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 $0$. $<v,v> $ 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?



  • $\begingroup$ You need to choose an $a$ that is big enough, or convince the reader that such a choice exists. $\endgroup$ Mar 29, 2016 at 18:59
  • $\begingroup$ @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? $\endgroup$
    – Alan
    Mar 29, 2016 at 19:02
  • $\begingroup$ Well, it needs to be big enough and positve. For example, what if $A = -I$? $\endgroup$ Mar 29, 2016 at 19:51
  • $\begingroup$ @Omnomnomnom I see. Do you think the way I moved through the problem was good? Do you know how to continue from here? $\endgroup$
    – Alan
    Mar 29, 2016 at 19:52

1 Answer 1


Another (perhaps simpler) approach is to work with eigenvalues.

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .