Mixing
Eigenvalue of (some) $ 4 times 4 $ symmetric matrices
$begingroup$
$$A=pmatrix{
0 & 3 & 2 & 0 \
3 & 0 & 0 & 2 \
2 & 0 & 0 & 3 \
0 & 2 & 3 & 0 \
}$$
Is there a quicker way to compute eigenvalues of this matrix other than to do it the long way? And what are the strategies for similar matrices?
matrices eigenvalues-eigenvectors
edited Jan 15 at 5:52
user549397
1,5101418
asked Jan 15 at 5:05
5Six5Six
84
$endgroup$
add a comment |
$begingroup$
$$A=pmatrix{
0 & 3 & 2 & 0 \
3 & 0 & 0 & 2 \
2 & 0 & 0 & 3 \
0 & 2 & 3 & 0 \
}$$
Is there a quicker way to compute eigenvalues of this matrix other than to do it the long way? And what are the strategies for similar matrices?
matrices eigenvalues-eigenvectors
edited Jan 15 at 5:52
user549397
1,5101418
asked Jan 15 at 5:05
5Six5Six
84
$endgroup$
$begingroup$
I can quickly eyeball this and notice that the matrix has a constant row sum of $5$. This means that $(1, 1, 1, 1)$ is an eigenvector, corresponding to eigenvalue $5$. That at least tells you that $lambda - 5$ is a factor of the characteristic polynomial.
$endgroup$
– Theo Bendit
Jan 15 at 5:13
$begingroup$
Note that $ A $ is a symmetric matrix. You can always diagonalize a symmetric matrix by a series of elementary matrices acting on the rows and the columns correspondingly without changing the eigenvalues.
$endgroup$
– user549397
Jan 15 at 5:27
add a comment |
$begingroup$
$$A=pmatrix{
0 & 3 & 2 & 0 \
3 & 0 & 0 & 2 \
2 & 0 & 0 & 3 \
0 & 2 & 3 & 0 \
}$$
Is there a quicker way to compute eigenvalues of this matrix other than to do it the long way? And what are the strategies for similar matrices?
matrices eigenvalues-eigenvectors
edited Jan 15 at 5:52
user549397
1,5101418
asked Jan 15 at 5:05
5Six5Six
84
$endgroup$
$$A=pmatrix{
0 & 3 & 2 & 0 \
3 & 0 & 0 & 2 \
2 & 0 & 0 & 3 \
0 & 2 & 3 & 0 \
}$$
Is there a quicker way to compute eigenvalues of this matrix other than to do it the long way? And what are the strategies for similar matrices?
matrices eigenvalues-eigenvectors
matrices eigenvalues-eigenvectors
edited Jan 15 at 5:52
user549397
1,5101418
asked Jan 15 at 5:05
5Six5Six
84
edited Jan 15 at 5:52
user549397
1,5101418
asked Jan 15 at 5:05
5Six5Six
84
edited Jan 15 at 5:52
user549397
1,5101418
edited Jan 15 at 5:52
user549397
1,5101418
edited Jan 15 at 5:52
user549397
1,5101418
1,5101418
asked Jan 15 at 5:05
5Six5Six
84
asked Jan 15 at 5:05
5Six5Six
84
asked Jan 15 at 5:05
5Six5Six
84
84
$begingroup$
I can quickly eyeball this and notice that the matrix has a constant row sum of $5$. This means that $(1, 1, 1, 1)$ is an eigenvector, corresponding to eigenvalue $5$. That at least tells you that $lambda - 5$ is a factor of the characteristic polynomial.
$endgroup$
– Theo Bendit
Jan 15 at 5:13
$begingroup$
Note that $ A $ is a symmetric matrix. You can always diagonalize a symmetric matrix by a series of elementary matrices acting on the rows and the columns correspondingly without changing the eigenvalues.
$endgroup$
– user549397
Jan 15 at 5:27
add a comment |
$begingroup$
I can quickly eyeball this and notice that the matrix has a constant row sum of $5$. This means that $(1, 1, 1, 1)$ is an eigenvector, corresponding to eigenvalue $5$. That at least tells you that $lambda - 5$ is a factor of the characteristic polynomial.
