How do I prove that for an nxn matrix M, with all eigenvalues being positive that (M+Id) is or is not invertible? So far I've tried creating a regressive definition for the determinant of an nxn matrix to create an inequality. I know that Det(M) is positive and I think Det(I+M) is positive, I've also tried creating counterexamples. Does anyone know if assuming its not invertible for a 2x2 matrix and proving that at least one eigenvalue has to be negative for it not to be invertible would hold to scrutiny?
 A: If $M+I$ is singular then its determinant is zero.
By definition, $-1$ is therefore an eigenvalue of $M$. This contradicts the fact that all eigenvalues are positive.
Therefore $M+I$ is invertible.
A: Let $\sigma(A)$ denote the set of eigenvalues of a matrix $A$. For any polynomial $p$ you have $\sigma(p(M)) = p(\sigma(M))$ (spectral mapping theorem). Now, choose $p(x) = x+1$. Then $p(M) = M+I$ and so $\sigma(M+I) = p(\sigma(M)) = \{1+\lambda : \lambda\in\sigma(M)\}$. So, the eigenvalues of $M+I$ are those of $M$, translated by one to the right. The matrix $M+I$ is thus invertible (as $0\notin\sigma(M+I)$).
A: If $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $M$, then the eigenvalue of $I+M$ would be $1+\lambda_1,\dots,1+\lambda_n$ which are all greater than zero. Hence $M$ is invertible.
A: The eigenvalues of $A$ are the values of $\lambda$ that make $A-\lambda I$ a singular matrix. All other values of $\lambda$ make $A-\lambda I$ invertible.
If all of the eigenvalues of $A$ are positive, then $A+d I$ must always be invertible because the eigenvalues all correspond to negative values of $d$.
I have assumed that your symbol, $d$, is nonnegative. If not, then $A+d I$ is invertible if and only if $d$ is the negative of an eigenvalues.
