Cayley-Hamilton says that evaluating an endomorphism's characteristic polynomial over that endomorphism gives...
$begingroup$
Given an endomorphism $f$ of a vector space $V$, its characteristic polynomial, say $P(x)$, is defined as follows: $P(x) = det(f -xI)$, where $I$ is the identity endomorphism. It is well known that, when we evaluate over $f$, we get $P(f) = 0$, as a consequence of Cayley-Hamilton's theorem.
But, working with the expression directly, we get
$$P(f) = det(f - fI) = det(f - f) = det(0) = 0$$
even without Cayley-Hamilton. What's wrong about this?
linear-algebra alternative-proof cayley-hamilton
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
asked Jan 5 at 16:59
math_noobmath_noob
13
$endgroup$
add a comment |
$begingroup$
Given an endomorphism $f$ of a vector space $V$, its characteristic polynomial, say $P(x)$, is defined as follows: $P(x) = det(f -xI)$, where $I$ is the identity endomorphism. It is well known that, when we evaluate over $f$, we get $P(f) = 0$, as a consequence of Cayley-Hamilton's theorem.
But, working with the expression directly, we get
$$P(f) = det(f - fI) = det(f - f) = det(0) = 0$$
even without Cayley-Hamilton. What's wrong about this?
linear-algebra alternative-proof cayley-hamilton
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
asked Jan 5 at 16:59
math_noobmath_noob
13
$endgroup$
2
$begingroup$
see Wikipedia
$endgroup$
– 0x539
Jan 5 at 17:11
$begingroup$
Note: This argument is not too far from being correct(able) though. Working with matrices: The argument above is nearly all that is required to show that CH holds for an upper triangular matrix. The general case is obtained by extending the ground field to an algebraic closure of it. When the field is algebraically closed, all matrices are similar to upper triangular ones, and CH follows easily.
$endgroup$
– Ittay Weiss
Jan 5 at 18:21
$begingroup$
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
$endgroup$
– user1551
Jan 5 at 18:53
$begingroup$
@user1551 Sorry, I was perhaps unclear. When the characteristic polynomial splits, computing f(A) for an upper triangular matrix is just plugging in $A$ for $x$ in the decomposition of $f$ and computing. That fact that you get $0$ is easy. I see some resemblance to OP's faulty argument here: You just substitute $A$ for $x$, and the computation is nearly just evaluating at each eigenvalue.
$endgroup$
– Ittay Weiss
Jan 5 at 19:09
add a comment |
$begingroup$
Given an endomorphism $f$ of a vector space $V$, its characteristic polynomial, say $P(x)$, is defined as follows: $P(x) = det(f -xI)$, where $I$ is the identity endomorphism. It is well known that, when we evaluate over $f$, we get $P(f) = 0$, as a consequence of Cayley-Hamilton's theorem.
But, working with the expression directly, we get
$$P(f) = det(f - fI) = det(f - f) = det(0) = 0$$
even without Cayley-Hamilton. What's wrong about this?
linear-algebra alternative-proof cayley-hamilton
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
asked Jan 5 at 16:59
math_noobmath_noob
13
$endgroup$
Given an endomorphism $f$ of a vector space $V$, its characteristic polynomial, say $P(x)$, is defined as follows: $P(x) = det(f -xI)$, where $I$ is the identity endomorphism. It is well known that, when we evaluate over $f$, we get $P(f) = 0$, as a consequence of Cayley-Hamilton's theorem.
