Mixing
Finding out Eigenvalues and Eigenvectors
Clash Royale CLAN TAG#URR8PPP
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Question:
$A=beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$
I am trying to find out the Eigenvalues and Eigenvector for this question. During my working out (I use the cover-up method- First row/First column, Second row/Second column etc.), I've gone until $(λ-1)[(λ-1)(λ-2)]+(1)[(0)(λ-2)]-(2)[(0)(λ-1)]$. Is the working out right so far? Also not sure what $1[(0)cdot(λ-2)]$ equals to?
Mainly the eigenvalues and I'm having issues with.
matrices eigenvalues-eigenvectors
edited Aug 2 at 13:47
mrtaurho
619117
asked Aug 2 at 13:39
SGcipher
6
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up vote
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down vote
favorite
Question:
$A=beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$
I am trying to find out the Eigenvalues and Eigenvector for this question. During my working out (I use the cover-up method- First row/First column, Second row/Second column etc.), I've gone until $(λ-1)[(λ-1)(λ-2)]+(1)[(0)(λ-2)]-(2)[(0)(λ-1)]$. Is the working out right so far? Also not sure what $1[(0)cdot(λ-2)]$ equals to?
Mainly the eigenvalues and I'm having issues with.
matrices eigenvalues-eigenvectors
edited Aug 2 at 13:47
mrtaurho
619117
asked Aug 2 at 13:39
SGcipher
6
1
$1.[0.(lambda-2)]=0$ simply
– Isham
Aug 2 at 13:50
I found the eigenvectors as well
– RHowe
Aug 2 at 20:57
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question:
$A=beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$
I am trying to find out the Eigenvalues and Eigenvector for this question. During my working out (I use the cover-up method- First row/First column, Second row/Second column etc.), I've gone until $(λ-1)[(λ-1)(λ-2)]+(1)[(0)(λ-2)]-(2)[(0)(λ-1)]$. Is the working out right so far? Also not sure what $1[(0)cdot(λ-2)]$ equals to?
Mainly the eigenvalues and I'm having issues with.
matrices eigenvalues-eigenvectors
edited Aug 2 at 13:47
mrtaurho
619117
asked Aug 2 at 13:39
SGcipher
6
Question:
$A=beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$
I am trying to find out the Eigenvalues and Eigenvector for this question. During my working out (I use the cover-up method- First row/First column, Second row/Second column etc.), I've gone until $(λ-1)[(λ-1)(λ-2)]+(1)[(0)(λ-2)]-(2)[(0)(λ-1)]$. Is the working out right so far? Also not sure what $1[(0)cdot(λ-2)]$ equals to?
Mainly the eigenvalues and I'm having issues with.
matrices eigenvalues-eigenvectors
edited Aug 2 at 13:47
mrtaurho
619117
asked Aug 2 at 13:39
SGcipher
6
edited Aug 2 at 13:47
mrtaurho
619117
edited Aug 2 at 13:47
mrtaurho
619117
edited Aug 2 at 13:47
mrtaurho
619117
619117
asked Aug 2 at 13:39
SGcipher
6
asked Aug 2 at 13:39
SGcipher
6
asked Aug 2 at 13:39
SGcipher
6
6
1
$1.[0.(lambda-2)]=0$ simply
– Isham
Aug 2 at 13:50
I found the eigenvectors as well
– RHowe
Aug 2 at 20:57
add a comment |Â
1
$1.[0.(lambda-2)]=0$ simply
– Isham
Aug 2 at 13:50
I found the eigenvectors as well
– RHowe
Aug 2 at 20:57
1
1
$1.[0.(lambda-2)]=0$ simply
– Isham
Aug 2 at 13:50
$1.[0.(lambda-2)]=0$ simply
– Isham
Aug 2 at 13:50
I found the eigenvectors as well
– RHowe
Aug 2 at 20:57
I found the eigenvectors as well
– RHowe
Aug 2 at 20:57
add a comment |Â
5 Answers
5
active
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up vote
3
down vote
From here
$$|A-lambda I|=beginvmatrix
1-lambda & 1 & 2\
0 & 1-lambda & 3\
0 & 0 & 2-lambdaendvmatrix=(1-lambda)^2(2-lambda)=0$$
we can see that for a up triangular matrix eigenvalues are the diagonal entries.
