Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are the following matrices invertible?

(1) $A= (a_{ij})_{2003 \times 2003}$, where $a_{ii}=2003, a_{ij}=1$ for $i \not=j$.

(2) $B= (b_{ij})_{n \times n }$ with $b_{ii}= \pi$ and $b_{ij} \in \mathbb{Q}$ for $i \not= j$.

Thank you so much.

share|cite|improve this question

Hints: (1) $A$ is strictly diagonally dominant. (2) Let $C=B-\pi I$. If $B$ is singular, then $\lambda=-\pi$ is a zero of the characteristic polynomial $p(\lambda)=\det(C-\lambda I)$. However, $\pi$ is a transcendental number.

share|cite|improve this answer


For the first one, $$A = 2002 I + uu^T$$ where $u$ is a vector of $1$'s. In fact, you can get the inverse exactly. See Sherman Morrison Woodbury formula for more details.

For the second one, $$B = \pi I + C$$ where $C_{ij} \in \mathbb{Q}$. If $B$ is not invertible, then there is $x \in \mathbb{R}^n$ such that $Bx = 0$ i.e. $\pi x + Cx = 0$. This means $-\pi$ is an eigenvalue of $C$. Is that possible?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.