But, working with the expression directly, we get
$$P(f) = det(f - fI) = det(f - f) = det(0) = 0$$
even without Cayley-Hamilton. What's wrong about this?
linear-algebra alternative-proof cayley-hamilton
linear-algebra alternative-proof cayley-hamilton
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
asked Jan 5 at 16:59
math_noobmath_noob
13
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
asked Jan 5 at 16:59
math_noobmath_noob
13
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
edited Jan 5 at 18:11
José Carlos Santos
173k23133241
173k23133241
asked Jan 5 at 16:59
math_noobmath_noob
13
asked Jan 5 at 16:59
math_noobmath_noob
13
asked Jan 5 at 16:59
math_noobmath_noob
13
13
2
$begingroup$
see Wikipedia
$endgroup$
– 0x539
Jan 5 at 17:11
$begingroup$
Note: This argument is not too far from being correct(able) though. Working with matrices: The argument above is nearly all that is required to show that CH holds for an upper triangular matrix. The general case is obtained by extending the ground field to an algebraic closure of it. When the field is algebraically closed, all matrices are similar to upper triangular ones, and CH follows easily.
$endgroup$
– Ittay Weiss
Jan 5 at 18:21
$begingroup$
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
$endgroup$
– user1551
Jan 5 at 18:53
$begingroup$
@user1551 Sorry, I was perhaps unclear. When the characteristic polynomial splits, computing f(A) for an upper triangular matrix is just plugging in $A$ for $x$ in the decomposition of $f$ and computing. That fact that you get $0$ is easy. I see some resemblance to OP's faulty argument here: You just substitute $A$ for $x$, and the computation is nearly just evaluating at each eigenvalue.
$endgroup$
– Ittay Weiss
Jan 5 at 19:09
add a comment |
2
$begingroup$
see Wikipedia
$endgroup$
– 0x539
Jan 5 at 17:11
$begingroup$
Note: This argument is not too far from being correct(able) though. Working with matrices: The argument above is nearly all that is required to show that CH holds for an upper triangular matrix. The general case is obtained by extending the ground field to an algebraic closure of it. When the field is algebraically closed, all matrices are similar to upper triangular ones, and CH follows easily.
$endgroup$
– Ittay Weiss
Jan 5 at 18:21
$begingroup$
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
$endgroup$
– user1551
Jan 5 at 18:53
$begingroup$
@user1551 Sorry, I was perhaps unclear. When the characteristic polynomial splits, computing f(A) for an upper triangular matrix is just plugging in $A$ for $x$ in the decomposition of $f$ and computing. That fact that you get $0$ is easy. I see some resemblance to OP's faulty argument here: You just substitute $A$ for $x$, and the computation is nearly just evaluating at each eigenvalue.
$endgroup$
– Ittay Weiss
Jan 5 at 19:09
2
2
$begingroup$
see Wikipedia
$endgroup$
– 0x539
Jan 5 at 17:11
$begingroup$
see Wikipedia
$endgroup$
– 0x539
Jan 5 at 17:11
$begingroup$
Note: This argument is not too far from being correct(able) though. Working with matrices: The argument above is nearly all that is required to show that CH holds for an upper triangular matrix. The general case is obtained by extending the ground field to an algebraic closure of it. When the field is algebraically closed, all matrices are similar to upper triangular ones, and CH follows easily.
$endgroup$
– Ittay Weiss
Jan 5 at 18:21
$begingroup$
Note: This argument is not too far from being correct(able) though. Working with matrices: The argument above is nearly all that is required to show that CH holds for an upper triangular matrix. The general case is obtained by extending the ground field to an algebraic closure of it. When the field is algebraically closed, all matrices are similar to upper triangular ones, and CH follows easily.
$endgroup$
– Ittay Weiss
Jan 5 at 18:21
$begingroup$
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
$endgroup$
– user1551
Jan 5 at 18:53
$begingroup$
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
$endgroup$
– user1551
Jan 5 at 18:53
$begingroup$
@user1551 Sorry, I was perhaps unclear. When the characteristic polynomial splits, computing f(A) for an upper triangular matrix is just plugging in $A$ for $x$ in the decomposition of $f$ and computing. That fact that you get $0$ is easy. I see some resemblance to OP's faulty argument here: You just substitute $A$ for $x$, and the computation is nearly just evaluating at each eigenvalue.
