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38 votes
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If two matrices have the same characteristic polynomial then do they have the same determinant?

If $\chi_A$ denotes the characteristic polynomial of $A$, then $\chi_A(0)=\det(A)$, so the answer is yes.
TheSilverDoe's user avatar
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23 votes
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Can we deduce the characteristic polynomial for this matrix?

The answer is no. You are given a degree $n$ polynomial $p(x)$ that an $n\times n$ matrix $A$ satisfies. This is not enough information to find the characteristic polynomial $c_A(x)$, although you ...
user729424's user avatar
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22 votes
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Is a characteristic polynomial we consider in Linear Algebra a polynomial or a polynomial function?

Nice question! In many cases, that distinction is irrelevant, but in some cases it matters. And, when it matters, you are not right: it is a polynomial, not a polynomial function. For instance, ...
José Carlos Santos's user avatar
20 votes

If two matrices have the same characteristic polynomial then do they have the same determinant?

If they have the same characteristic polynomial then they have the same eigenvalues and by this fact alone they have the same determinant because the determinant is equal to the product of the matrix'...
Localth's user avatar
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16 votes

Do matrices $ AB $ and $ BA $ have the same minimal and characteristic polynomials?

Yes, $AB$ and $BA$ have the same characteristic polynomial. Basic facts: $\det(A^T) = \det(A)$, $\det(AB) = \det(A) \det(B)$ $A$ and $A^T$ share the same characteristic polynomial. \begin{align*} \...
Bio's user avatar
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12 votes

If two matrices have the same characteristic polynomial then do they have the same determinant?

For a square matrix $n \times n$ matrix $A$ the characteristic polynomial will have constant term $(-1)^n \det A$ and the coefficient of the $t^{n-1}$ term will be $- \text{tr} A$. So if two matrices ...
CyclotomicField's user avatar
11 votes
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Minimal polynomial of direct sum

If you apply a polynomial to $A \oplus B$ you get $f(A \oplus B) = f(A) \oplus f(B)$. If $f(A \oplus B) = 0$ then $f(A) = f(B) = 0$ so $\mu_A \mid f$ and $\mu_B \mid f$. Conversely, if $\mu_A \mid f$ ...
Sera Gunn's user avatar
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10 votes

Do matrices $ AB $ and $ BA $ have the same minimal and characteristic polynomials?

For square matrices, the characteristic polynomials are same, but for $A$ a matrix of size $m \times n$ and $B$ a matrix of size $n \times m$ we have $x^{m}C_{BA}(x)=x^{n}C_{AB}(x)$. This implies that ...
user499117's user avatar
10 votes

Is $A$ the $2 × 2$ identity matrix?

Counterexample: $$A=\begin{pmatrix}e^{\frac{2}{3}\pi i} & 0\\ 0 & e^{\frac{4}{3}\pi i}\end{pmatrix}$$ and $$A^{2}=\begin{pmatrix}e^{\frac{4}{3}\pi i} & 0\\ 0 & e^{\frac{2}{3}\pi i}\end{...
Floris Claassens's user avatar
9 votes

Do matrices $ AB $ and $ BA $ have the same minimal and characteristic polynomials?

There are a lot of proofs for characteristic polynomials to be same. I want to provide mine. It may be more complicated, but it is less "consider magic product of matrices". Let $\chi_M(x)$ ...
Александр Тряпицын's user avatar
8 votes
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Is $A$ the $2 × 2$ identity matrix?

Your proof would be correct if $A$ had only one eigenvalue. But if it has two distinct eigenvalues $\alpha$ and $\beta$, it can happen that $\{\alpha^2,\beta^2\}=\{\alpha,\beta\}$; all you need is to ...
José Carlos Santos's user avatar
8 votes
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If $A^3+2A^2+2A+I_n=0_n$, then $\det(A)=-1$, where $A$ is a square matrix of odd size, with real entries

Since each eigenvalue of $A$ solves $(\lambda+1)(\lambda^2+\lambda+1)=0$, the possible eigenvalues are $-1,\,\exp\frac{\pm2\pi i}{3}$. Since $\det A\in\Bbb R$, each non-real eigenvalue is used equally ...
J.G.'s user avatar
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8 votes
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Prove $\det((AB)^{n}-(BA)^{n})$ is a perfect cube.

Since $AB$ is an integer matrix, its characteristic polynomial $\det(xI-AB)$ has integer coefficients. Let us write it as $$ x^3-a_3x^2-b_3x-c_3. $$ So, for every $n\ge3$ we have $$ (AB)^n=a_n(AB)^2+...
user1551's user avatar
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7 votes

Coefficients of characteristic polynomial of a matrix

$\newcommand\sgn{\operatorname{sgn}}$ I learned of the following proof from @J_P's answer to what effectively is the same question. It arises from expanding the usual definition $\det A=\sum_{\sigma\...
ho boon suan's user avatar
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7 votes
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If $A,B$ are diagonalisable, does $AB$ diagonalisable imply $BA$ diagonalisable?

