Proof of the following inequality

Question

let $\lambda_1,\lambda_2,...,\lambda_n$ be positive numbers such that $\sum_{i=1}^n\lambda_i=n$, we denote $S=\text{diag}(\lambda_1,\lambda_2,...,\lambda_n)$. For a given set of $m\geq n$ vectors $f_1,f_2,...,f_m$ in $\mathbb{R}^n$ such that $$\sum_{i=1}^mf_if_i^T=S$$ I need to prove that $$∥S^{−1}f_i∥∥f_i∥\leq1$$ can anyone help me to prove it plz. In case $\lambda_1=\lambda_2=...=\lambda_n$ I've already proved it and also $$∥S^{−1/2}f_i∥\leq1$$

Answer 1

I think I missed something :

Let $n=m=2$, $f_1 =(\frac{\sqrt{3}}{4} , \frac{\sqrt{7}}{4})$, $f_2 =(\frac{\sqrt{21}}{4},-\frac{1}{4})$, $\lambda_1 = 3/2$ and $\lambda_2 = 1/2$. We have that $\|S^{-1} f_1\|\|f_1\|>1$.

Isn't that a counterexample ?

Answer 2

Take a look i think there is an error in your Exercise, to prove that i will consider the case in which $m=n$, here S is a Positive definite matrix, so i will consider Dot product associate to S and i will prove that $(f_1,...,f_n)$ is an orthonormal basis for this product, So from the given equality, and for any vector x we have : $$\sum_{i=1}^nf_i<f_i,x>=Sx⇒\sum_{i=1}^nS^{-1}f_i<f_i,x>=x$$ Then we put $e_i=S^{-1}f_i$, we have : $$\sum_{i=1}^n<Se_i,x>e_i=\sum_{i=1}^n<e_i,x>_Se_i=x$$ And this for any x, Which means that $(e_1,...,e_n)$ is an orthonormal basis for the Dot product associate to S (Which we note $< , >_S$), Particularly We have : $$<e_i,e_i>_S=<S^{-1}f_i,f_i>=1$$ And By Cauchy-Schwarz inequality we have : $$1=<S^{-1}f_i,f_i>⩽\Vert{S^{-1}f_i}\Vert\Vert{f_i}\Vert$$ You See !!