Now for each $lambda$ solve $(A-lambda I)x=0$ to find the corresponding eigenvectors.
answered Aug 2 at 13:45
gimusi
63.8k73480
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2
down vote
Hint: All Eigenvalues of a upper-tringular matrix are the items on it's diagonal so they are $1,2$.
answered Aug 2 at 13:50
user 108128
18.8k41544
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up vote
1
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$$A~=~beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$$
By plugging in the eigenvalues and compute the determinate we get
$$A~=~beginvmatrix1-lambda&1&2\0&1-lambda&3\0&0&2-lambdaendvmatrix=(1-lambda)cdotbeginvmatrix1-lambda&3\0&2-lambdaendvmatrix+1cdotbeginvmatrix3&0\2-lambda&0endvmatrix+2cdotbeginvmatrix0&1-lambda\0&0endvmatrix=(1-lambda)^2(2-lambda)+1cdot(0)+2cdot(0)$$
Where two of the three summands vanish because a product with $0$ is always $0$ aswell and so only $(1-lambda)^2(2-lambda)$ remains which you have to set equal to zero. This term is already in the form where you can just see the roots, which are $lambda_1,2=1,lambda_3=2$.
answered Aug 2 at 13:49
mrtaurho
619117
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up vote
1
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If $A$ is a triangular matrix then the determinant of $A$ is $det(A) = prod_i=1^n a_ii$
It follows that
$$ det(A-lambda I) = prod_i=1^n a_ii - lambda$$
then if
$$ A = beginbmatrix 1& 1 & 2 \ 0 & 1 & 3 \ 0 & 0 & 2endbmatrix $$
$$det(A- lambda I) = prod_i=1^n a_ii - lambda = (1-lambda)(1-lambda)(2-lambda) $$
Then we know
$$lambda_1,lambda_2 = 1, lambda_3 = 2 $$
To determine the eigenvectors we solve the linear equation $(A-lambda_jI)x =mathbb0 $ for $ j=1,2,3$ using $lambda_3 =2$ we have
$$(A-I)x = beginbmatrix -x_1+ x_2 + 2x_3 \ -x_2 + 3x_3 \ 0 x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ -x_2 + x_3 = 0 implies x_2 = 3x_3$$
$$ -x_1 + x_2 +2x_3 implies -x_1 + 3x_3 +2x_2 =0 implies x_1 = 5x_3 $$
$$ u_3 = beginbmatrix 1 \ frac35 \ frac15endbmatrix$$
$lambda_1 = 1 $
$$ | u_3| = sqrt1+ (frac35)^2 + (frac15)^2 $$
$$ | u_3| = sqrt1+ frac925 + frac125 $$
$$ | u_3| = sqrt1+ frac1025 $$
$$ | u_3| = sqrt frac3525 $$
$$ | u_3| = fracsqrt355 $$
$$ v_3 = fracu_3 u_3 = [frac1fracsqrt355 , fracfrac35fracsqrt355, fracfrac15fracsqrt355 ]$$
now $lambda_1 = 1 $
$$(A-I)x = beginbmatrix x_2 + 2x_3 \ x_2 + 3x_3 \ x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ x_3 = 0 implies x_2 + 3x_3 = x_2 + 0 = 0 implies x_2 = 0 $$
our assumption was $x_1 =1$
$$u_1 = beginbmatrix1 \ 0 \0 endbmatrix $$
$$u_2 = beginbmatrix-1 \ 0 \0 endbmatrix $$
testing this in computer
A = np.array([[1,1,2],[0,1,3],[0,0,2]])
vals, vecs = LA.eig(A)
test = np.sqrt(35)/5
test1 = [1/test, (3/5)/test, (1/5)/test]
Out[2]:
array([[ 1.00000000e+00, -1.00000000e+00, 8.45154255e-01],
[ 0.00000000e+00, 2.22044605e-16, 5.07092553e-01],
[ 0.00000000e+00, 0.00000000e+00, 1.69030851e-01]])
test1
Out[3]: [0.8451542547285166, 0.50709255283711, 0.16903085094570333]
answered Aug 2 at 13:54
RHowe
825715
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0
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There is actually just one genuine eigenvector for the eigenvalue $1.$ There are "generalized" eigenvectors, which lead to the Jordan Normal Form. For the middle column of my matrix $R,$ I chose from among the vectors $v$ for which $(A-I)^2 v = 0$ but $(A-I) v neq 0.$ Then the first column, a genuine eigenvector, is $u =(A-I) v $
$$
R =
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
$$
The Jordan form is $ R^-1 A R = J$ and has a special shape.
