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?

$$V$$ 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!

• You don't need commuting matrices. All you need is a lower bound for $Pv\cdot v$, which you can get from $\|v\|$ and the smallest eigenvalue of $P$, and a lower bound for $Tv\cdot v$, which again you can get from $\|v\|$ and the most negative eigenvalue of $T$. Apr 9, 2022 at 3:37

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.
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.
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