$endgroup$
– Theo Bendit
Jan 15 at 5:13
$begingroup$
Note that $ A $ is a symmetric matrix. You can always diagonalize a symmetric matrix by a series of elementary matrices acting on the rows and the columns correspondingly without changing the eigenvalues.
$endgroup$
– user549397
Jan 15 at 5:27
$begingroup$
I can quickly eyeball this and notice that the matrix has a constant row sum of $5$. This means that $(1, 1, 1, 1)$ is an eigenvector, corresponding to eigenvalue $5$. That at least tells you that $lambda - 5$ is a factor of the characteristic polynomial.
$endgroup$
– Theo Bendit
Jan 15 at 5:13
$begingroup$
I can quickly eyeball this and notice that the matrix has a constant row sum of $5$. This means that $(1, 1, 1, 1)$ is an eigenvector, corresponding to eigenvalue $5$. That at least tells you that $lambda - 5$ is a factor of the characteristic polynomial.
$endgroup$
– Theo Bendit
Jan 15 at 5:13
$begingroup$
Note that $ A $ is a symmetric matrix. You can always diagonalize a symmetric matrix by a series of elementary matrices acting on the rows and the columns correspondingly without changing the eigenvalues.
$endgroup$
– user549397
Jan 15 at 5:27
$begingroup$
Note that $ A $ is a symmetric matrix. You can always diagonalize a symmetric matrix by a series of elementary matrices acting on the rows and the columns correspondingly without changing the eigenvalues.
$endgroup$
– user549397
Jan 15 at 5:27
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Observe
begin{align}
M=
begin{pmatrix}
A & B\
B & A
end{pmatrix}
end{align}
where $B = 2I_2$ and $A=begin{pmatrix}0 & 3\ 3 & 0 end{pmatrix}$ and $A$ & $B$ commute. Then we see that
begin{align}
detleft(M-lambda I_4 right) = det ((A-lambda I_2)^2-B^2)
end{align}
where we used the determinant formula for block matrices.
Note that
begin{align}
(A-lambda I_2)^2=
begin{pmatrix}
lambda^2+9 & -6lambda\
-6lambda & lambda^2+9
end{pmatrix}
end{align}
which means
begin{align}
det (M-lambda I_4) = (lambda^2+5)^2-36lambda^2 = (lambda^2-1^2).(lambda^2-5^2)
end{align}
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
$endgroup$
1
$begingroup$
There's a problem here - that's the wrong answer. The characteristic polynomial is $(lambda^2-5^2)(lambda^2-1^2)=(lambda-5)(lambda-1)(lambda+1)(lambda+5)$. Fortunately, everything but that final entry looks right.
$endgroup$
– jmerry
Jan 15 at 7:13
$begingroup$
@jmerry Thanks for catching the typo.
$endgroup$
– Jacky Chong
Jan 15 at 21:13
add a comment |
$begingroup$
I don't have a general strategy.
But here, all the rows sum the same, so $(1,1,1,1)$ is an eigenvector for $lambda=5$. Similarly, the alternating sums of the rows are $1$ and $-1$, and so $(1,-1,1,-1)$ is an eigenvector for $lambda=-1$.
With similar ideas we see that $(1,1,-1,-1)$ is an eigenevector for $lambda=1$.
If we don't have the imagination to find the last eigenvector/eigenvalue, we may notice that the trace is zero, so the last eigenvalues is $lambda=-5$. The eigenvector is $(1,-1,-1,1)$.
