Two new lower bounds for the smallest singular value

Shun Xu¹¹1Email: xushun@mail.ustc.edu.cn

Abstract

In this paper, we obtain two new lower bounds for the smallest singular value of nonsingular matrices which is better than the bound presented by Zou [1], Lin and Xie [2] under certain circumstances.

keywords:

Singular values, Frobenius norm, determinant.

^†^†journal: Applied Mathematics and Computation

\affiliation

[inst1]organization=School of Mathematical Sciences, addressline=University of Science and Technology of China, city=Hefei, postcode=230026, country=China

1 Introduction

Let $M_{n}(n\geqslant 2)$ be the space of $n\times n$ complex matrices. Let $\sigma_{i}$ $(i=1,\cdots,n)$ be the singular values of $A\in M_{n}$ which is nonsingular and suppose that $\sigma_{1}\geqslant\sigma_{2}\geqslant\cdots\geqslant\sigma_{n-1}\geqslant\sigma_{n}>0$ . For $A=\left[a_{ij}\right]\in M_{n}$ , the Frobenius norm of $A$ is defined by

\|A\|_{F}=\left(\sum_{i,j=1}^{n}\left|a_{ij}\right|^{2}\right)^{1/2}={\rm tr}\left(A^{H}A\right)^{\frac{1}{2}}

where $A^{H}$ is the conjugate transpose of $A$ . The relationship between the Frobenius norm and singular values is

\|A\|_{F}^{2}=\sigma_{1}^{2}+\sigma_{2}^{2}+\cdots+\sigma_{n}^{2}

It is well known that lower bounds for the smallest singular value $\sigma_{n}$ of a nonsingular matrix $A\in M_{n}$ have many potential theoretical and practical applications [3, 4]. Yu and Gu [5] obtained a lower bound for $\sigma_{n}$ as follows:

\sigma_{n}\geqslant|\det A|\cdot\left(\frac{n-1}{\|A\|_{F}^{2}}\right)^{(n-1)/2}=l>0

The above inequality is also shown in [6]. In [1], Zou improved the above inequality by showing that

\sigma_{n}\geqslant|{\det}A|\left(\frac{n-1}{\|A\|_{F}^{2}-l^{2}}\right)^{(n-1)/2}=l_{0}

In [2], Lin, Minghua and Xie, Mengyan improve a lower bound for smallest singular value of matrices by showing that $a$ is the smallest positive solution to the equation

x^{2}\left(\|A\|_{F}^{2}-x^{2}\right)^{n-1}=|{\det}A|^{2}(n-1)^{n-1}.

and $\sigma\geqslant a>l_{0}$ .

In this paper, we obtain two new lower bounds for the smallest singular value of nonsingular matrices. We give some numerical examples which will show that our result is better than $l_{0}$ and $a$ under certain circumstances.

2 Main results

Lemma 1.

Let

l_{0}=|{\det}A|\left(\frac{n-1}{\|A\|_{F}^{2}-l^{2}}\right)^{(n-1)/2}

then $\sigma_{n}>l_{0}$ .

Proof.

In [1], we have

\sigma_{n}\geqslant|{\det}A|\left(\frac{n-1}{\|A\|_{F}^{2}-\sigma_{n}^{2}}\right)^{(n-1)/2}

since $\sigma_{n}\geqslant l_{0}>l$ , thus

	$\displaystyle\sigma$	$\displaystyle\geqslant\|{\det}A\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-\sigma_{n}^{2}}\right)^{(n-1)/2}$
		$\displaystyle\geqslant\|{\det}A\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-l_{0}^{2}}\right)^{(n-1)/2}$
		$\displaystyle>\|{\det}A\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-l^{2}}\right)^{(n-1)/2}=l_{0}$

so $\sigma_{n}>l_{0}$ . ∎

Theorem 1.

Let $A\in M_{n}$ be nonsingular. Then

\left(l_{0}^{2}+|\det(l_{0}^{2}I_{n}-A^{H}A)|\left(\frac{n-1}{\|A\|_{F}^{2}-nl_{0}^{2}}\right)^{n-1}\right)^{1/2}=l_{1}

then $\sigma_{n}\geqslant l_{1}$ , where

l=\left|\det A\right|\left(\frac{n-1}{\|A\|_{F}^{2}}\right)^{\frac{n{-1}}{2}},l_{0}=\left|\det A\right|\left(\frac{n-1}{\|A\|_{F}^{2}-l^{2}}\right)^{\frac{n{-1}}{2}}

Proof.

Let $0<\lambda<\sigma_{n}^{2}$ , denote

\left|\left(\lambda-\sigma_{1}^{2}\right)\left(\lambda-\sigma_{2}^{2}\right)\cdots\left(\lambda-\sigma_{n-1}^{2}\right)\right|\leqslant\left(\frac{\sigma_{1}^{2}+\cdots+\sigma_{n-1}^{2}-(n-1)\lambda}{n-1}\right)^{n-1}

Since

	$\displaystyle\left\|\left(\lambda-\sigma_{1}^{2}\right)\left(\lambda-\sigma_{2}^{2}\right)\cdots\left(\lambda-\sigma_{n-1}^{2}\right)\right\|$	$\displaystyle=\frac{\left\|\left(\lambda-\sigma_{1}^{2}\right)\left(\lambda-\sigma_{2}^{2}\right)\cdots\left(\lambda-\sigma_{n}^{2}\right)\right\|}{\sigma_{n}^{2}-\lambda}$
		$\displaystyle=\frac{\|\det(\lambda I_{n}-A^{H}A)\|}{\sigma_{n}^{2}-\lambda}$

then

\frac{|\det(\lambda I_{n}-A^{H}A)|}{\sigma_{n}^{2}-\lambda}\leqslant\left(\frac{\sigma_{1}^{2}+\cdots+\sigma_{n-1}^{2}-(n-1)\lambda}{n-1}\right)^{n-1}

\sigma_{n}^{2}\geqslant\lambda+|\det(\lambda I_{n}-A^{H}A)|\left(\frac{n-1}{\sigma_{1}^{2}+\cdots+\sigma_{n-1}^{2}-(n-1)\lambda}\right)^{n-1}

\sigma_{n}\geqslant\left(\lambda+|\det(\lambda I_{n}-A^{H}A)|\left(\frac{n-1}{\sigma_{1}^{2}+\cdots+\sigma_{n-1}^{2}-(n-1)\lambda}\right)^{n-1}\right)^{1/2}

By Lemma 1, $l_{0}<\sigma_{n}$ , $l_{0}^{2}<\sigma_{n}^{2}$ , let $\lambda=l_{0}^{2}$ , then

\sigma_{n}\geqslant\left(l_{0}^{2}+|\det(l_{0}^{2}I_{n}-A^{H}A)|\left(\frac{n-1}{\|A\|_{F}^{2}-\sigma_{n}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}

(1)

Therefore

\sigma_{n}\geqslant\left(l_{0}^{2}+|\det(l_{0}^{2}I_{n}-A^{H}A)|\left(\frac{n-1}{\|A\|_{F}^{2}-nl_{0}^{2}}\right)^{n-1}\right)^{1/2}

∎

Theorem 2.

Let $A\in M_{n}$ be nonsingular. Let

b_{k+1}=\left(l_{0}^{2}+|\det(l_{0}^{2}I_{n}-A^{H}A)|\left(\frac{n-1}{\|A\|_{F}^{2}-(n-1)l_{0}^{2}-b_{k}^{2}}\right)^{n-1}\right)^{1/2},k=1,2,\cdots

with $l=\left|\det A\right|\left(\frac{n-1}{\|A\|_{F}^{2}}\right)^{\frac{n{-1}}{2}},l_{0}=\left|\det A\right|\left(\frac{n-1}{\|A\|_{F}^{2}-l^{2}}\right)^{\frac{n{-1}}{2}}$

b_{1}=\left(l_{0}^{2}+|\det(l_{0}^{2}I_{n}-A^{H}A)|\left(\frac{n-1}{\|A\|_{F}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}

then $0<b_{k}<b_{k+1}\leqslant\sigma_{n},k=1,2,\cdots$ , $\lim_{k\to\infty}b_{k}$ exists.

