# $A$ self adjoint matrix. Prove there is a real $c$ such that $cI+A$ is positive

This was partially asked before (1. There exists a real number $c$ such that $A+cI$ is positive when $A$ is symmetric, 2. Proving $aI+A$ is Positive Definite) but I'd like revision of this proof using forms.

Let $A$ be a self-adjoint $n \times n$ matrix. Prove that there is a real number $c$ such that the matrix $cI+A$ is positive

This is my attempt:

Let be $f$ the form with matrix $A$ in the canonical basis. Since $A$ is self adjoint there exists a unitary matrix $Q$ such that $$Q^*AQ = D$$ where $D$ is a diagonal matrix. But this is just the matrix of the form $f$ in the basis composed of the column vectors of $Q$. Moreover, since $A$ is self adjoint it follows that every entry of the diagonal matrix D is real.
So, consider the matrix $A+cI$ wich is also self adjoint and the consider the form $f'$ associated to this matrix. Again, there exists $P$ unitary such that $P^*(A+cI)P = D'$ is diagonal. But
$$P^*(A+cI)P = P^*AP + cP^*P = P^*AP + cI = D'$$ Choose $c$ so the entries of $D'$ are positive (for example, $c = 2\max \left\{ |(P^*AP)_{ii}|: i \in \left\{1,\ldots,n\right\}\right\}$) and then obviously $$X^*D'X > 0 \quad\forall X \in V$$ so the form $f'$ is positive and then $A+cI$ is positive (using the fact that if a form $f$ is positive then its matrix in every ordered basis is a positive matrix)
Thanks a lot for your feedback!

Looks good! I do have one stylistic note: you use $D$ to represent two different diagonal matrices, which is a little confusing. Instead, you can assert that $Q^* A Q$ is diagonal, without naming it $D$ as well.
• Oh, and one more comment, it's probably worth mentioning explicitly that $D$ has real entries, so that the maximum along the diagonal makes sense. Nov 26, 2017 at 2:47