$$
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)=
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
$$
For applications such as finding $e^At$ it is the reverse identity that is important, $RJR^-1 = A,$
$$
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right) =
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
$$
answered Aug 2 at 18:09
Will Jagy
96.7k594195
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
From here
$$|A-lambda I|=beginvmatrix
1-lambda & 1 & 2\
0 & 1-lambda & 3\
0 & 0 & 2-lambdaendvmatrix=(1-lambda)^2(2-lambda)=0$$
we can see that for a up triangular matrix eigenvalues are the diagonal entries.
Now for each $lambda$ solve $(A-lambda I)x=0$ to find the corresponding eigenvectors.
answered Aug 2 at 13:45
gimusi
63.8k73480
add a comment |Â
up vote
3
down vote
From here
$$|A-lambda I|=beginvmatrix
1-lambda & 1 & 2\
0 & 1-lambda & 3\
0 & 0 & 2-lambdaendvmatrix=(1-lambda)^2(2-lambda)=0$$
we can see that for a up triangular matrix eigenvalues are the diagonal entries.
Now for each $lambda$ solve $(A-lambda I)x=0$ to find the corresponding eigenvectors.
answered Aug 2 at 13:45
gimusi
63.8k73480
add a comment |Â
up vote
3
down vote
up vote
3
down vote
From here
$$|A-lambda I|=beginvmatrix
1-lambda & 1 & 2\
0 & 1-lambda & 3\
0 & 0 & 2-lambdaendvmatrix=(1-lambda)^2(2-lambda)=0$$
we can see that for a up triangular matrix eigenvalues are the diagonal entries.
Now for each $lambda$ solve $(A-lambda I)x=0$ to find the corresponding eigenvectors.
answered Aug 2 at 13:45
gimusi
63.8k73480
From here
$$|A-lambda I|=beginvmatrix
1-lambda & 1 & 2\
0 & 1-lambda & 3\
0 & 0 & 2-lambdaendvmatrix=(1-lambda)^2(2-lambda)=0$$
we can see that for a up triangular matrix eigenvalues are the diagonal entries.
Now for each $lambda$ solve $(A-lambda I)x=0$ to find the corresponding eigenvectors.
answered Aug 2 at 13:45
gimusi
63.8k73480
answered Aug 2 at 13:45
gimusi
63.8k73480
answered Aug 2 at 13:45
gimusi
63.8k73480
answered Aug 2 at 13:45
gimusi
63.8k73480
63.8k73480
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint: All Eigenvalues of a upper-tringular matrix are the items on it's diagonal so they are $1,2$.
answered Aug 2 at 13:50
user 108128
18.8k41544
add a comment |Â
up vote
2
down vote
Hint: All Eigenvalues of a upper-tringular matrix are the items on it's diagonal so they are $1,2$.
answered Aug 2 at 13:50
user 108128
18.8k41544
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint: All Eigenvalues of a upper-tringular matrix are the items on it's diagonal so they are $1,2$.
