How to find Eigenvalue with Characteristic Equation?
up vote
0
down vote
favorite
When my book explains using the characteristic equation to find eigenvalues, it gives this example.
Find the eigenvalues and eigenvectors of
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$
= $(lambda - 2)^3$
It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)
linear-algebra
asked Dec 4 at 19:37
Evan Kim
758
add a comment |
up vote
0
down vote
favorite
When my book explains using the characteristic equation to find eigenvalues, it gives this example.
Find the eigenvalues and eigenvectors of
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$
= $(lambda - 2)^3$
It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)
linear-algebra
asked Dec 4 at 19:37
Evan Kim
758
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
– Federico
Dec 4 at 19:56
fixed, thank you
– Evan Kim
Dec 4 at 19:58
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When my book explains using the characteristic equation to find eigenvalues, it gives this example.
Find the eigenvalues and eigenvectors of
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$
= $(lambda - 2)^3$
It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)
linear-algebra
asked Dec 4 at 19:37
Evan Kim
758
When my book explains using the characteristic equation to find eigenvalues, it gives this example.
Find the eigenvalues and eigenvectors of
A =
$begin{bmatrix}
2 & 1 & 0 \
0 & 2 & 0 \
0 & 0 & 2 \
end{bmatrix}$
$|lambda I - A| =
begin{bmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{bmatrix}$
= $(lambda - 2)^3$
It doesn't show any work for as how it got to $(lambda - 2)^3$. Can someone fill me in to how the book gets this result? (using quadratic equation?)
linear-algebra
linear-algebra
asked Dec 4 at 19:37
Evan Kim
758
asked Dec 4 at 19:37
Evan Kim
758
edited Dec 4 at 19:58
asked Dec 4 at 19:37
Evan Kim
758
asked Dec 4 at 19:37
Evan Kim
758
asked Dec 4 at 19:37
Evan Kim
758
758
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
– Federico
Dec 4 at 19:56
fixed, thank you
– Evan Kim
Dec 4 at 19:58
add a comment |
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
– Federico
Dec 4 at 19:56
fixed, thank you
– Evan Kim
Dec 4 at 19:58
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
– Federico
Dec 4 at 19:56
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
– Federico
Dec 4 at 19:56
fixed, thank you
– Evan Kim
Dec 4 at 19:58
fixed, thank you
– Evan Kim
Dec 4 at 19:58
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.
Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.
answered Dec 4 at 19:41
Federico
4,203512
1
So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
– Evan Kim
Dec 4 at 19:52
You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
– Federico
Dec 4 at 19:53
@Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
– user376343
Dec 4 at 19:54
add a comment |
up vote
1
down vote
A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :
$$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$
If you miss on that, calculate it the explicit way, by :
$$|lambda I - A| =
begin{vmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$
$$Rightarrow$$
$$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$
The eigenvalues then, would simply be :
$$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$
answered Dec 4 at 19:44
Rebellos
13.7k31243
Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
– Evan Kim
Dec 4 at 19:59
@EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
– Rebellos
Dec 4 at 20:03
add a comment |
up vote
0
down vote
In fact, your matrix is special. it makes the computation easier.
Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
$$(x,y,z)ne (0,0,0)$$
and
$$2x+y=kx$$
$$2y=ky$$
$$2z=kz$$
but
$$kne 2implies z=y=x=0$$
which is in contradiction with the assumption.
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
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
up vote
1
down vote
The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.
Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.
answered Dec 4 at 19:41
Federico
4,203512
1
So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
– Evan Kim
Dec 4 at 19:52
You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
– Federico
Dec 4 at 19:53
@Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
– user376343
Dec 4 at 19:54
add a comment |
up vote
1
down vote
The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.
Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.
answered Dec 4 at 19:41
Federico
4,203512
1
So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
– Evan Kim
Dec 4 at 19:52
You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
– Federico
Dec 4 at 19:53
@Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
– user376343
Dec 4 at 19:54
add a comment |
up vote
1
down vote
up vote
1
down vote
The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.
Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.
answered Dec 4 at 19:41
Federico
4,203512
The matrix $lambda I-A$ is upper triangular, so the determinant is the product of the entries on the diagonal.
Anyway, you should be able to compute the determinant of a generic matrix, even when it isn't triangular.
answered Dec 4 at 19:41
Federico
4,203512
edited Dec 4 at 19:54
answered Dec 4 at 19:41
Federico
4,203512
answered Dec 4 at 19:41
Federico
4,203512
answered Dec 4 at 19:41
Federico
4,203512
4,203512
1
So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
– Evan Kim
Dec 4 at 19:52
You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
– Federico
Dec 4 at 19:53
@Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
– user376343
Dec 4 at 19:54
add a comment |
1
So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
– Evan Kim
Dec 4 at 19:52
You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
– Federico
Dec 4 at 19:53
@Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
– user376343
Dec 4 at 19:54
1
1
So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
– Evan Kim
Dec 4 at 19:52
So in a non-triangular matrix, you just compute the determinant of the matrix and that will give you the roots?
– Evan Kim
Dec 4 at 19:52
You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
– Federico
Dec 4 at 19:53
You compute the determinant of $lambda I-A$ and that will give you the characteristic polynomial $P_A(lambda)$. The eigenvalues of $A$ are the roots of $P_A$.
– Federico
Dec 4 at 19:53
@Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
– user376343
Dec 4 at 19:54
@Evan Yes, it is a determinant. You wrote wrongly the matrix instead, it seems that it was confusing.
