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Let $A\in \mathbb{R}^{m\times k}\ (k<m)$ be a real matrix. Suppose, $\forall x \in \mathbb{R}^{k}$, and for a $\delta \in (0,1)$, the following inequality holds: \begin{equation} (1-\delta) \|x\|_2^2 \leq \|Ax\|_2^2 \leq (1+\delta)\|x\|_2^2.\ \ \ \ (1). \end{equation} Let, $\sigma_{max}$, and $\sigma_{min}$ denote the largest and smallest singular values of the matrix $A$, respectively.

Then I have the following statements:

(1). Rank of the matrix $A$ is $k$.

(2). $\sigma_{min}\geq \sqrt{1-\delta}$.

(3). $\sigma_{max}\leq \sqrt{1+\delta}$.

I want to know if all the above three statements are true. Can anybody confirm that all the above three statements are true?

~ As Robert Israel asked me, I am going to explain what I have tried: ~

The $\sigma_{max}$ of a matrix $A\in\mathbb{R}^{m\times k}$ is defined as: \begin{equation} \sigma_{max} = \max_{x\in\mathbb{R}^{k}} \frac{\|Ax\|_2}{\|x\|_2} \end{equation} Now, suppose that the above expression is maximized by the vector $\hat{x}$. Then for an arbitrary vector $x\in\mathbb{R}^{k}$, the following inequality holds: \begin{equation} \sigma_{max} \geq \frac{\|Ax\|_2}{\|x\|_2}. \end{equation} The above relationship can be written as: \begin{equation} \sigma_{max}^2 \|x\|_2^2\geq \|Ax\|_2^2. \end{equation}

Now, on comparing the above inequality with the inequality given in (1), we can say that $\sigma_{max}\leq \sqrt{1+\delta}$. This proves that my statement number (3) is right.

Now, I am stuck on how I can prove that: \begin{equation} \sigma_{min}= \min_{x\in\mathbb{R}^k, x\neq 0} \frac{\|Ax\|_2}{\|x\|_2}. \end{equation} If I am able to prove the above relationship, I will be able to prove my statements number (1) and (2).

Any help would be very much appreciated!

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  • $\begingroup$ Do you have any thoughts on your question? Do you know the definitions of rank and the largest/smallest singular values? $\endgroup$
    – JKL
    Nov 23, 2022 at 17:00
  • $\begingroup$ @JKL ; Yes, I do. $\endgroup$ Nov 23, 2022 at 17:01
  • $\begingroup$ Care to share those thoughts with us? The point is that we do not want "please do my homework for me" questions. If you tell us what you've tried and where you are stuck, we can help. $\endgroup$ Nov 23, 2022 at 18:18
  • $\begingroup$ @RobertIsrael ; I have shared what I have tried to prove my statements. Please look at my edited question. $\endgroup$ Nov 24, 2022 at 12:34

1 Answer 1

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The singular values are by definition the square roots of the nonzero eigenvalues of $A^T A$. $A^T A$ is a real symmetric matrix, so its eigenvalues are real and the corresponding eigenvectors may be chosen to be real. Note that for any $x \in \mathbb R^k$, $x^T A^T A x = \|Ax\|^2$, and so if $\sigma^2$ is an eigenvalue of $A^T A$ corresponding to eigenvector $x$, $ \|A x\|^2 = x^T (\sigma^2 x) = \sigma^2 \|x\|^2$. Thus your inequality (1) implies $1 - \delta \le \sigma^2 \le 1 + \delta$.

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