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I was unable to prove the following statement. Let $A,B$ are both matrices of $n\times n$ of real numbers. We know that the matrix $ A$ is invertible. Then, we have to prove that there are $n$ diferent $\lambda_i$ so that the matrix $C=\lambda_iA+B $is invertible. ($\lambda_i$ are real numbers)

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  • $\begingroup$ Perhaps you meant $C$ is non invertible? $\endgroup$ – user1551 Jul 11 '13 at 15:51
  • $\begingroup$ This problem was in an exam i took, and i'm pretty sure it said that C was invertible, but maybe there was a mistake in the question... $\endgroup$ – Felipe Jul 11 '13 at 15:57
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The question seems to have some issues:

If $C$ is supposed to be invertible: It suffices that $\lambda I + BA^{-1}$ is invertible. As long as $\lambda$ is not negative of any of the eigenvalues of $BA^{-1}$ this matrix is invertible. So there are infinitely many valid $\lambda$s.

If $C$ is supposed to be non-invertible: Take $A=B=I$, then any $C$ is a multiple of the identity and it can only be non-invertible when $\lambda = -1$. So there is only one $\lambda$ satisfying the claim not $n$ different $\lambda$s.

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Uzing the QZ (generalized Schur) decomposition, we see that $$A = QSZ^*, \quad B = QTZ^*,$$ for some unitary $Q$ and $Z$, and upper triangular $S$ and $T$. Then $$C = \lambda_i A + B = Q(\lambda_iS + T)Z^*.$$ Edit: As egreg corrected my initial silly statement, $C$ is invertible if and only if $X := \lambda_iS+T$ is invertible (since unitary matrices $Q$ and $Z$ are invertible), and $X$ is invertible if and only if all its diagonal elements (which also happen to be its eigenvalues, because $X$ is upper triangular) are non-zero.

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  • $\begingroup$ I don't see why the diagonal elements of $\lambda_iS+T$ are the eigenvalues of $C$, unless $Q=Z$. However, $C$ is invertible if and only if $\lambda_iS+T$ is. $\endgroup$ – egreg Jul 11 '13 at 16:25
  • $\begingroup$ @egreg Let's attribute my temporary insanity to today's heat. <blush> Thank you. $\endgroup$ – Vedran Šego Jul 11 '13 at 16:41

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