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Let $A\in \mathbb{R}^{n\times m}$ be an arbitrary matrix with any positive integer $n$ and $m$. I would like to show that there exists a positive constant $a>0$ such that $$x^T(A^TAA^TA - aA^TA)x\geq 0$$ for all $x\in \mathbb{R}^m$.

I have zero ideas on it, but, since we know that $A^TA$ has always nonnegative eigenvalue, $a=0$ works anyway...which is not our case. I guess the key should be showing that $Ax$ is not an eigenvector with zero eigenvalue..? I have tested with some matrices and it seems true. Any idea would be appreciated. Thank you in advance!

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2 Answers 2

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$A^{T}A$ is semi-definite, we can find an orthogonal matrix $P$ such that $$A^{T}A=P^{T}\mathrm{diag}\left\{ \lambda _1,\lambda _2,\cdots ,\lambda _m \right\}P$$

where $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_s>\lambda_{s+1}=\cdots=\lambda_m= 0$.

Let $x=Py$, we can get $x^T\left( A^TAA^TA-aA^TA \right) x=y^T\left( \mathrm{diag}\left\{ \lambda _{1}^{2},\lambda _{2}^{2},\cdots ,\lambda _{m}^{2} \right\} -\mathrm{diag}\left\{ a\lambda _1,a\lambda _2,\cdots ,a\lambda _m \right\} \right) y$.

We only need $\mathrm{diag}\left\{ \lambda _{1}^{2},\lambda _{2}^{2},\cdots ,\lambda _{m}^{2} \right\} -\mathrm{diag}\left\{ a\lambda _1,a\lambda _2,\cdots ,a\lambda _m \right\} $is semi-definite, i.e. $$\lambda _{i}^{2}-a\lambda _i\geqslant 0~~(1\le i\le m).$$

Thus we can get $0<a\le \lambda_s$

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  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Lev Bahn
    Commented Jul 19, 2023 at 11:21
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Given a real matrix $A$, we can perform the Singular Value Decomposition (SVD) to obtain $A = U \Sigma V^T$, where $U$ and $V$ are orthogonal matrices and $\Sigma$ is a diagonal matrix containing the singular values of $A$. The singular values are the square roots of the eigenvalues of $A^TA$.

In terms of the SVD, we have $A^TA = V \Sigma^2 V^T$ and $A^TAA^TA = V \Sigma^4 V^T$. The eigenvalues of $A^TA$ are the squares of the singular values of $A$, and the eigenvalues of $A^TAA^TA$ are the fourth powers of the singular values of $A$.

We want to find a positive constant $a$ such that $A^TAA^TA - aA^TA$ is positive semi-definite. In terms of the SVD, this becomes $V (\Sigma^4 - a\Sigma^2) V^T$. For this matrix to be positive semi-definite, all the eigenvalues of $\Sigma^4 - a\Sigma^2$ must be nonnegative. These eigenvalues are the fourth powers of the singular values of $A$ minus $a$ times the squares of the singular values. Therefore, we can choose $a$ to be larger than the maximum fourth power of the singular values of $A$. This would ensure that all the eigenvalues of $\Sigma^4 - a\Sigma^2$ are nonpositive, which would make $V (\Sigma^4 - a\Sigma^2) V^T$ positive semi-definite.

Therefore, for any real matrix $A$, there exists a positive constant $a$ such that $A^TAA^TA - aA^TA$ is positive semi-definite.

This proof uses the properties of the SVD, the properties of positive semi-definite matrices, and the properties of eigenvalues to show that such an $a$ exists. However, it's important to note that this argument assumes that all the singular values of $A$ are real. If $A$ has complex singular values, the situation could be more complex. In the case where $A$ has complex singular values, we would need to consider the complex conjugate transpose $A^H$ instead of the transpose $A^T$, and the proof would involve the properties of Hermitian matrices, which generalize the properties of real symmetric matrices to the complex case.

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