Cayley-Hamilton says that evaluating an endomorphism's characteristic polynomial over that endomorphism gives...
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 2 days ago
José Carlos Santos
152k22123226
asked 2 days ago
math_noobmath_noob
12
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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 2 days ago
José Carlos Santos
152k22123226
asked 2 days ago
math_noobmath_noob
12
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
see Wikipedia
– 0x539
2 days ago
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.
– Ittay Weiss
2 days ago
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
– user1551
2 days ago
@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.
– Ittay Weiss
2 days ago
add a comment |
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 2 days ago
José Carlos Santos
152k22123226
asked 2 days ago
math_noobmath_noob
12
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
José Carlos Santos
152k22123226
asked 2 days ago
math_noobmath_noob
12
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
José Carlos Santos
152k22123226
asked 2 days ago
math_noobmath_noob
12
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
José Carlos Santos
152k22123226
edited 2 days ago
José Carlos Santos
152k22123226
edited 2 days ago
José Carlos Santos
152k22123226
152k22123226
asked 2 days ago
math_noobmath_noob
12
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
math_noobmath_noob
12
asked 2 days ago
math_noobmath_noob
12
12
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
see Wikipedia
– 0x539
2 days ago
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.
– Ittay Weiss
2 days ago
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
– user1551
2 days ago
@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.
– Ittay Weiss
2 days ago
add a comment |
2
see Wikipedia
– 0x539
2 days ago
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.
– Ittay Weiss
2 days ago
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
– user1551
2 days ago
@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.
– Ittay Weiss
2 days ago
2
2
see Wikipedia
– 0x539
2 days ago
see Wikipedia
– 0x539
2 days ago
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.
– Ittay Weiss
2 days ago
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.
– Ittay Weiss
2 days ago
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
– user1551
2 days ago
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
– user1551
2 days ago
@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.
– Ittay Weiss
2 days ago
@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.
– Ittay Weiss
2 days ago
add a comment |
1 Answer
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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 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
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.
– Ted
2 days ago
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
– José Carlos Santos
2 days ago
add a comment |
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1 Answer
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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 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
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.
– Ted
2 days ago
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
– José Carlos Santos
2 days ago
add a comment |
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 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
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.
– Ted
2 days ago
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
– José Carlos Santos
2 days ago
add a comment |
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 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
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 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
edited 2 days ago
answered 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
answered 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
answered 2 days ago
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
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.
– Ted
2 days ago
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
– José Carlos Santos
2 days ago
add a comment |
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.
– Ted
2 days ago
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
– José Carlos Santos
2 days ago
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.
– Ted
2 days ago
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.
– Ted
2 days ago
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
– José Carlos Santos
2 days ago
@Ted Since I defined $Q(x)$ as $operatorname{tr}(f-xoperatorname{Id})$ and since $operatorname{tr}$ is linear and $operatorname{tr}(operatorname{Id})=n$, $Q(x)=operatorname{tr}(f)-nx$.
– José Carlos Santos
2 days ago
add a comment |
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2
see Wikipedia
– 0x539
2 days ago
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.
– Ittay Weiss
2 days ago
@IttayWeiss I don't follow. Yes, CH follows easily if the matrix is triangular, but how is this related to the argument in question?
– user1551
2 days ago
@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.
– Ittay Weiss
2 days ago