Invertible matrix

Linear Algebra
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

Invertible matrix

#1

Post by Tolaso J Kos »

Consider the matrices $A \in \mathcal{M}_{m \times n}$ and $B \in \mathcal{M}_{n \times m}$. If $AB +\mathbb{I}_m$ is invertible prove that $BA+\mathbb{I}_n$ is also invertible.
(Romania, 2012)
Imagination is much more important than knowledge.
User avatar
Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: Invertible matrix

#2

Post by Riemann »

What the question basically asks is if $-1$ is a zero of the essentially same characteristic polynomials. $AB$ and $BA$ have quite similar characteristic polynomials. In fact if denote $p(x)$ the polynomial of $AB$, then the polynomial of $BA$ will be $q(x)= x^{n-m} p(x)$. It is easy to see that $-1$ cannot be an eigenvalue of the $AB$ matrix, otherwise it wouldn't be invertible. Now, let us assume that $BA$ is not invertible. Then it must have an eigenvalue of $-1$ and let $\mathbf{x}$ be the corresponding eigenvector. Hence:

$$\left ( BA \right )\mathbf{x}= -\mathbf{x} \Rightarrow AB \left ( A \mathbf{x} \right )= -A\mathbf{x}$$

meaning that $AB$ has an eigenvalue of $-1$ which is a contradiction. The result follows.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 11 guests