Matrix Eigenvalue formula $|lambda I-A |$ or $|A-lambda I|$?
my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.
Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?
Thanks
matrices eigenvalues-eigenvectors
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
add a comment |
my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.
Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?
Thanks
matrices eigenvalues-eigenvectors
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
2
That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44
$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46
okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49
4
@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50
add a comment |
my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.
Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?
Thanks
matrices eigenvalues-eigenvectors
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
my lecturer wrote a solution for our exam, and he wrote the eigenvalue formula as $|lambda I-A |$ instead of $|A-lambda I|$. What I understand is we use the latter to get the eigenvalue of a matrix.
Is this a typo or it doesn't matter whether it's $|A-lambda I|$ or $|lambda I-A |$?
Thanks
matrices eigenvalues-eigenvectors
matrices eigenvalues-eigenvectors
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
asked Nov 29 '18 at 22:38
JIM BOYJIM BOY
356
356
2
That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44
$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46
okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49
4
@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50
add a comment |
2
That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44
$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46
okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49
4
@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50
2
2
That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44
That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44
$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46
$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46
okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46
okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49
4
4
@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50
@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50
add a comment |
3 Answers
3
active
oldest
votes
Both are actually equivalent!
Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:
$$Avec{x} = lambda vec{x}$$
Then,
$$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$
Then we try to find $lambda$ such that $det(A - lambda I) = 0$.
But wait! We can do this a different way, as:
$$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$
and thus we seek $lambda$ such that $det(lambda I - A) = 0$
Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
add a comment |
If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:50
They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
– José Carlos Santos
Nov 29 '18 at 22:51
Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
– Eevee Trainer
Nov 29 '18 at 22:52
add a comment |
Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
add a comment |
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3 Answers
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3 Answers
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Both are actually equivalent!
Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:
$$Avec{x} = lambda vec{x}$$
Then,
$$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$
Then we try to find $lambda$ such that $det(A - lambda I) = 0$.
But wait! We can do this a different way, as:
$$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$
and thus we seek $lambda$ such that $det(lambda I - A) = 0$
Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
add a comment |
Both are actually equivalent!
Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:
$$Avec{x} = lambda vec{x}$$
Then,
$$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$
Then we try to find $lambda$ such that $det(A - lambda I) = 0$.
But wait! We can do this a different way, as:
$$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$
and thus we seek $lambda$ such that $det(lambda I - A) = 0$
Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
add a comment |
Both are actually equivalent!
Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:
$$Avec{x} = lambda vec{x}$$
Then,
$$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$
Then we try to find $lambda$ such that $det(A - lambda I) = 0$.
But wait! We can do this a different way, as:
$$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$
and thus we seek $lambda$ such that $det(lambda I - A) = 0$
Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
Both are actually equivalent!
Let us recall how we derive the notion of eigenvalues and such. That is, for a square matrix $A$ and eigenvalues $lambda$ of $A$, we have:
$$Avec{x} = lambda vec{x}$$
Then,
$$Avec{x} - lambda vec{x} = 0 ;;; Rightarrow ;;; (A-lambda I) vec{x} = 0$$
Then we try to find $lambda$ such that $det(A - lambda I) = 0$.
But wait! We can do this a different way, as:
$$Avec{x} = lambda vec{x} ;;; Rightarrow ;;; 0 = lambda vec{x} - A vec{x} ;;; Rightarrow ;;; 0 = (lambda I - A) vec{x}$$
and thus we seek $lambda$ such that $det(lambda I - A) = 0$
Thus, as a result, you can use either equation - $det(lambda I - A) = 0$ or $det(A - lambda I) = 0$. Which you use is a matter of personal taste; I was actually taught the former but found the latter more intuitive.
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
answered Nov 29 '18 at 22:45
Eevee TrainerEevee Trainer
5,3811836
5,3811836
add a comment |
add a comment |
If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:50
They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
– José Carlos Santos
Nov 29 '18 at 22:51
Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
– Eevee Trainer
Nov 29 '18 at 22:52
add a comment |
If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:50
They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
– José Carlos Santos
Nov 29 '18 at 22:51
Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
– Eevee Trainer
Nov 29 '18 at 22:52
add a comment |
If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
If $A$ is a $ntimes n$ matrix, then the polynomials $det(A-lambdaoperatorname{Id})$ and $det(lambdaoperatorname{Id}-A)$ are equal if $n$ is even and symmetric otherwise. Therefore, they have the same zeros. So, as far as searching for eigenvalues is concerned, both provide the same information.
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
answered Nov 29 '18 at 22:48
José Carlos SantosJosé Carlos Santos
152k22123225
152k22123225
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:50
They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
– José Carlos Santos
Nov 29 '18 at 22:51
Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
– Eevee Trainer
Nov 29 '18 at 22:52
add a comment |
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:50
They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
– José Carlos Santos
Nov 29 '18 at 22:51
Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
– Eevee Trainer
Nov 29 '18 at 22:52
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:50
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A−B|$ will not be the same as $|B−A|$ right? There's no such a thing as $|A−B|=|B−A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:50
They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
– José Carlos Santos
Nov 29 '18 at 22:51
They are no equal, but they are symmetric: if $C$ is a $3times3$ matrix, $det(-C)=-det(C)$.
– José Carlos Santos
Nov 29 '18 at 22:51
Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
– Eevee Trainer
Nov 29 '18 at 22:52
Technically, there is a property that is somewhat related. Note: $$det(A-B) = det(-1 cdot (B-A)) = (-1)^n det(B-A)$$ where $n$ is the size of the matrices (i.e. $A,B$ are $n times n$ matrices). Notice what happens if $n$ is even. Generally, this property is stated for any scalar $k$: $$det(kA) = k^n det(A)$$
– Eevee Trainer
Nov 29 '18 at 22:52
add a comment |
Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
add a comment |
Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
add a comment |
Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
Just note that $det (A-lambda I) =(-1)^ndet (lambda I-A)$, so $lambda$ solves $ det (A-lambda I) =0$ if and only if it solves $det (lambda I-A)=0$, therefore you can calculate the eigenvalues of $A$ by solving $det (lambda I-A)=0$ or $det (A-lambda I)=0$. The solutions are the same.
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
answered Nov 29 '18 at 22:54
FJ.marsanFJ.marsan
414
414
add a comment |
add a comment |
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2
That's the same equation, simply multiplied by $(-1)^n$, but solutions are just the same. It's like $$x+3=0$$ and $$-x-3=0$$ they are equivalent equations.
– Crostul
Nov 29 '18 at 22:44
$|lambda I - A| = |A- lambda I|$.
– amWhy
Nov 29 '18 at 22:46
okay thanks, just realized that
– JIM BOY
Nov 29 '18 at 22:46
But if we have two 3x3 matrices, A & B, both have different values in their elements, the result of $|A-B|$ will not be the same as $|B-A|$ right? There's no such a thing as $|A-B|=|B-A|$, am I right?
– JIM BOY
Nov 29 '18 at 22:49
4
@amWhy Not for odd-dimensional matrices. Those are determinants, not absolute values.
– Arthur
Nov 29 '18 at 22:50