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
We can look at the matrices $A=begin{bmatrix}0&1&0&0\1&0&0&0\0&0&0&1\0&0&1&0end{bmatrix}$ and $B=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix}$. Each of these can be split into two $begin{bmatrix}0&1\1&0end{bmatrix}$ reflection blocks, so they have eigenvalues $1$ and $-1$ each with multiplicity two. Specifically, the eigenvectors are $begin{bmatrix}1\1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\1end{bmatrix}$ for $A$ and the eigenvalue $1$, $begin{bmatrix}1\-1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\-1end{bmatrix}$ for $A$ and the eigenvalue $-1$, $begin{bmatrix}1\0\1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\1end{bmatrix}$ for $B$ and the eigenvalue $1$, and finally $begin{bmatrix}1\0\-1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\-1end{bmatrix}$ for $B$ and the eigenvalue $-1$. Linear combinations of each pair will also work, of course, such as $begin{bmatrix}1\1\1\1end{bmatrix}$ for the eigenvalue $1$ in both $A$ and $B$, or $begin{bmatrix}1\-1\-1\1end{bmatrix}$ for the eigenvalue $-1$ in both $A$ and $B$.
The matrix we care about is a linear combination $M=3A+2B$, so the two common eigenvectors $begin{bmatrix}1\1\1\1end{bmatrix}$ and $begin{bmatrix}1\-1\-1\1end{bmatrix}$ we found are still eigenvectors, with eigenvalues $3+2=5$ and $-3-2=-5$ respectively. We need two more - and it turns out we can get more out of our eigenvector lists. $begin{bmatrix}1\1\-1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $1$, and an eigenvector for $B$ with eigenvalue $-1$, so it's an eigenvector for $M$ with eigenvalue $3-2=1$. Similarly, $begin{bmatrix}1\-1\1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $-1$ and for $B$ with eigenvalue $1$, so it's an eigenvector for $M$ with eigenvalue $-3+2=-1$.
We have found four linearly independent eigenvectors, with four different eigenvalues. As $M$ is a $4times 4$ matrix, that's everything.
The strategy implied here isn't something that will work very reliably, but when it does and we can decompose a matrix into pieces with known eigenvectors, we can do a lot.
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe
begin{align}
M=
begin{pmatrix}
A & B\
B & A
end{pmatrix}
end{align}
where $B = 2I_2$ and $A=begin{pmatrix}0 & 3\ 3 & 0 end{pmatrix}$ and $A$ & $B$ commute. Then we see that
begin{align}
detleft(M-lambda I_4 right) = det ((A-lambda I_2)^2-B^2)
end{align}
where we used the determinant formula for block matrices.
Note that
begin{align}
(A-lambda I_2)^2=
begin{pmatrix}
lambda^2+9 & -6lambda\
-6lambda & lambda^2+9
end{pmatrix}
end{align}
which means
begin{align}
det (M-lambda I_4) = (lambda^2+5)^2-36lambda^2 = (lambda^2-1^2).(lambda^2-5^2)
end{align}
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
$endgroup$
1
$begingroup$
There's a problem here - that's the wrong answer. The characteristic polynomial is $(lambda^2-5^2)(lambda^2-1^2)=(lambda-5)(lambda-1)(lambda+1)(lambda+5)$. Fortunately, everything but that final entry looks right.
$endgroup$
– jmerry
Jan 15 at 7:13
$begingroup$
@jmerry Thanks for catching the typo.
$endgroup$
– Jacky Chong
Jan 15 at 21:13
add a comment |
$begingroup$
Observe
begin{align}
M=
begin{pmatrix}
A & B\
B & A
end{pmatrix}
end{align}
where $B = 2I_2$ and $A=begin{pmatrix}0 & 3\ 3 & 0 end{pmatrix}$ and $A$ & $B$ commute. Then we see that
begin{align}
detleft(M-lambda I_4 right) = det ((A-lambda I_2)^2-B^2)
end{align}
where we used the determinant formula for block matrices.
Note that
begin{align}
(A-lambda I_2)^2=
begin{pmatrix}
lambda^2+9 & -6lambda\
-6lambda & lambda^2+9
end{pmatrix}
end{align}
which means
begin{align}
det (M-lambda I_4) = (lambda^2+5)^2-36lambda^2 = (lambda^2-1^2).(lambda^2-5^2)
end{align}
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
$endgroup$
1
$begingroup$
There's a problem here - that's the wrong answer. The characteristic polynomial is $(lambda^2-5^2)(lambda^2-1^2)=(lambda-5)(lambda-1)(lambda+1)(lambda+5)$. Fortunately, everything but that final entry looks right.
