigenvectors as a basis for space
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)Find an orthogonal basis consisting of eigenvectorsLooking for orthogonal basis of eigenvectors using Gram Schmidt processChange of basis help turning a sub-matrix from complex to realConflicting answers for eigenvectorsBasis of the null spaceEigenvectors for prime numbers matricesSymmetric Matrix , Eigenvectors are not orthogonal to the same eigenvalue.diagonalizable matrices using eigenvectorsBasis consisting of eigenvectorsFind basis of fundamental subspaces with given eigenvalues and eigenvectors
$begingroup$
Consider the following matrix
$A=beginbmatrix0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0endbmatrix$
Which $a,b,c$ are real numbers
What conditions are required for $a,b,c$ such that $mathbbR^4$ has a basis made of A eigenvectors?
linear-algebra
$endgroup$
add a comment |
$begingroup$
Consider the following matrix
$A=beginbmatrix0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0endbmatrix$
Which $a,b,c$ are real numbers
What conditions are required for $a,b,c$ such that $mathbbR^4$ has a basis made of A eigenvectors?
linear-algebra
$endgroup$
add a comment |
$begingroup$
Consider the following matrix
$A=beginbmatrix0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0endbmatrix$
Which $a,b,c$ are real numbers
What conditions are required for $a,b,c$ such that $mathbbR^4$ has a basis made of A eigenvectors?
linear-algebra
$endgroup$
Consider the following matrix
$A=beginbmatrix0&0&0&0\a&0&0&0\0&b&0&0\0&0&c&0endbmatrix$
Which $a,b,c$ are real numbers
What conditions are required for $a,b,c$ such that $mathbbR^4$ has a basis made of A eigenvectors?
linear-algebra
linear-algebra
asked Mar 25 at 9:53
Alireza HosseiniAlireza Hosseini
233
233
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2 Answers
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$begingroup$
The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.
If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.
Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.
So the necessary and sufficient condition is that $a=b=c=0$.
$endgroup$
add a comment |
$begingroup$
$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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active
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votes
$begingroup$
The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.
If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.
Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.
So the necessary and sufficient condition is that $a=b=c=0$.
$endgroup$
add a comment |
$begingroup$
The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.
If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.
Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.
So the necessary and sufficient condition is that $a=b=c=0$.
$endgroup$
add a comment |
$begingroup$
The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.
If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.
Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.
So the necessary and sufficient condition is that $a=b=c=0$.
$endgroup$
The characteristic polynomial of $A$ is $X^4$, so the only eigenvalue is $0$.
If $A$ is diagonalizable then, it means that $A=0$, so $a=b=c=0$.
Conversely, if $a=b=c=0$, all vectors are eigenvectors, so of course you can find a basis made of eigenvectors.
So the necessary and sufficient condition is that $a=b=c=0$.
answered Mar 25 at 9:59
TheSilverDoeTheSilverDoe
5,566216
5,566216
add a comment |
add a comment |
$begingroup$
$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.
$endgroup$
add a comment |
$begingroup$
$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.
$endgroup$
add a comment |
$begingroup$
$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.
$endgroup$
$lambda = 0$ is the only eigenvalue (it has algebraic multiplicity 4). The only way the system $Av = 0$ has 4 degrees of freedom is if $A=0$, i.e. $a=b=c=0$.
answered Mar 25 at 10:03
PierreCarrePierreCarre
2,178215
2,178215
add a comment |
add a comment |
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