$B$ symmetric, then exists $\lambda\ge 0$ such that $B+\lambda I$ is positive definite

Question

Show that if $B$ is symmetric, then there exists $\lambda\ge 0$ such that $B+\lambda I$ is positive definite

I'm reading a proof of this:

And I have a problem in the part

$$B+\lambda I = Q(A+\epsilon I)Q^T$$

I know that $B+\lambda I = QAQ^T + \lambda I$, and we can certainly do a decomposition $I(\lambda I)I^T$ because $\lambda I$ is already in the desired form. But I don't think we can do $\lambda I = Q(\lambda I)Q^T$ as in the solution

Answer 1

This does not require anything like diagonalization. If $\lambda > \|B\|$ then $\langle Bx, x \rangle +\lambda \langle x, x \rangle \geq \lambda \|x\|^{2}-\|B\|\|x\|^{2} \geq 0$.