answered Aug 2 at 13:50
user 108128
18.8k41544
Hint: All Eigenvalues of a upper-tringular matrix are the items on it's diagonal so they are $1,2$.
answered Aug 2 at 13:50
user 108128
18.8k41544
answered Aug 2 at 13:50
user 108128
18.8k41544
answered Aug 2 at 13:50
user 108128
18.8k41544
answered Aug 2 at 13:50
user 108128
18.8k41544
18.8k41544
add a comment |Â
add a comment |Â
up vote
1
down vote
$$A~=~beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$$
By plugging in the eigenvalues and compute the determinate we get
$$A~=~beginvmatrix1-lambda&1&2\0&1-lambda&3\0&0&2-lambdaendvmatrix=(1-lambda)cdotbeginvmatrix1-lambda&3\0&2-lambdaendvmatrix+1cdotbeginvmatrix3&0\2-lambda&0endvmatrix+2cdotbeginvmatrix0&1-lambda\0&0endvmatrix=(1-lambda)^2(2-lambda)+1cdot(0)+2cdot(0)$$
Where two of the three summands vanish because a product with $0$ is always $0$ aswell and so only $(1-lambda)^2(2-lambda)$ remains which you have to set equal to zero. This term is already in the form where you can just see the roots, which are $lambda_1,2=1,lambda_3=2$.
answered Aug 2 at 13:49
mrtaurho
619117
add a comment |Â
up vote
1
down vote
$$A~=~beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$$
By plugging in the eigenvalues and compute the determinate we get
$$A~=~beginvmatrix1-lambda&1&2\0&1-lambda&3\0&0&2-lambdaendvmatrix=(1-lambda)cdotbeginvmatrix1-lambda&3\0&2-lambdaendvmatrix+1cdotbeginvmatrix3&0\2-lambda&0endvmatrix+2cdotbeginvmatrix0&1-lambda\0&0endvmatrix=(1-lambda)^2(2-lambda)+1cdot(0)+2cdot(0)$$
Where two of the three summands vanish because a product with $0$ is always $0$ aswell and so only $(1-lambda)^2(2-lambda)$ remains which you have to set equal to zero. This term is already in the form where you can just see the roots, which are $lambda_1,2=1,lambda_3=2$.
answered Aug 2 at 13:49
mrtaurho
619117
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$A~=~beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$$
By plugging in the eigenvalues and compute the determinate we get
$$A~=~beginvmatrix1-lambda&1&2\0&1-lambda&3\0&0&2-lambdaendvmatrix=(1-lambda)cdotbeginvmatrix1-lambda&3\0&2-lambdaendvmatrix+1cdotbeginvmatrix3&0\2-lambda&0endvmatrix+2cdotbeginvmatrix0&1-lambda\0&0endvmatrix=(1-lambda)^2(2-lambda)+1cdot(0)+2cdot(0)$$
Where two of the three summands vanish because a product with $0$ is always $0$ aswell and so only $(1-lambda)^2(2-lambda)$ remains which you have to set equal to zero. This term is already in the form where you can just see the roots, which are $lambda_1,2=1,lambda_3=2$.
answered Aug 2 at 13:49
mrtaurho
619117
$$A~=~beginpmatrix1&1&2\0&1&3\0&0&2endpmatrix$$
By plugging in the eigenvalues and compute the determinate we get
$$A~=~beginvmatrix1-lambda&1&2\0&1-lambda&3\0&0&2-lambdaendvmatrix=(1-lambda)cdotbeginvmatrix1-lambda&3\0&2-lambdaendvmatrix+1cdotbeginvmatrix3&0\2-lambda&0endvmatrix+2cdotbeginvmatrix0&1-lambda\0&0endvmatrix=(1-lambda)^2(2-lambda)+1cdot(0)+2cdot(0)$$
Where two of the three summands vanish because a product with $0$ is always $0$ aswell and so only $(1-lambda)^2(2-lambda)$ remains which you have to set equal to zero. This term is already in the form where you can just see the roots, which are $lambda_1,2=1,lambda_3=2$.
answered Aug 2 at 13:49
mrtaurho
619117
answered Aug 2 at 13:49
mrtaurho
619117
answered Aug 2 at 13:49
mrtaurho
619117
answered Aug 2 at 13:49
mrtaurho
619117
619117
add a comment |Â
add a comment |Â
up vote
1
down vote
If $A$ is a triangular matrix then the determinant of $A$ is $det(A) = prod_i=1^n a_ii$
It follows that
$$ det(A-lambda I) = prod_i=1^n a_ii - lambda$$
then if
$$ A = beginbmatrix 1& 1 & 2 \ 0 & 1 & 3 \ 0 & 0 & 2endbmatrix $$
$$det(A- lambda I) = prod_i=1^n a_ii - lambda = (1-lambda)(1-lambda)(2-lambda) $$
Then we know
$$lambda_1,lambda_2 = 1, lambda_3 = 2 $$
To determine the eigenvectors we solve the linear equation $(A-lambda_jI)x =mathbb0 $ for $ j=1,2,3$ using $lambda_3 =2$ we have
$$(A-I)x = beginbmatrix -x_1+ x_2 + 2x_3 \ -x_2 + 3x_3 \ 0 x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ -x_2 + x_3 = 0 implies x_2 = 3x_3$$
$$ -x_1 + x_2 +2x_3 implies -x_1 + 3x_3 +2x_2 =0 implies x_1 = 5x_3 $$
$$ u_3 = beginbmatrix 1 \ frac35 \ frac15endbmatrix$$
$lambda_1 = 1 $
$$ | u_3| = sqrt1+ (frac35)^2 + (frac15)^2 $$
$$ | u_3| = sqrt1+ frac925 + frac125 $$
$$ | u_3| = sqrt1+ frac1025 $$
$$ | u_3| = sqrt frac3525 $$
$$ | u_3| = fracsqrt355 $$
$$ v_3 = fracu_3 u_3 = [frac1fracsqrt355 , fracfrac35fracsqrt355, fracfrac15fracsqrt355 ]$$
now $lambda_1 = 1 $
$$(A-I)x = beginbmatrix x_2 + 2x_3 \ x_2 + 3x_3 \ x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ x_3 = 0 implies x_2 + 3x_3 = x_2 + 0 = 0 implies x_2 = 0 $$
our assumption was $x_1 =1$
$$u_1 = beginbmatrix1 \ 0 \0 endbmatrix $$
$$u_2 = beginbmatrix-1 \ 0 \0 endbmatrix $$
testing this in computer
A = np.array([[1,1,2],[0,1,3],[0,0,2]])
vals, vecs = LA.eig(A)
test = np.sqrt(35)/5
test1 = [1/test, (3/5)/test, (1/5)/test]
Out[2]:
array([[ 1.00000000e+00, -1.00000000e+00, 8.45154255e-01],
[ 0.00000000e+00, 2.22044605e-16, 5.07092553e-01],
[ 0.00000000e+00, 0.00000000e+00, 1.69030851e-01]])
test1
Out[3]: [0.8451542547285166, 0.50709255283711, 0.16903085094570333]
answered Aug 2 at 13:54
RHowe
825715
add a comment |Â
up vote
1
down vote
If $A$ is a triangular matrix then the determinant of $A$ is $det(A) = prod_i=1^n a_ii$
It follows that
$$ det(A-lambda I) = prod_i=1^n a_ii - lambda$$
then if
$$ A = beginbmatrix 1& 1 & 2 \ 0 & 1 & 3 \ 0 & 0 & 2endbmatrix $$
$$det(A- lambda I) = prod_i=1^n a_ii - lambda = (1-lambda)(1-lambda)(2-lambda) $$
Then we know
$$lambda_1,lambda_2 = 1, lambda_3 = 2 $$
To determine the eigenvectors we solve the linear equation $(A-lambda_jI)x =mathbb0 $ for $ j=1,2,3$ using $lambda_3 =2$ we have
$$(A-I)x = beginbmatrix -x_1+ x_2 + 2x_3 \ -x_2 + 3x_3 \ 0 x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ -x_2 + x_3 = 0 implies x_2 = 3x_3$$
