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).


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 $\lambda_1\geq\ldots\geq\lambda_k>0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.