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Let $A$ be an $n\times n$ matrix with real entries and $I$ denotes the $n\times n$ identity matrix. Suppose all roots (include complex roots) of $$\operatorname{char} A = \det(A-\lambda I)$$ the characteristic polynomial of $A$, have absolute value strictly less than $1$. Prove that the matrix $I-A$ is invertible.

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    $\begingroup$ It's helpful to include what you have tried so far. Hint: how can you characterize a singular matrix in terms of its eigenvalues? $\endgroup$ – Eric Auld Nov 17 '13 at 6:55
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    $\begingroup$ $\det(A - I)\ne 0 \implies A-I$ invertible already. How elementary do you want your proof???? $\endgroup$ – achille hui Nov 17 '13 at 7:33
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Hint: if $\lambda$ is an eigenvalue of$~A$, then it means that some matrix is not invertible; do you recall which matrix? Then use the contrapositive. You need only a small part of the given information about char$~A$.

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If $1$ is not an eigenvalue, then there is no vector $v$ such that $Av=v$, ie. $(A-I)v\ne 0$ for any $v$. Hence $A-I$ is injective, hence surjective since we are dealing with finite dimensions, thus it is invertible.

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All roots (include complex roots) of char $A=\operatorname{det}(A−λI)$,the characteristic polynomial of A, have absolute value strictly less than $1$,so $1$ is not a charactristic value,and then det $(I-A)$ is not zero,namely $I-A$ is invertible.

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