Prove that for each real number $r$ we can find an $n \times n$ square matrix $A$ with real entries such that determinant of $A$ is $r$.

Asked Mar 24 '16 at 15:02

Active Mar 24 '16 at 15:27

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I have proved that for each positive real number we can find an $n \times n$ diagonal matrix with each diagonal entries $r^{1/n}$ such $\det A=r.$ But how to prove for negative real numbers?

edited Mar 24 '16 at 15:27

asked Mar 24 '16 at 15:02

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Just use $|r|^{1/n}$ and change the sign of one diagonal entry. – Brian M. Scott Mar 24 '16 at 15:04
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Concerning your title question: even easier is to take the diagonal matrix $(1,1,1,\ldots, r)$. No need to take $n$-th roots. – Dietrich Burde Mar 24 '16 at 15:07

Prove that for each real number $r$ we can find an $n \times n$ square matrix $A$ with real entries such that determinant of $A$ is $r$.

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