$$ -x_1 + x_2 +2x_3 implies -x_1 + 3x_3 +2x_2 =0 implies x_1 = 5x_3 $$
$$ u_3 = beginbmatrix 1 \ frac35 \ frac15endbmatrix$$
$lambda_1 = 1 $
$$ | u_3| = sqrt1+ (frac35)^2 + (frac15)^2 $$
$$ | u_3| = sqrt1+ frac925 + frac125 $$
$$ | u_3| = sqrt1+ frac1025 $$
$$ | u_3| = sqrt frac3525 $$
$$ | u_3| = fracsqrt355 $$
$$ v_3 = fracu_3 u_3 = [frac1fracsqrt355 , fracfrac35fracsqrt355, fracfrac15fracsqrt355 ]$$
now $lambda_1 = 1 $
$$(A-I)x = beginbmatrix x_2 + 2x_3 \ x_2 + 3x_3 \ x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ x_3 = 0 implies x_2 + 3x_3 = x_2 + 0 = 0 implies x_2 = 0 $$
our assumption was $x_1 =1$
$$u_1 = beginbmatrix1 \ 0 \0 endbmatrix $$
$$u_2 = beginbmatrix-1 \ 0 \0 endbmatrix $$
testing this in computer
A = np.array([[1,1,2],[0,1,3],[0,0,2]])
vals, vecs = LA.eig(A)
test = np.sqrt(35)/5
test1 = [1/test, (3/5)/test, (1/5)/test]
Out[2]:
array([[ 1.00000000e+00, -1.00000000e+00, 8.45154255e-01],
[ 0.00000000e+00, 2.22044605e-16, 5.07092553e-01],
[ 0.00000000e+00, 0.00000000e+00, 1.69030851e-01]])
test1
Out[3]: [0.8451542547285166, 0.50709255283711, 0.16903085094570333]
answered Aug 2 at 13:54
RHowe
825715
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $A$ is a triangular matrix then the determinant of $A$ is $det(A) = prod_i=1^n a_ii$
It follows that
$$ det(A-lambda I) = prod_i=1^n a_ii - lambda$$
then if
$$ A = beginbmatrix 1& 1 & 2 \ 0 & 1 & 3 \ 0 & 0 & 2endbmatrix $$
$$det(A- lambda I) = prod_i=1^n a_ii - lambda = (1-lambda)(1-lambda)(2-lambda) $$
Then we know
$$lambda_1,lambda_2 = 1, lambda_3 = 2 $$
To determine the eigenvectors we solve the linear equation $(A-lambda_jI)x =mathbb0 $ for $ j=1,2,3$ using $lambda_3 =2$ we have
$$(A-I)x = beginbmatrix -x_1+ x_2 + 2x_3 \ -x_2 + 3x_3 \ 0 x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ -x_2 + x_3 = 0 implies x_2 = 3x_3$$
$$ -x_1 + x_2 +2x_3 implies -x_1 + 3x_3 +2x_2 =0 implies x_1 = 5x_3 $$
$$ u_3 = beginbmatrix 1 \ frac35 \ frac15endbmatrix$$
$lambda_1 = 1 $
$$ | u_3| = sqrt1+ (frac35)^2 + (frac15)^2 $$
$$ | u_3| = sqrt1+ frac925 + frac125 $$
$$ | u_3| = sqrt1+ frac1025 $$
$$ | u_3| = sqrt frac3525 $$
$$ | u_3| = fracsqrt355 $$
$$ v_3 = fracu_3 u_3 = [frac1fracsqrt355 , fracfrac35fracsqrt355, fracfrac15fracsqrt355 ]$$
now $lambda_1 = 1 $
$$(A-I)x = beginbmatrix x_2 + 2x_3 \ x_2 + 3x_3 \ x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ x_3 = 0 implies x_2 + 3x_3 = x_2 + 0 = 0 implies x_2 = 0 $$
our assumption was $x_1 =1$
$$u_1 = beginbmatrix1 \ 0 \0 endbmatrix $$
$$u_2 = beginbmatrix-1 \ 0 \0 endbmatrix $$
testing this in computer
A = np.array([[1,1,2],[0,1,3],[0,0,2]])
vals, vecs = LA.eig(A)
test = np.sqrt(35)/5
test1 = [1/test, (3/5)/test, (1/5)/test]
Out[2]:
