2
  • Caclulate the characteristic & the minimal polynomial of the matrix:

$$A\in\mathbb C^{n\times n}:a_{ij}=i\cdot j ,\forall i,j=1,..,n$$

$$\text{i.e for $n=3$, } $$ $$A=\left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\\3 & 6 & 9\end{matrix}\right]$$

$$\text{I've calculate $X_A(x),m_A(x)$ for the following cases:}$$

$$ \left[\begin{matrix}1 & 2\\2 & 4\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3\\2 & 4 & 6\\3 & 6 & 9\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4\\2 & 4 & 6 & 8\\3 & 6 & 9 & 12\\4 & 8 & 12 & 16\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4 & 5\\2 & 4 & 6 & 8 & 10\\3 & 6 & 9 & 12 & 15\\4 & 8 & 12 & 16 & 20\\5 & 10 & 15 & 20 & 25\end{matrix}\right], \quad \left[\begin{matrix}1 & 2 & 3 & 4 & 5 & 6\\2 & 4 & 6 & 8 & 10 & 12\\3 & 6 & 9 & 12 & 15 & 18\\4 & 8 & 12 & 16 & 20 & 24\\5 & 10 & 15 & 20 & 25 & 30\\6 & 12 & 18 & 24 & 30 & 36\end{matrix}\right]$$

$$\xrightarrow{X_A(x)} \left [ x \left(x - 5\right), \quad - x^{2} \left(x - 14\right), \quad x^{3} \left(x - 30\right), \quad - x^{4} \left(x - 55\right), \quad x^{5} \left(x - 91\right)\right ]$$

$$\xrightarrow{m_A(x)} \left [ x \left(x - 5\right), \quad x \left(x - 14\right), \quad x \left(x - 30\right), \quad x \left(x - 55\right), \quad x \left(x - 91\right)\right ]$$

$$\text{Hence, i assumed that if }A\in\mathbb{C}^{n\times n}:$$ $$\fbox{$X_A(x)=(-1)^nx^{n-1}\big( x-\operatorname{tr}(A)\big)$},$$ $$\fbox{$m_A(x)=x\big( x-\operatorname{tr}(A)\big)$} .$$

How can i prove this in general case?
any suggestions?

Marc van Leeuwen
  • 115,048
Rick Sanchez C-666
  • 1,010

1 Answers1

1

Let $\mathbf{v}:=(1,2,\ldots,n)$ and $\mathbf{w}_i:=(i,0,\ldots,0,-1,0,\ldots)$ for $i=2,\ldots,n$. These vectors are non-zero orthogonal and hence form a basis of $\mathbb{R}^n$. They are also eigenvectors of $A$ since $$A\mathbf{v}=|\mathbf{v}|^2\mathbf{v},\qquad A\mathbf{w}_i=\mathbf{0}$$ Note that $|\mathbf{v}|^2=\mathrm{trace}(A)$.
Hence $A$ has eigenvalues $|\mathbf{v}|^2$ and $0$ repeated $(n-1)$ times. The characteristic polynomial is thus $$X_A(x)=(-1)^n(x-\lambda_1)\cdots(x-\lambda_n)=(-1)^nx^{n-1}(x-|\mathbf{v}|^2)$$ The minimal polynomial can be found by multiplying $A$ with itself: $$A^2=\begin{bmatrix}\mathbf{v}\\ 2\mathbf{v}\\ \vdots\\n\mathbf{v}\end{bmatrix}\begin{bmatrix}\mathbf{v}&2\mathbf{v}&\cdots&n\mathbf{v}\end{bmatrix}=|\mathbf{v}|^2A$$ $$\therefore\ m_A(x)=x^2-|\mathbf{v}|^2x=x(x-|\mathbf{v}|^2)$$ $m_A$ is the minimal polynomial since it is the least degree polynomial containing both eigenvalues.

These results for $X_A$ and $m_A$ generalize to any matrix of the type $A={\mathbf{v}\mathbf{v}}^\top$.

Chrystomath
  • 10,798