# Questions tagged [characteristic-polynomial]

The characteristic polynomial of a square matrix is a polynomial that is invariant under matrix similarity and has the eigenvalues as roots.

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### $A ∈ GL_n(K)$. Show that there exists a polynomial $P(X) ∈ K[X]$ of degree $< n$ such that $A^{−1} = P(A)$ [duplicate]

Hey I am having problems with this exercise. Can someone help me? Let $A ∈ GL_n(K)$. Show that there exists a polynomial $P(X) ∈ K[X]$ of degree $< n$ such that $A^{−1} = P(A)$. To show that there ...
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### Why is the following "proof" false? [duplicate]

Exercise: Why is the following "proof" false? $\text{ch}_A(xI-A)=\det(xI-A)$ Substitude $A$ for $x$, and obtain $\det(A-A)=0=\text{ch}_A(A)$. Solution: Explanation We cannot substitude $A$ ...
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### How to find ${\rm rank}(2I_n-A)$ where A is a square matrix of size $n$ and $A^3 - 6A^2 + 12A = 0_n$?

I think the rank has to be $n$ since anything else would be impossible to prove with so little information about the matrix. $$\det(2I_n-A) = -P_A(2) = -\det(A - 2I_n) \ .$$ So, if I can show the ...
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### Best (or correct) approach on finding characteristic polynomial of this $4 \times 4$ matrix

Let $f: \mathbb{R}^4 \to \mathbb{R}^4$ be a linear transformation with matrix representation \begin{align*} A = \begin{pmatrix} -5 & -5 & -9 & 7 \\ 8 & 9 & 18 &...
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### Show that $A+B=AB+BA$ iff $\text{Tr}(A)=\text{Tr}(B)=\text{Tr}(AB)=1$

We have $A,B$ $(2×2)$ matrices with complex entries. We know $AB≠BA$. Show that $A+B=AB+BA$ if and only if $\text{Tr}(A)=\text{Tr}(B)=\text{Tr}(AB)=1$. I tried writing $A=X+Y$ and $B=X-Y$ so we can ...
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### Show that $\det(xA+yB+zI_{n})=\det(yA+xB+zI_{n})$

We have $A$ and $B$ $(n×n)$ matrices with complex entries. We know that $A-B=AB-BA$. Show that $$\det(xA+yB+zI_{n})=\det(yA+xB+zI_{n})$$ for every $x,y,z$ complex numbers with $x+y≠0$. We can see that ...
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### Coefficients of characteristic polynomial and eigenvalues

Let $A$ be a $3x3$ symmetric (thus, orthogonally diagonalizable) real matrix. We know that its characteristic polynomial is (in the variable $λ$) $$-λ^3 + tr(A)λ^2-cλ+det(A)$$ where $c\in \mathbb{R}$ ...
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### Show that equation $\det(A+xB)=0$ has real solutions if and only if $\det(A^{2}+B^{2})\geq(\det(A)+\det(B))^{2}$

We have $A,B$ two $2×2$ matrices with real values and we know $\det(AB-BA)=0$. Show that equation $\det(A+xB)=0$ has real solutions if and only if $$\det(A^{2}+B^{2})\geq(\det(A)+\det(B))^{2}.$$ I ...
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### Range Space and Null Space of Projection Matrix

I'm studying my midterm exam and solving the problem set. Unfortunately, there is no solution manual for this set. I will show what I did and I will ask my specific question. Firstly, the question is ...
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### How is the matrix $(sI - A)$ called whose determinant is the characteristic polynomial of a (real valued) matrix $A$?

Context: When calculating eigenvalues of a real valued matrix $A$ one often constructs an auxiliary polynomial matrix $M(s):= (sI - A)$ and then calculates its determinant $d(s):=\det\Big(M(s)\Big)$ – ...
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### Minimal polynomial and characteristic polynomial of $T$

Let 𝑽 be a vector space over the field 𝑭 and 𝑻 be a linear operator on 𝑽. Then all eigen values of 𝑻 are zeros of the minimal polynomial of 𝑻. Minimal polynomial divides the characteristic ...
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### Show that characteristic polynomial of $Ty = ay$ is power of minimal polynomial of $a$

Let $K/F$ be a finite, separable, algebraic field extension and let $T: K\to K, Ty = ay$. Show that $p = m^n$ where $p$ is $T$'s characteristic polynomial and $m$ is $a$'s minimal polynomial. $m$ ...
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### Prove that A^k = 0 implies A^2 = 0 for A_2x2 [duplicate]

I shall prove the statement that if A is a 2x2 matrix and $A^2 \neq 0$, then it follows that $A^k \neq 0$ for $k > 2$ as well. I intended to prove the contraposition, but I’m not quite sure how to ...
### Showing $\lambda I_V$ diagonalizable and has only one eigenvalue
Problem: Let $V$ be a finite-dimensional vector space, and let $\lambda$ be any scalar. For any ordered basis $\beta$ for $V$, prove that $[\lambda I_V]_{\beta}=\lambda I$. Then compute the ...