# Proving $aI+A$ is Positive Definite

Let $$A \in M_{n \times n}^\Bbb C$$ be a self adjoint matrix. Prove that there is $$a \in \Bbb R$$ such that $$aI+A$$ is a positive definite matrix.

What I did so far

Let $$v$$ be a vector in an orthonormal basis of $$A$$ which exists since $$A$$ is self adjoint.$$A$$ is also diagonalizable so $$Av=\lambda v$$ where $$\lambda$$ is an eigenvalue of $$A$$. I now use the definition of the inner product:

$$\langle(aI+A)v,v\rangle=\langle(av+Av),v\rangle=\langle a v,v\rangle+\langle Av,v\rangle=a\langle v,v\rangle+\lambda\langle v,v\rangle$$.

Now I think I have to prove that this term is bigger than $$0$$. $$$$ is bigger from the inner product definition, but how do I deal with this scalar $$a$$? can I just say that it's positive since I have to prove there is a positive one?

Thanks,

Alan

• You need to choose an $a$ that is big enough, or convince the reader that such a choice exists. Mar 29, 2016 at 18:59
• @Omnomnomnom is there a reason there wouldn't be one? They tell me $a$ is some real number. Why can't I choose an arbitrary real number which is positive?
– Alan
Mar 29, 2016 at 19:02
• Well, it needs to be big enough and positve. For example, what if $A = -I$? Mar 29, 2016 at 19:51
• @Omnomnomnom I see. Do you think the way I moved through the problem was good? Do you know how to continue from here?
– Alan
Mar 29, 2016 at 19:52

First, $$A$$ has real eigenvalues. Second, if $$\lambda_i$$ is an eigenvalue of $$A$$, then the corresponding eigenvalue of $$B=A+aI$$ is $$\lambda_i +a$$. Third, a self adjoint matrix is positive definite iff the eigenvalues are positive.
Putting all together, by picking any $$a > \min(\lambda_i)$$ , $$B=A+aI$$ results positive definite.