# For any $A\in \mathbb{R}^{n\times m}$, there exists a constant $a>0$ such that $A^TAA^TA-aA^TA \geq 0$.

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!

## 2 Answers

$$A^{T}A$$ 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

• Thank you for your answer! Commented Jul 19, 2023 at 11:21

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.