array([[ 1.00000000e+00, -1.00000000e+00, 8.45154255e-01],
[ 0.00000000e+00, 2.22044605e-16, 5.07092553e-01],
[ 0.00000000e+00, 0.00000000e+00, 1.69030851e-01]])
test1
Out[3]: [0.8451542547285166, 0.50709255283711, 0.16903085094570333]
answered Aug 2 at 13:54
RHowe
825715
If $A$ is a triangular matrix then the determinant of $A$ is $det(A) = prod_i=1^n a_ii$
It follows that
$$ det(A-lambda I) = prod_i=1^n a_ii - lambda$$
then if
$$ A = beginbmatrix 1& 1 & 2 \ 0 & 1 & 3 \ 0 & 0 & 2endbmatrix $$
$$det(A- lambda I) = prod_i=1^n a_ii - lambda = (1-lambda)(1-lambda)(2-lambda) $$
Then we know
$$lambda_1,lambda_2 = 1, lambda_3 = 2 $$
To determine the eigenvectors we solve the linear equation $(A-lambda_jI)x =mathbb0 $ for $ j=1,2,3$ using $lambda_3 =2$ we have
$$(A-I)x = beginbmatrix -x_1+ x_2 + 2x_3 \ -x_2 + 3x_3 \ 0 x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ -x_2 + x_3 = 0 implies x_2 = 3x_3$$
$$ -x_1 + x_2 +2x_3 implies -x_1 + 3x_3 +2x_2 =0 implies x_1 = 5x_3 $$
$$ u_3 = beginbmatrix 1 \ frac35 \ frac15endbmatrix$$
$lambda_1 = 1 $
$$ | u_3| = sqrt1+ (frac35)^2 + (frac15)^2 $$
$$ | u_3| = sqrt1+ frac925 + frac125 $$
$$ | u_3| = sqrt1+ frac1025 $$
$$ | u_3| = sqrt frac3525 $$
$$ | u_3| = fracsqrt355 $$
$$ v_3 = fracu_3 u_3 = [frac1fracsqrt355 , fracfrac35fracsqrt355, fracfrac15fracsqrt355 ]$$
now $lambda_1 = 1 $
$$(A-I)x = beginbmatrix x_2 + 2x_3 \ x_2 + 3x_3 \ x_3 endbmatrix = beginbmatrix 0\ 0\ 0 endbmatrix $$
$$ x_3 = 0 implies x_2 + 3x_3 = x_2 + 0 = 0 implies x_2 = 0 $$
our assumption was $x_1 =1$
$$u_1 = beginbmatrix1 \ 0 \0 endbmatrix $$
$$u_2 = beginbmatrix-1 \ 0 \0 endbmatrix $$
testing this in computer
A = np.array([[1,1,2],[0,1,3],[0,0,2]])
vals, vecs = LA.eig(A)
test = np.sqrt(35)/5
test1 = [1/test, (3/5)/test, (1/5)/test]
Out[2]:
array([[ 1.00000000e+00, -1.00000000e+00, 8.45154255e-01],
[ 0.00000000e+00, 2.22044605e-16, 5.07092553e-01],
[ 0.00000000e+00, 0.00000000e+00, 1.69030851e-01]])
test1
Out[3]: [0.8451542547285166, 0.50709255283711, 0.16903085094570333]
answered Aug 2 at 13:54
RHowe
825715
edited Aug 3 at 18:33
answered Aug 2 at 13:54
RHowe
825715
answered Aug 2 at 13:54
RHowe
825715
answered Aug 2 at 13:54
RHowe
825715
825715
add a comment |Â
add a comment |Â
up vote
0
down vote
There is actually just one genuine eigenvector for the eigenvalue $1.$ There are "generalized" eigenvectors, which lead to the Jordan Normal Form. For the middle column of my matrix $R,$ I chose from among the vectors $v$ for which $(A-I)^2 v = 0$ but $(A-I) v neq 0.$ Then the first column, a genuine eigenvector, is $u =(A-I) v $
$$
R =
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
$$
The Jordan form is $ R^-1 A R = J$ and has a special shape.