$endgroup$
– Ittay Weiss
Jan 5 at 19:09
$begingroup$
@user1551 Sorry, I was perhaps unclear. When the characteristic polynomial splits, computing f(A) for an upper triangular matrix is just plugging in $A$ for $x$ in the decomposition of $f$ and computing. That fact that you get $0$ is easy. I see some resemblance to OP's faulty argument here: You just substitute $A$ for $x$, and the computation is nearly just evaluating at each eigenvalue.
$endgroup$
– Ittay Weiss
Jan 5 at 19:09
add a comment |
1 Answer
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$begingroup$
What's wrong is that $P(x)=det(f-xoperatorname{Id})$ is a polynomial. Yes, you can compute $P(x_0)$ for a number $x_0$ by computing $det(f-x_0operatorname{Id})$, but $f$ is not a number; it's an endomorphism.
Conseder, for instance, $Q(x)=operatorname{tr}(f-xoperatorname{Id})$. By your argument, $Q(f)=0$ too. But, in fact, since $Q(x)=operatorname{tr}(f)-nx$, you actually have $Q(f)=(1-n)operatorname{tr}(f)$.
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
A slight correction to this answer: it should be $Q(f) = tr(f) - nf$. @math_noob: If you substitute $f$ first then take the determinant, then the answer you get will be a scalar, but if you first take the determinant to get a polynomial in $x$, then substitute $f$, the answer is a matrix. So there is no reason to expect them to be the same.
$endgroup$
– Ted
Jan 5 at 17:42
$begingroup$
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
$endgroup$
– José Carlos Santos
Jan 5 at 18:10
add a comment |
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1 Answer
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$begingroup$
What's wrong is that $P(x)=det(f-xoperatorname{Id})$ is a polynomial. Yes, you can compute $P(x_0)$ for a number $x_0$ by computing $det(f-x_0operatorname{Id})$, but $f$ is not a number; it's an endomorphism.
Conseder, for instance, $Q(x)=operatorname{tr}(f-xoperatorname{Id})$. By your argument, $Q(f)=0$ too. But, in fact, since $Q(x)=operatorname{tr}(f)-nx$, you actually have $Q(f)=(1-n)operatorname{tr}(f)$.
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
A slight correction to this answer: it should be $Q(f) = tr(f) - nf$. @math_noob: If you substitute $f$ first then take the determinant, then the answer you get will be a scalar, but if you first take the determinant to get a polynomial in $x$, then substitute $f$, the answer is a matrix. So there is no reason to expect them to be the same.
$endgroup$
– Ted
Jan 5 at 17:42
$begingroup$
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
$endgroup$
– José Carlos Santos
Jan 5 at 18:10
add a comment |
$begingroup$
What's wrong is that $P(x)=det(f-xoperatorname{Id})$ is a polynomial. Yes, you can compute $P(x_0)$ for a number $x_0$ by computing $det(f-x_0operatorname{Id})$, but $f$ is not a number; it's an endomorphism.
Conseder, for instance, $Q(x)=operatorname{tr}(f-xoperatorname{Id})$. By your argument, $Q(f)=0$ too. But, in fact, since $Q(x)=operatorname{tr}(f)-nx$, you actually have $Q(f)=(1-n)operatorname{tr}(f)$.
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
$begingroup$
A slight correction to this answer: it should be $Q(f) = tr(f) - nf$. @math_noob: If you substitute $f$ first then take the determinant, then the answer you get will be a scalar, but if you first take the determinant to get a polynomial in $x$, then substitute $f$, the answer is a matrix. So there is no reason to expect them to be the same.
$endgroup$
– Ted
Jan 5 at 17:42
$begingroup$
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
$endgroup$
– José Carlos Santos
Jan 5 at 18:10
add a comment |
$begingroup$
What's wrong is that $P(x)=det(f-xoperatorname{Id})$ is a polynomial. Yes, you can compute $P(x_0)$ for a number $x_0$ by computing $det(f-x_0operatorname{Id})$, but $f$ is not a number; it's an endomorphism.