$endgroup$
– jmerry
Jan 15 at 7:13
$begingroup$
@jmerry Thanks for catching the typo.
$endgroup$
– Jacky Chong
Jan 15 at 21:13
add a comment |
$begingroup$
Observe
begin{align}
M=
begin{pmatrix}
A & B\
B & A
end{pmatrix}
end{align}
where $B = 2I_2$ and $A=begin{pmatrix}0 & 3\ 3 & 0 end{pmatrix}$ and $A$ & $B$ commute. Then we see that
begin{align}
detleft(M-lambda I_4 right) = det ((A-lambda I_2)^2-B^2)
end{align}
where we used the determinant formula for block matrices.
Note that
begin{align}
(A-lambda I_2)^2=
begin{pmatrix}
lambda^2+9 & -6lambda\
-6lambda & lambda^2+9
end{pmatrix}
end{align}
which means
begin{align}
det (M-lambda I_4) = (lambda^2+5)^2-36lambda^2 = (lambda^2-1^2).(lambda^2-5^2)
end{align}
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
$endgroup$
Observe
begin{align}
M=
begin{pmatrix}
A & B\
B & A
end{pmatrix}
end{align}
where $B = 2I_2$ and $A=begin{pmatrix}0 & 3\ 3 & 0 end{pmatrix}$ and $A$ & $B$ commute. Then we see that
begin{align}
detleft(M-lambda I_4 right) = det ((A-lambda I_2)^2-B^2)
end{align}
where we used the determinant formula for block matrices.
Note that
begin{align}
(A-lambda I_2)^2=
begin{pmatrix}
lambda^2+9 & -6lambda\
-6lambda & lambda^2+9
end{pmatrix}
end{align}
which means
begin{align}
det (M-lambda I_4) = (lambda^2+5)^2-36lambda^2 = (lambda^2-1^2).(lambda^2-5^2)
end{align}
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
edited Jan 15 at 21:14
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
answered Jan 15 at 5:21
Jacky ChongJacky Chong
18.7k21128
18.7k21128
1
$begingroup$
There's a problem here - that's the wrong answer. The characteristic polynomial is $(lambda^2-5^2)(lambda^2-1^2)=(lambda-5)(lambda-1)(lambda+1)(lambda+5)$. Fortunately, everything but that final entry looks right.
$endgroup$
– jmerry
Jan 15 at 7:13
$begingroup$
@jmerry Thanks for catching the typo.
$endgroup$
– Jacky Chong
Jan 15 at 21:13
add a comment |
1
$begingroup$
There's a problem here - that's the wrong answer. The characteristic polynomial is $(lambda^2-5^2)(lambda^2-1^2)=(lambda-5)(lambda-1)(lambda+1)(lambda+5)$. Fortunately, everything but that final entry looks right.
$endgroup$
– jmerry
Jan 15 at 7:13
$begingroup$
@jmerry Thanks for catching the typo.
$endgroup$
– Jacky Chong
Jan 15 at 21:13
1
1
$begingroup$
There's a problem here - that's the wrong answer. The characteristic polynomial is $(lambda^2-5^2)(lambda^2-1^2)=(lambda-5)(lambda-1)(lambda+1)(lambda+5)$. Fortunately, everything but that final entry looks right.
$endgroup$
– jmerry
Jan 15 at 7:13
$begingroup$
There's a problem here - that's the wrong answer. The characteristic polynomial is $(lambda^2-5^2)(lambda^2-1^2)=(lambda-5)(lambda-1)(lambda+1)(lambda+5)$. Fortunately, everything but that final entry looks right.
$endgroup$
– jmerry
Jan 15 at 7:13
$begingroup$
@jmerry Thanks for catching the typo.
$endgroup$
– Jacky Chong
Jan 15 at 21:13
$begingroup$
@jmerry Thanks for catching the typo.
$endgroup$
– Jacky Chong
Jan 15 at 21:13
add a comment |
$begingroup$
I don't have a general strategy.
