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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.

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2 Answers

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.

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HINT

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?

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