– user376343
Dec 4 at 19:54
add a comment |
up vote
1
down vote
A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :
$$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$
If you miss on that, calculate it the explicit way, by :
$$|lambda I - A| =
begin{vmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$
$$Rightarrow$$
$$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$
The eigenvalues then, would simply be :
$$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$
answered Dec 4 at 19:44
Rebellos
13.7k31243
Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
– Evan Kim
Dec 4 at 19:59
@EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
– Rebellos
Dec 4 at 20:03
add a comment |
up vote
1
down vote
A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :
$$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$
If you miss on that, calculate it the explicit way, by :
$$|lambda I - A| =
begin{vmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$
$$Rightarrow$$
$$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$
The eigenvalues then, would simply be :
$$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$
answered Dec 4 at 19:44
Rebellos
13.7k31243
Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
– Evan Kim
Dec 4 at 19:59
@EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
– Rebellos
Dec 4 at 20:03
add a comment |
up vote
1
down vote
up vote
1
down vote
A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :
$$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$
If you miss on that, calculate it the explicit way, by :
$$|lambda I - A| =
begin{vmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$
$$Rightarrow$$
$$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$
The eigenvalues then, would simply be :
$$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$
answered Dec 4 at 19:44
Rebellos
13.7k31243
A fast approach hint : The matrix given is in upper triangular form, thus the determinant is simply the product of the diagonal elements. Thus :
$$|lambda I - A | = prod_{n=1}^3 text{diag}_i(lambda I-A) = (lambda-2)^3$$
If you miss on that, calculate it the explicit way, by :
$$|lambda I - A| =
begin{vmatrix}
lambda - 2 & -1 & 0 \
0 & lambda - 2 & 0 \
0 & 0 & lambda - 2 \
end{vmatrix} = (λ-2)cdotbegin{vmatrix}λ-2 & 0 \ 0 & λ-2end{vmatrix} + 1cdot begin{vmatrix}0 & 0 \ 0 & λ-2end{vmatrix} + 0 cdot begin{vmatrix} 0 & λ-2 \ 0 & 0end{vmatrix}$$
$$Rightarrow$$
$$|lambda I - A| = (λ-2) cdot (λ-2)^2 + 0 + 0 = (λ-2)^3$$
The eigenvalues then, would simply be :
$$|lambda I - A| = 0 Rightarrow (lambda-2)^3 = 0 Leftrightarrow lambda = 2, ; text{with multiplicity of 3}$$
answered Dec 4 at 19:44
Rebellos
13.7k31243
edited Dec 4 at 19:59
answered Dec 4 at 19:44
Rebellos
13.7k31243
answered Dec 4 at 19:44
Rebellos
13.7k31243
answered Dec 4 at 19:44
Rebellos
13.7k31243
13.7k31243
Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
– Evan Kim
Dec 4 at 19:59
@EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
– Rebellos
Dec 4 at 20:03
add a comment |
Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
– Evan Kim
Dec 4 at 19:59
@EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
– Rebellos
Dec 4 at 20:03
Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
– Evan Kim
Dec 4 at 19:59
Would a valid strategy to find eigenvalues be to convert a non-triangular matrix into a triangular matrix to make finding the determinant easier?
– Evan Kim
Dec 4 at 19:59
@EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
– Rebellos
Dec 4 at 20:03
@EvanKim Well, that really depends on the given problem. If, for example, you have a $4 times 4$ or bigger dimension matrix, then if it's properly fixed so that it can be converted to upper triangular, it may be an easier way out then determinants since it takes a lot of time to calculate them explicitly. Other than that, on $2 times 2$ or $3 times 3$ matrices, the procedure to calculate the determinant is really fast anyway, so it wouldn't make much difference, except in the case that the given matrix is already in upper or lower triangular form (like your case).
– Rebellos
Dec 4 at 20:03
add a comment |
up vote
0
down vote
In fact, your matrix is special. it makes the computation easier.
Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
$$(x,y,z)ne (0,0,0)$$
and
$$2x+y=kx$$
$$2y=ky$$
$$2z=kz$$
but
$$kne 2implies z=y=x=0$$
which is in contradiction with the assumption.
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
add a comment |
up vote
0
down vote
In fact, your matrix is special. it makes the computation easier.
Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
$$(x,y,z)ne (0,0,0)$$
and
$$2x+y=kx$$
$$2y=ky$$
$$2z=kz$$
but
$$kne 2implies z=y=x=0$$
which is in contradiction with the assumption.
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
add a comment |
up vote
0
down vote
up vote
0
down vote
In fact, your matrix is special. it makes the computation easier.
Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
$$(x,y,z)ne (0,0,0)$$
and
$$2x+y=kx$$
$$2y=ky$$
$$2z=kz$$
but
$$kne 2implies z=y=x=0$$
which is in contradiction with the assumption.
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
In fact, your matrix is special. it makes the computation easier.
Assume $kne 2$ is an other eigenvalue. then there will exist $x,y,z$ satisfying
$$(x,y,z)ne (0,0,0)$$
and
$$2x+y=kx$$
$$2y=ky$$
$$2z=kz$$
but
$$kne 2implies z=y=x=0$$
which is in contradiction with the assumption.
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
answered Dec 4 at 19:46
hamam_Abdallah
37.7k21634
37.7k21634
add a comment |
add a comment |
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Required, but never shown
Also note that in $lambda I-A$ the entry $(1,2)$ should be $-1$, not $1$
– Federico
Dec 4 at 19:56
fixed, thank you
– Evan Kim
Dec 4 at 19:58