3
$\begingroup$

There is the Jordan block matrix

$J_\lambda(n):=\begin{pmatrix} \lambda & 1 & & & \\ & \lambda & 1 \\ & & ... & ... \\ & & & \lambda & 1 \\ & & & & \lambda \end{pmatrix} \in \mathbb{C^{n \times n}}$

How to find the inverse of this matrix?

I tried with the Gauss Jordan Elimination and got

$J_\lambda(n)^{-1} = \begin{pmatrix} \frac{1}{\lambda} & 0 & & & \\ & \frac{1}{\lambda} & 0 \\ & & ... & ... \\ & & & \frac{1}{\lambda} & 0 \\ & & & & \frac{1}{\lambda} \end{pmatrix}$

But i don't know if this works.

$\endgroup$
1
  • 2
    $\begingroup$ The solution already has been provided. - But the wrongness of your proposal would have been obvious just by calculating the matrix product of those 2 matrices of yours. $\endgroup$ Commented May 12, 2018 at 11:22

1 Answer 1

7
$\begingroup$

Your matrix $J_\lambda(n)=\lambda I+N$ where $$N=\pmatrix{0&1&0&\cdots&0&0\\ 0&0&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&1&0\\ 0&0&0&\cdots&0&1\\ 0&0&0&\cdots&0&0 }.$$ Then $N$ is nilpotent: $N^n=0$ and so $I+tN$ will have the inverse $I-tN+t^2N^2-\cdots +(-t)^{n-1}N^{n-1}$. Then $$J_\lambda(n)^{-1}=\lambda^{-1}(1+\lambda^{-1}N)^{-1} =\lambda^{-1}(I-\lambda^{-1} N+\lambda^{-2}N^2-\cdots +(-\lambda)^{-n+1}N^{n-1}).$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .