If the operator $P$ is positive and $T$ is self-adjoint, then is there exists a positive number $n$ such that $nP+T$ is positive?

Question

$V$ is a finite-dimensional complex inner product space and $P, T\in L(V)$. If the operator $P$ is positive and $T$ is self-adjoint, then is there exists a positive number $n$ such that $nP+T$ is positive?

I don't know if this is right or wrong, but the followings are my thoughts:

I initially tried to use inner product $\langle (nP+T)v,v\rangle =n\langle Pv,v\rangle+\langle Tv,v\rangle $

Then since $T$ is only self-adjoint, $\langle Tv,v\rangle$ can sometimes be negative, so as long as $n$ is "large enough", then the statement will be right. But since $n$ isn't dependent on the vectors $v$, I need to exactly know "how negative" $\langle Tv,v\rangle$ can be, so I can choose the right $n$ to let $ n\langle Pv,v\rangle+\langle Tv,v\rangle $ always be positive, but I don't know how to do this

Then I think of this by matrix since based on the spectral theorem. both positive and self-adjoint operators are diagonalizable. However, I find that this will be right if $P$ and $T$ are commutable. Since for commutable and diagonalizable operators, there exists a basis that makes both of them diagonalized. If they are commutable, and since $P$ and $T$ are fixed, I can find a $n$ such that the entries of $nP+T$ on the diagonal are nonnegative. However, commutable isn't given by the question.

Thus basically, I am still not sure whether this statement is right or wrong. Any help on this? Thanks!

Answer 1

Hint: It suffices to consider $\langle (nP+T)v,v\rangle =n\langle Pv,v\rangle+\langle Tv,v\rangle$ when $v$ is a unit vector.

More details are hidden below:

When $v$ is a unit vector, there is a bound on how negative $\langle Tv,v\rangle$ can be, since it is a continuous function of $v$ and the unit sphere in $V$ is compact so $\langle Tv,v\rangle$ is bounded on it. Similarly, $\langle Pv,v\rangle$ has a strictly positive lower bound for $v$ a unit vector, again by compactness of the unit sphere. So you can find an $n$ that works for all unit vectors $v$, and it follows that it works for any $v$ since scaling $v$ multiplies $\langle (nP+T)v,v\rangle$ by a positive number.

Answer 2

The counterexample is to let $P$ be a zero operator and $T$ be a self-adjoint but not positive operator. Then, $nP+T=T$ can't be a positive operator