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Consider this matrix :

$$A = \left( \begin{matrix} 1 & - 2 & 0 & \cdots & 0 & 0 & - 2 \\ 2 & 1 & - 2 & \cdots & 0 & 0 & 0 \\ 0 & 2 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & - 2 & 0 \\ 0 & 0 & 0 & \cdots & 2 & 1 & - 2 \\ 2 & 0 & 0 & \cdots & 0 & 2 & 1 \end{matrix}\right)_{2023\times 2023}$$

How to prove $A$ is invertible?

My attempt:

Consider this matrix :
$$B=\left( \begin{matrix} 1 & - 2 & \cdots & 0 & 0 & 0 \\ 2 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & - 2 & 0 \\ 0 & 0 & \cdots & 2 & 1 & - 2 \\ 0 & 0 & \cdots & 0 & 2 & 1 \end{matrix}\right)$$

Inductively, we see $$\det B>0$$

Let $e=(1,0,\cdots,1)'$, then , using row elimination method , we get
$$A=\begin{pmatrix}1&-2e'\\ 2e&B\end{pmatrix}$$

I could not see the $\det A$ is $\neq 0$ here ...
Help. Thanks.

Prem
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xldd
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    The matrix is like 1+Something antisymmetric. Maybe try from there – Confuse-ray30 Jun 28 '24 at 10:17
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    basically a duplicate of https://math.stackexchange.com/questions/4435559/general-argument-to-argue-t-is-an-isomorphism-from-its-matrix/ . When you look at your matrix modulo $2$ it is the identity matrix, so the determinant is odd. – user8675309 Jun 28 '24 at 16:30

1 Answers1

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We can write A as the sum of an antisymmetric and identity matrix( Verify this for yourself). There is a general result which says that if a matrix A is antisymmetric, then the matrix $A+I$ is invertible. The proof is as follows :

Let $x\in \mathbb{R}^{2023}$ and assume $x\in ker(A+I)$, then we have : $$0=(A+I)(x)=Ax+Ix=x^T(Ax+Ix)=x^TAx+x^TIx=0+||x||^2$$ The last equality follows from the fact A is anti-symmetric.

This gives us $||x||^2=0$ which implies $x=0$. As a result the kernel is trivial and our matrix $A+I$ is invertible.

muhammed gunes
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