Cayley-Hamilton says that evaluating an endomorphism's characteristic polynomial over that endomorphism gives...












0














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







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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
















0














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







share|cite|improve this question









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














0












0








0







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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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math_noob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









José Carlos Santos

152k22123226




152k22123226






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asked 2 days ago









math_noobmath_noob

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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














  • 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










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 $Q(f)=(1-n)operatorname{tr}(f)$.






share|cite|improve this answer























  • 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













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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 $Q(f)=(1-n)operatorname{tr}(f)$.






share|cite|improve this answer























  • 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


















3














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 $Q(f)=(1-n)operatorname{tr}(f)$.






share|cite|improve this answer























  • 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
















3












3








3






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 $Q(f)=(1-n)operatorname{tr}(f)$.






share|cite|improve this answer














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 $Q(f)=(1-n)operatorname{tr}(f)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























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




















  • 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












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