But here, all the rows sum the same, so $(1,1,1,1)$ is an eigenvector for $lambda=5$. Similarly, the alternating sums of the rows are $1$ and $-1$, and so $(1,-1,1,-1)$ is an eigenvector for $lambda=-1$.
With similar ideas we see that $(1,1,-1,-1)$ is an eigenevector for $lambda=1$.
If we don't have the imagination to find the last eigenvector/eigenvalue, we may notice that the trace is zero, so the last eigenvalues is $lambda=-5$. The eigenvector is $(1,-1,-1,1)$.
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
I don't have a general strategy.
But here, all the rows sum the same, so $(1,1,1,1)$ is an eigenvector for $lambda=5$. Similarly, the alternating sums of the rows are $1$ and $-1$, and so $(1,-1,1,-1)$ is an eigenvector for $lambda=-1$.
With similar ideas we see that $(1,1,-1,-1)$ is an eigenevector for $lambda=1$.
If we don't have the imagination to find the last eigenvector/eigenvalue, we may notice that the trace is zero, so the last eigenvalues is $lambda=-5$. The eigenvector is $(1,-1,-1,1)$.
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
I don't have a general strategy.
But here, all the rows sum the same, so $(1,1,1,1)$ is an eigenvector for $lambda=5$. Similarly, the alternating sums of the rows are $1$ and $-1$, and so $(1,-1,1,-1)$ is an eigenvector for $lambda=-1$.
With similar ideas we see that $(1,1,-1,-1)$ is an eigenevector for $lambda=1$.
If we don't have the imagination to find the last eigenvector/eigenvalue, we may notice that the trace is zero, so the last eigenvalues is $lambda=-5$. The eigenvector is $(1,-1,-1,1)$.
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
I don't have a general strategy.
But here, all the rows sum the same, so $(1,1,1,1)$ is an eigenvector for $lambda=5$. Similarly, the alternating sums of the rows are $1$ and $-1$, and so $(1,-1,1,-1)$ is an eigenvector for $lambda=-1$.
With similar ideas we see that $(1,1,-1,-1)$ is an eigenevector for $lambda=1$.
If we don't have the imagination to find the last eigenvector/eigenvalue, we may notice that the trace is zero, so the last eigenvalues is $lambda=-5$. The eigenvector is $(1,-1,-1,1)$.
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 15 at 5:30
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
add a comment |
$begingroup$
We can look at the matrices $A=begin{bmatrix}0&1&0&0\1&0&0&0\0&0&0&1\0&0&1&0end{bmatrix}$ and $B=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix}$. Each of these can be split into two $begin{bmatrix}0&1\1&0end{bmatrix}$ reflection blocks, so they have eigenvalues $1$ and $-1$ each with multiplicity two. Specifically, the eigenvectors are $begin{bmatrix}1\1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\1end{bmatrix}$ for $A$ and the eigenvalue $1$, $begin{bmatrix}1\-1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\-1end{bmatrix}$ for $A$ and the eigenvalue $-1$, $begin{bmatrix}1\0\1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\1end{bmatrix}$ for $B$ and the eigenvalue $1$, and finally $begin{bmatrix}1\0\-1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\-1end{bmatrix}$ for $B$ and the eigenvalue $-1$. Linear combinations of each pair will also work, of course, such as $begin{bmatrix}1\1\1\1end{bmatrix}$ for the eigenvalue $1$ in both $A$ and $B$, or $begin{bmatrix}1\-1\-1\1end{bmatrix}$ for the eigenvalue $-1$ in both $A$ and $B$.
The matrix we care about is a linear combination $M=3A+2B$, so the two common eigenvectors $begin{bmatrix}1\1\1\1end{bmatrix}$ and $begin{bmatrix}1\-1\-1\1end{bmatrix}$ we found are still eigenvectors, with eigenvalues $3+2=5$ and $-3-2=-5$ respectively. We need two more - and it turns out we can get more out of our eigenvector lists. $begin{bmatrix}1\1\-1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $1$, and an eigenvector for $B$ with eigenvalue $-1$, so it's an eigenvector for $M$ with eigenvalue $3-2=1$. Similarly, $begin{bmatrix}1\-1\1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $-1$ and for $B$ with eigenvalue $1$, so it's an eigenvector for $M$ with eigenvalue $-3+2=-1$.
