For every square matrix $U$ there exist a diagonal matrix $E$ with $e_{i,i}=-1 , 1 $ such that $E U – I_n$ is non-singular

linear algebramatrices

For every square matrix $U$ there exist a diagonal matrix $E$ with $e_{i,i} = \pm 1$ such that

  1. $E U – I_n$ is non-singular.
  2. If $U$ is unitary then $EU$ is also unitary

Note that all computation is over coefficient modulo $3$.

I want to prove this statement, but have no idea how and where to start.

Best Answer

To prove $EU-I_n$ non singular matrix it is sufficient to show that $0$ can't be an eigenvalue of $EU-I_n.$ or det$(EU-I_n) \ne 0$

Clearly $E^2=I_n,$ $EU-I_n$ $\Rightarrow$ $EU-E^2$

$E(U-E)$ (by distributive law of matrices),

$det(E(U-E))\ne 0$ $\Rightarrow$ $det(E).det(U-E)\ne 0$ and clearly $det(E)=\pm1,$ $det(U-E)\ne 0$

Claim: $0$ is not an eigenvalue of $U-E$.

Suppose

Case1: $1$ is an eigenvalue of $U$ with (A.M=$r_1$) then in that case take $-1$ with $r_1$ times entries in $E$ so that $0$ can't be eigenvalue of $E-U.$

case2: $-1$ is an eigenvalue of $U$ with (A.M=$r_1$) then in that case take $1$ with $r_1$ times entries in $E$ so that $0$ can't be eigenvalue of $E-U.$

case 3: $1$ (A.M.=$r_1$) and $-1$ (A.M=$r_2$) then take $-1$ (A.M.=$r_1$) and $1$ (A.M=$r_2$) entries (or which are same as eigenvalues of $E$)

and for any other eigenvalues except $\pm1$ of $U$ always gives nonzero eigenvalue of $U-E.$

Hence $EU-I_n$ is non singular matrix.

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