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I have an $n\times n$ matrix where the diagonals are equal to one and the off diagonals all have magnitude smaller than one. Is this matrix invertible? What is the quickest way to prove it?

sonicboom
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    I know a very special case https://en.wikipedia.org/wiki/Diagonally_dominant_matrix – Sumanta Sep 24 '20 at 15:18
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    Sorry, I confused diagonal and off-diagonal - see this post. Just a question: is the exercise perhaps with absolute value smaller than one? – Dietrich Burde Sep 24 '20 at 15:19
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    $\pmatrix{1&-1/2&-1/2\-1/2&1&-1/2\-1/2&-1/2&1}$ is not invertible. – kimchi lover Sep 24 '20 at 15:19
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    No, take $$\begin{pmatrix}1 &-0.5&0.5\-0.5&1&0.5\0.5&0.5&1\end{pmatrix}$$ – player3236 Sep 24 '20 at 15:19
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    in addition, OP should consider $\mathbf X$ which has n columns each with norm 1 but only $k\lt n$ of them are linearly independent (and no column is equal to another column or equal to another rescaled by -1). Then $\mathbf X^T\mathbf X$ is a matrix with diagonals of 1 and each off diagonal element has modulus less than 1 (e.g. apply Cauchy Schwarz or law of cosines) yet $\text{rank}\big(\mathbf X^T\mathbf X\big) =\text{rank}\big(\mathbf X\big) = k\lt n$ – user8675309 Sep 24 '20 at 16:00

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