Mixing
How to determine eigenvectors of symmetric circulant matrix {{A,B,B},{B,A,B},{B,B,A}}?
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I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .
I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .
But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.
From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:
$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$
Any help very appreciated.
linear-algebra eigenvalues-eigenvectors
asked Jan 16 at 20:06
fazanfazan
417
$endgroup$
add a comment |
$begingroup$
I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .
I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .
But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.
From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:
$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$
Any help very appreciated.
linear-algebra eigenvalues-eigenvectors
asked Jan 16 at 20:06
fazanfazan
417
$endgroup$
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06
add a comment |
$begingroup$
I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .
I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .
But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.
From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:
$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$
Any help very appreciated.
linear-algebra eigenvalues-eigenvectors
asked Jan 16 at 20:06
fazanfazan
417
$endgroup$
I'm trying to find the eigenvectors for the matrix $$begin{bmatrix} A & B & B \ B & A & B \ B & B & A end{bmatrix} $$ .
I determined the eigenvalues to be $lambda_1=lambda_2=A-B$ and $lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=frac{1}{sqrt{3}}begin{bmatrix} 1 \1\1 end{bmatrix}$ .
But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$begin{bmatrix} 1 & 1 & 1 \ 0& 0 & 0\ 0& 0 & 0 end{bmatrix} $$.
From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:
$$v_1+v_2+v_3=0$$
$$v_1neq v_2neq v_3neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$
Any help very appreciated.
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Jan 16 at 20:06
fazanfazan
417
asked Jan 16 at 20:06
fazanfazan
417
asked Jan 16 at 20:06
fazanfazan
417
asked Jan 16 at 20:06
fazanfazan
417
asked Jan 16 at 20:06
fazanfazan
417
417
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06
add a comment |
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06
$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06
add a comment |
1 Answer
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We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.
In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.
We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
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1 Answer
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$begingroup$
We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.
In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.
We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
$endgroup$
add a comment |
$begingroup$
We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.
In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.
We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
$endgroup$
add a comment |
$begingroup$
We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.
In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.
We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
$endgroup$
We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.
In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.
We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
answered Jan 16 at 20:47
I like SerenaI like Serena
4,2221722
4,2221722
add a comment |
add a comment |
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$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of math.stackexchange.com/q/2177457/265466 and many other questions.
$endgroup$
– amd
Jan 16 at 23:06