Counterexample: \begin{aligned} AB&=\pmatrix{1&0&0\\ 0&0&0\\ 0&0&1}\pmatrix{0&0&0\\ 1&1&0\\ 0&0&1}=\pmatrix{0&0&0\\ 0&0&0\\ 0&0&...
user1551's user avatar
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7 votes
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Any smart approach to finding a matrix given the characteristic polynomial?

Hint: Find the companion matrix of the polynomial.
cqfd's user avatar
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7 votes

Products of matrices in either order have the same characteristic polynomial

Here is a classical solution. Step 1 If $B$ is invertible. Then \begin{align}P_{AB}(x)&= \det(xI-AB)\\&=\det(xB^{-1}B-AB)\\&= \det(xB^{-1}-A) \det(B)\\&=\det(B) \det(xB^{-1}-A)\\& =...
N. S.'s user avatar
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7 votes

Is a characteristic polynomial we consider in Linear Algebra a polynomial or a polynomial function?

The characteristic polynomial of $T$ (either a matrix or a linear transformation, depending on your preference) is a polynomial, not a function. What we really care about are its coefficients. For ...
Ittay Weiss's user avatar
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7 votes
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Is the Jordan normal form uniquely determined by the characteristic and minimal polynomial?

Generally knowing only the characteristic polynomial and the minimal polynomial is not enough to determine uniquely the Jordan normal form, like you showed in the question. I think that the only times ...
Sant97's user avatar
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7 votes
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Finding polynomial to the power of 2020

All you need to do is multiply $(X-5)(X+1)^2$ by any polynomial of degree $2017$.
ancient mathematician's user avatar
7 votes
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Is there any intuitive explanation to $A^2=\operatorname{tr}(A)A-\det(A)I_{2\times 2}$ for $A \in \mathbb{R}^{2\times 2}$?

The characteristic polynomial of a $2 \times 2$ matrix $A$ is $$f(t) = t^2 - \operatorname{tr}(A)t + \operatorname{det}(A).$$ So your identity is a special case of the Cayley-Hamilton theorem. Some ...
spin's user avatar
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6 votes
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Dimension of eigenspace of a transpose

This is because of the Rank nullity theorem: as a matrix and its transpose have the same rank, we have $$\DeclareMathOperator\rk{rank} \rk(A-\lambda I)=\rk{}{}^\mathrm t\mkern-1.5mu(A-\lambda I)\iff \...
Bernard's user avatar
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6 votes

Let $A$ be a $3\times 3$ matrix with characteristic polynomial $x^3-3x+a$, for what values of $a$ given matrix must be diagonalizable.

Let $p(x)=x^3-3x+a$. If $p(x)$ has a multiple root, then that root will also be a root of $p'(x)$. But the roots of $p'(x)$ are $\pm1$. So,\begin{align}p(x)\text{ has a multiple root}&\iff\pm1\...
José Carlos Santos's user avatar
6 votes

Can we deduce the characteristic polynomial for this matrix?

Take matrix $$ A = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right) $$ Then $$ A^4 - ...
Will Jagy's user avatar
  • 141k
6 votes
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$ A^2 - B^2 = I_{2n+1} \implies det(AB-BA)=0 $ where A,B are complex matrices of odd size

Let $P=A+B$ and $Q=A-B$. Then $A=\frac 12(P+Q)$ and $B=\frac 12(P-Q)$. Hence the hypothesis reads as $PQ+QP=2I$ and we have to prove that $\det(QP-PQ)=0$. We have $I-PQ=-(I-QP)$. Therefore $\det(I-PQ)...
timon92's user avatar
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6 votes
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How to show that $det(AB-xI)=det(BA-xI)$ ,for any $x\in \mathbb F$.

A bit tricky way, but which it even can be applied when $A$ and $B$ are not square matrices, so $A$ is $n\times m$ and $B$ is $m\times n$, is the following: Consider the identity of the product of ...
Nulhomologous's user avatar
6 votes

Two permutation matrices represent conjugate permutations iff they have same characteristic polynomial.

We'll first prove that the characteristic polynomial of a permutation matrix representing a permutation $\pi\in S_{n}$ having the cycle structure $1^{c_{1}}2^{c_{2}}...n^{c_{n}}$ is of the form $\...
ImBatman's user avatar
  • 378
6 votes
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Prove that $P^k=P$ for any $k \in \mathbb{N}$ where $1$ is the only eigenvalue of $P$ implies $P=I$.

A proof that uses some of the ideas you tried to: Let $f(x)$ be the minimal polynomial of $P$. The equation $P^k = P$ implies that $f(x)$ divides $x^k -x$. As you stated, the characteristic polynomial ...
Dionel Jaime's user avatar
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6 votes

Relation among Chern classes

One may consider $(c_1,c_2)$ as a pair of integers by choosing generators of $H^{2}$ and $H^4$ (both isomorphic to $\mathbb{Z}$). Then it was proved by Schwarzenberger that any pair $(c_1,c_2) \in \...
Nick L's user avatar
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6 votes
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Can a matrix have no eigenvectors?

If $A$ is a square matrix of order $n$ its characteristic polynomial has the form $$ \det(A-tI_n)=(-1)^nt^n+\text{terms of degree $<n$}.$$ Such a polynomial has always a root $\tau$ over the ...
Andrea Mori's user avatar
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