What can $\alpha$ be in $k-\alpha\frac{x^TAx}{x^Tx}$ such that it becomes negative?

Question

Given scalars $k$ and $\alpha$, matrix $A \in \mathbb{R}^{n \times n}$ and

$$k-\alpha\frac{x^TAx}{x^Tx}$$

I wish to find the value of $\alpha$ such that this expression in $x \in \mathbb{R}^n$ becomes negative.

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Now, I have already found the answer, however I want to know if my reasoning is correct or if anyone can improve upon the answer?

\begin{equation} k-\alpha\frac{x^TAx}{x^Tx} \leq k-\alpha\lambda_{\min}(A) \end{equation}

Therefore, in order for the RHS to be negative,

\begin{equation} \alpha > \frac{k}{\lambda_{\min}(A)} \end{equation}

Now, I want to go a step further and ask, can a value of $\alpha$ be obtained that does not depend on $A$ explicitly? Like in this case it does depend on the minimum eigenvalue of $A$. Any insight or a better proof would be really helpful! Thanks!

Answer 1

Your argument depends on $A$ is positive definite. In the event that it is positive semidefinite, replace the term with the smallest positive eigenvalue.

• Suppose $A=-I$, then $$k- \alpha \frac{x^TAx}{x^Tx}=k+\alpha$$

For it to be negative, we need $k+\alpha <0$, we need $\alpha < -k$.

• Let's consider the case where $A =diag(1, -1)$, the case where it is indefinite. If $x=e_1$, then we have $k-\alpha <0 \iff \alpha>k$ and if $x=e_2$, then $k+\alpha <0 \iff \alpha < -k$. If $k$ is positive, then no such $\alpha$ exists.

Now, suppose that you assume that your matrix is positive definite.

Let $v$ be an eigenvector corresponding to the smallest eigenvalue $\lambda_{\min}(A)$.

$$k-\alpha \frac{v^TAv}{v^Tv}=k-\alpha \lambda_{\min}(A)<0$$

We can construct matrix $A$ such that the smallest eigenvalue is arbitraryly small and hence an $\alpha$ that is independent of $A$ can't exist.