Conseder, for instance, $Q(x)=operatorname{tr}(f-xoperatorname{Id})$. By your argument, $Q(f)=0$ too. But, in fact, since $Q(x)=operatorname{tr}(f)-nx$, you actually have $Q(f)=(1-n)operatorname{tr}(f)$.
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
$endgroup$
What's wrong is that $P(x)=det(f-xoperatorname{Id})$ is a polynomial. Yes, you can compute $P(x_0)$ for a number $x_0$ by computing $det(f-x_0operatorname{Id})$, but $f$ is not a number; it's an endomorphism.
Conseder, for instance, $Q(x)=operatorname{tr}(f-xoperatorname{Id})$. By your argument, $Q(f)=0$ too. But, in fact, since $Q(x)=operatorname{tr}(f)-nx$, you actually have $Q(f)=(1-n)operatorname{tr}(f)$.
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
edited Jan 5 at 17:10
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
answered Jan 5 at 17:04
José Carlos SantosJosé Carlos Santos
173k23133241
173k23133241
$begingroup$
A slight correction to this answer: it should be $Q(f) = tr(f) - nf$. @math_noob: If you substitute $f$ first then take the determinant, then the answer you get will be a scalar, but if you first take the determinant to get a polynomial in $x$, then substitute $f$, the answer is a matrix. So there is no reason to expect them to be the same.
$endgroup$
– Ted
Jan 5 at 17:42
$begingroup$
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
$endgroup$
– José Carlos Santos
Jan 5 at 18:10
add a comment |
$begingroup$
A slight correction to this answer: it should be $Q(f) = tr(f) - nf$. @math_noob: If you substitute $f$ first then take the determinant, then the answer you get will be a scalar, but if you first take the determinant to get a polynomial in $x$, then substitute $f$, the answer is a matrix. So there is no reason to expect them to be the same.
$endgroup$
– Ted
Jan 5 at 17:42
$begingroup$
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
$endgroup$
– José Carlos Santos
Jan 5 at 18:10
$begingroup$
A slight correction to this answer: it should be $Q(f) = tr(f) - nf$. @math_noob: If you substitute $f$ first then take the determinant, then the answer you get will be a scalar, but if you first take the determinant to get a polynomial in $x$, then substitute $f$, the answer is a matrix. So there is no reason to expect them to be the same.
$endgroup$
– Ted
Jan 5 at 17:42
$begingroup$
A slight correction to this answer: it should be $Q(f) = tr(f) - nf$. @math_noob: If you substitute $f$ first then take the determinant, then the answer you get will be a scalar, but if you first take the determinant to get a polynomial in $x$, then substitute $f$, the answer is a matrix. So there is no reason to expect them to be the same.
$endgroup$
– Ted
Jan 5 at 17:42
$begingroup$
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
$endgroup$
– José Carlos Santos
Jan 5 at 18:10
$begingroup$
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
$endgroup$
– José Carlos Santos
Jan 5 at 18:10
add a comment |
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$begingroup$
see Wikipedia
$endgroup$
– 0x539
Jan 5 at 17:11
$begingroup$
Note: This argument is not too far from being correct(able) though. Working with matrices: The argument above is nearly all that is required to show that CH holds for an upper triangular matrix. The general case is obtained by extending the ground field to an algebraic closure of it. When the field is algebraically closed, all matrices are similar to upper triangular ones, and CH follows easily.
$endgroup$
– Ittay Weiss
Jan 5 at 18:21
$begingroup$
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
$endgroup$
– user1551
Jan 5 at 18:53
$begingroup$
@user1551 Sorry, I was perhaps unclear. When the characteristic polynomial splits, computing f(A) for an upper triangular matrix is just plugging in $A$ for $x$ in the decomposition of $f$ and computing. That fact that you get $0$ is easy. I see some resemblance to OP's faulty argument here: You just substitute $A$ for $x$, and the computation is nearly just evaluating at each eigenvalue.
$endgroup$
– Ittay Weiss
Jan 5 at 19:09