$$
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)=
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
$$
For applications such as finding $e^At$ it is the reverse identity that is important, $RJR^-1 = A,$
$$
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right) =
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
$$
answered Aug 2 at 18:09
Will Jagy
96.7k594195
add a comment |Â
up vote
0
down vote
There is actually just one genuine eigenvector for the eigenvalue $1.$ There are "generalized" eigenvectors, which lead to the Jordan Normal Form. For the middle column of my matrix $R,$ I chose from among the vectors $v$ for which $(A-I)^2 v = 0$ but $(A-I) v neq 0.$ Then the first column, a genuine eigenvector, is $u =(A-I) v $
$$
R =
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
$$
The Jordan form is $ R^-1 A R = J$ and has a special shape.
$$
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)=
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
$$
For applications such as finding $e^At$ it is the reverse identity that is important, $RJR^-1 = A,$
$$
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right) =
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
$$
answered Aug 2 at 18:09
Will Jagy
96.7k594195
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There is actually just one genuine eigenvector for the eigenvalue $1.$ There are "generalized" eigenvectors, which lead to the Jordan Normal Form. For the middle column of my matrix $R,$ I chose from among the vectors $v$ for which $(A-I)^2 v = 0$ but $(A-I) v neq 0.$ Then the first column, a genuine eigenvector, is $u =(A-I) v $
$$
R =
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
$$
The Jordan form is $ R^-1 A R = J$ and has a special shape.
$$
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)=
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
$$
For applications such as finding $e^At$ it is the reverse identity that is important, $RJR^-1 = A,$
$$
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right) =
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
$$
answered Aug 2 at 18:09
Will Jagy
96.7k594195
There is actually just one genuine eigenvector for the eigenvalue $1.$ There are "generalized" eigenvectors, which lead to the Jordan Normal Form. For the middle column of my matrix $R,$ I chose from among the vectors $v$ for which $(A-I)^2 v = 0$ but $(A-I) v neq 0.$ Then the first column, a genuine eigenvector, is $u =(A-I) v $
$$
R =
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
$$
The Jordan form is $ R^-1 A R = J$ and has a special shape.
$$
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)=
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
$$
For applications such as finding $e^At$ it is the reverse identity that is important, $RJR^-1 = A,$
$$
left(
beginarrayccc
1 & 0 & 5 \
0 & 1 & 3 \
0 & 0 & 1
endarray
right)
left(
beginarrayccc
1 & 1 & 0 \
0 & 1 & 0 \
0 & 0 & 2
endarray
right)
left(
beginarrayccc
1 & 0 & -5 \
0 & 1 & -3 \
0 & 0 & 1
endarray
right) =
left(
beginarrayccc
1 & 1 & 2 \
0 & 1 & 3 \
0 & 0 & 2
endarray
right)
$$
answered Aug 2 at 18:09
Will Jagy
96.7k594195
answered Aug 2 at 18:09
Will Jagy
96.7k594195
answered Aug 2 at 18:09
Will Jagy
96.7k594195
answered Aug 2 at 18:09
Will Jagy
96.7k594195
96.7k594195
add a comment |Â
add a comment |Â
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1
$1.[0.(lambda-2)]=0$ simply
– Isham
Aug 2 at 13:50
I found the eigenvectors as well
– RHowe
Aug 2 at 20:57