We have found four linearly independent eigenvectors, with four different eigenvalues. As $M$ is a $4times 4$ matrix, that's everything.
The strategy implied here isn't something that will work very reliably, but when it does and we can decompose a matrix into pieces with known eigenvectors, we can do a lot.
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
$endgroup$
add a comment |
$begingroup$
We can look at the matrices $A=begin{bmatrix}0&1&0&0\1&0&0&0\0&0&0&1\0&0&1&0end{bmatrix}$ and $B=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix}$. Each of these can be split into two $begin{bmatrix}0&1\1&0end{bmatrix}$ reflection blocks, so they have eigenvalues $1$ and $-1$ each with multiplicity two. Specifically, the eigenvectors are $begin{bmatrix}1\1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\1end{bmatrix}$ for $A$ and the eigenvalue $1$, $begin{bmatrix}1\-1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\-1end{bmatrix}$ for $A$ and the eigenvalue $-1$, $begin{bmatrix}1\0\1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\1end{bmatrix}$ for $B$ and the eigenvalue $1$, and finally $begin{bmatrix}1\0\-1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\-1end{bmatrix}$ for $B$ and the eigenvalue $-1$. Linear combinations of each pair will also work, of course, such as $begin{bmatrix}1\1\1\1end{bmatrix}$ for the eigenvalue $1$ in both $A$ and $B$, or $begin{bmatrix}1\-1\-1\1end{bmatrix}$ for the eigenvalue $-1$ in both $A$ and $B$.
The matrix we care about is a linear combination $M=3A+2B$, so the two common eigenvectors $begin{bmatrix}1\1\1\1end{bmatrix}$ and $begin{bmatrix}1\-1\-1\1end{bmatrix}$ we found are still eigenvectors, with eigenvalues $3+2=5$ and $-3-2=-5$ respectively. We need two more - and it turns out we can get more out of our eigenvector lists. $begin{bmatrix}1\1\-1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $1$, and an eigenvector for $B$ with eigenvalue $-1$, so it's an eigenvector for $M$ with eigenvalue $3-2=1$. Similarly, $begin{bmatrix}1\-1\1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $-1$ and for $B$ with eigenvalue $1$, so it's an eigenvector for $M$ with eigenvalue $-3+2=-1$.
We have found four linearly independent eigenvectors, with four different eigenvalues. As $M$ is a $4times 4$ matrix, that's everything.
The strategy implied here isn't something that will work very reliably, but when it does and we can decompose a matrix into pieces with known eigenvectors, we can do a lot.
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
$endgroup$
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$begingroup$
We can look at the matrices $A=begin{bmatrix}0&1&0&0\1&0&0&0\0&0&0&1\0&0&1&0end{bmatrix}$ and $B=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix}$. Each of these can be split into two $begin{bmatrix}0&1\1&0end{bmatrix}$ reflection blocks, so they have eigenvalues $1$ and $-1$ each with multiplicity two. Specifically, the eigenvectors are $begin{bmatrix}1\1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\1end{bmatrix}$ for $A$ and the eigenvalue $1$, $begin{bmatrix}1\-1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\-1end{bmatrix}$ for $A$ and the eigenvalue $-1$, $begin{bmatrix}1\0\1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\1end{bmatrix}$ for $B$ and the eigenvalue $1$, and finally $begin{bmatrix}1\0\-1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\-1end{bmatrix}$ for $B$ and the eigenvalue $-1$. Linear combinations of each pair will also work, of course, such as $begin{bmatrix}1\1\1\1end{bmatrix}$ for the eigenvalue $1$ in both $A$ and $B$, or $begin{bmatrix}1\-1\-1\1end{bmatrix}$ for the eigenvalue $-1$ in both $A$ and $B$.
