# Let $T$ be a self-adjoint operator and$\langle T(w),w\rangle>0$ . If $\operatorname{dim}(W) = k$ then $T$, has at least $k$ positive eigenvalues

Qn: Let T be a finite-dimensional complex inner product space, and T a self-adjoint linear operator. Suppose there exists a subspace W of V such that $\langle T(w),w \rangle$ is positive for all non zero $w$. If $\operatorname{dim}(W) = k$, prove that $T$ has at least $k$ positive eigenvalues (counting algebraic multiplicities).

Below is my attempt:

Let $O$ be an orthonormal basis $O = \{v_1,\dots,v_n\}$, which are formed by eigenvectors of $T$. Since $T$ is self-adjoint, there some of the eigenvalues are positive, and others negative. Assume $\{\lambda_1,\dots,\lambda_m\}$ are positive eigenvalues and $\{\lambda_{m+1},...,\lambda_n\}$ are non-positive eigenvalues.

Let $U=\operatorname{span}\{v_{m+1},\dots,v_n\}$. Then $$\langle T(u), u \rangle = \langle u, T(u) \rangle= \lambda\langle u, u \rangle,$$ which is non-negative for all $u \in U$.

Does this imply that $W$ is a subset of $U$? (My plan is to show by proving something about the dimension but I am stuck).

## 1 Answer

Let $\lambda_1\geq\ldots\geq\lambda_k\geq\ldots\geq\lambda_n$ be the eigenvalues of a self-adjoint linear transformation $T$ acting on a $n-$dimensional vector space $V$.

Let $\{v_1,\ldots,v_n\}$ be an orthonormal basis of $V$ formed by eigenvectors of $T$ associated to $\{\lambda_1,\ldots,\lambda_n\}$.

Suppose $0\geq\lambda_k\geq\ldots\geq\lambda_n$.

Let $U=\text{span}\{v_{k},\ldots,v_n\}$. Notice that $\dim(U)=n-k+1$.

Let $W$ be your subspace. Since $\dim(W)=k$ and $\dim(U)+\dim(W)>n$ then exists $0\neq v\in\dim(U\cap W)$.

Thus, $v=\sum_{i=k}^na_iv_i$ and $0<\langle T(v),v \rangle=\sum_{i=k}^n\lambda_i|a_i|^2\leq 0$. This is a contradiction.

Therefore, $\lambda_k>0$ and $\lambda_1\geq\ldots\geq\lambda_k>0$.