The matrix we care about is a linear combination $M=3A+2B$, so the two common eigenvectors $begin{bmatrix}1\1\1\1end{bmatrix}$ and $begin{bmatrix}1\-1\-1\1end{bmatrix}$ we found are still eigenvectors, with eigenvalues $3+2=5$ and $-3-2=-5$ respectively. We need two more - and it turns out we can get more out of our eigenvector lists. $begin{bmatrix}1\1\-1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $1$, and an eigenvector for $B$ with eigenvalue $-1$, so it's an eigenvector for $M$ with eigenvalue $3-2=1$. Similarly, $begin{bmatrix}1\-1\1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $-1$ and for $B$ with eigenvalue $1$, so it's an eigenvector for $M$ with eigenvalue $-3+2=-1$.
We have found four linearly independent eigenvectors, with four different eigenvalues. As $M$ is a $4times 4$ matrix, that's everything.
The strategy implied here isn't something that will work very reliably, but when it does and we can decompose a matrix into pieces with known eigenvectors, we can do a lot.
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
$endgroup$
We can look at the matrices $A=begin{bmatrix}0&1&0&0\1&0&0&0\0&0&0&1\0&0&1&0end{bmatrix}$ and $B=begin{bmatrix}0&0&1&0\0&0&0&1\1&0&0&0\0&1&0&0end{bmatrix}$. Each of these can be split into two $begin{bmatrix}0&1\1&0end{bmatrix}$ reflection blocks, so they have eigenvalues $1$ and $-1$ each with multiplicity two. Specifically, the eigenvectors are $begin{bmatrix}1\1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\1end{bmatrix}$ for $A$ and the eigenvalue $1$, $begin{bmatrix}1\-1\0\0end{bmatrix}$ and $begin{bmatrix}0\0\1\-1end{bmatrix}$ for $A$ and the eigenvalue $-1$, $begin{bmatrix}1\0\1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\1end{bmatrix}$ for $B$ and the eigenvalue $1$, and finally $begin{bmatrix}1\0\-1\0end{bmatrix}$ and $begin{bmatrix}0\1\0\-1end{bmatrix}$ for $B$ and the eigenvalue $-1$. Linear combinations of each pair will also work, of course, such as $begin{bmatrix}1\1\1\1end{bmatrix}$ for the eigenvalue $1$ in both $A$ and $B$, or $begin{bmatrix}1\-1\-1\1end{bmatrix}$ for the eigenvalue $-1$ in both $A$ and $B$.
The matrix we care about is a linear combination $M=3A+2B$, so the two common eigenvectors $begin{bmatrix}1\1\1\1end{bmatrix}$ and $begin{bmatrix}1\-1\-1\1end{bmatrix}$ we found are still eigenvectors, with eigenvalues $3+2=5$ and $-3-2=-5$ respectively. We need two more - and it turns out we can get more out of our eigenvector lists. $begin{bmatrix}1\1\-1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $1$, and an eigenvector for $B$ with eigenvalue $-1$, so it's an eigenvector for $M$ with eigenvalue $3-2=1$. Similarly, $begin{bmatrix}1\-1\1\-1end{bmatrix}$ is an eigenvector for $A$ with eigenvalue $-1$ and for $B$ with eigenvalue $1$, so it's an eigenvector for $M$ with eigenvalue $-3+2=-1$.
We have found four linearly independent eigenvectors, with four different eigenvalues. As $M$ is a $4times 4$ matrix, that's everything.
The strategy implied here isn't something that will work very reliably, but when it does and we can decompose a matrix into pieces with known eigenvectors, we can do a lot.
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
answered Jan 15 at 5:35
jmerryjmerry
9,3431124
9,3431124
add a comment |
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$begingroup$
I can quickly eyeball this and notice that the matrix has a constant row sum of $5$. This means that $(1, 1, 1, 1)$ is an eigenvector, corresponding to eigenvalue $5$. That at least tells you that $lambda - 5$ is a factor of the characteristic polynomial.
$endgroup$
– Theo Bendit
Jan 15 at 5:13
$begingroup$
Note that $ A $ is a symmetric matrix. You can always diagonalize a symmetric matrix by a series of elementary matrices acting on the rows and the columns correspondingly without changing the eigenvalues.
$endgroup$
– user549397
Jan 15 at 5:27