Proof.

We show by induction on $k$ that

\sigma_{n}\geqslant b_{k+1}>b_{k}>0

By (1), we have

	$\displaystyle\sigma_{n}$	$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-\sigma_{n}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}$
		$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{1}$

so $\sigma_{n}\geqslant b_{1}$ , then

	$\displaystyle\sigma_{n}$	$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-\sigma_{n}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}$
		$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-(n-1)l_{0}^{2}-b_{1}^{2}}\right)^{n-1}\right)^{1/2}=b_{2}$
		$\displaystyle>\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{1}>0$

When $k=1$ , we have

\sigma_{n}\geqslant b_{2}>b_{1}>0

Assume that our claim is true for $k=m$ , that is $\sigma_{n}\geqslant b_{m+1}>b_{m}>0.$ Now we consider the case when $k=m+1$ . By (1), we have

	$\displaystyle\sigma_{n}$	$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-\sigma_{n}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}$
		$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-b_{m+1}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{m+2}$
		$\displaystyle>\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-b_{m}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{m+1}>0$

Hence $\sigma_{n}\geqslant b_{m+2}>b_{m+1}>0$ . This proves $\sigma_{n}\geqslant b_{k+1}>b_{k}>0,k=1,2,\cdots$ . By the well known monotone convergence theorem, $\lim_{k\to\infty}b_{k}$ exists. ∎

Theorem 3.

Let $b=\lim_{k\rightarrow\infty}b_{k}$ ,

f(x)=\left(l_{0}^{2}+|\det(l_{0}^{2}I_{n}-A^{H}A)|\left(\frac{n-1}{\|A\|_{F}^{2}-x^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}

then $b$ is the smallest positive solution to the equation $x=f(x)$ ,and $\sigma_{n}\geqslant b$ .

Proof.

Let $x_{0}$ is the smallest positive solution to the equation $x=f(x)$ , we show by induction on $k$ that $x_{0}>b_{k},k=1,2,\cdots$ . When $k=1$

	$\displaystyle x_{0}$	$\displaystyle=\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-x_{0}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}$
		$\displaystyle>\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{1}$

Assume that our claim is true for $k=m$ , that is $\sigma_{n}>b_{m}$ . Now we consider the case when $k=m+1$ .

	$\displaystyle x_{0}$	$\displaystyle=\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-x_{0}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}$
		$\displaystyle>\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-b_{m}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{m+1}$

Hence $x_{0}>b_{m+1}$ . This proves $x_{0}>b_{k},k=1,2,\cdots$ . Since $b$ is a positive solution to the equation $x=f(x)$ and $x_{0}>b_{k},k=1,2,\cdots$ , then $b=x_{0}$ . Therefore $b$ is the smallest positive solution to the equation $x=f(x)$ and $\sigma_{n}\geqslant b$ . ∎

Therefore we obtain two new lower bounds $l_{1}$ and $b$ for the smallest singular value of nonsingular matrices.

3 Numerical examples

We use Examples 1 and Example 2 to compare the values of $l,l_{0},l_{1}$ .

Example 1.

Let

A=\left[\begin{array}[]{ccc}4&-4&-3\\ 3&4&2\\ 4&1&0\end{array}\right]

Then $\sigma_{\min}=0.0231$ , and

l=0.0229885

l_{0}=0.0229886

Our result:

l_{1}=0.0230691

Example 2.

Let

A=\left[\begin{array}[]{ccc}4&0&0\\ -1&5&0\\ 0&5&4\end{array}\right]

Then

l=1.92771

l_{0}=2.01806

Our result:

l_{1}=2.31515

Next we use the following example to compare the values of $a,b,l_{1}$ .

Example 3.

Let

A=\left[\begin{array}[]{lll}3&2&0\\ 1&9&5\\ 0&5&7\end{array}\right]

Then

a=1.0367

Our result:

l_{1}=1.3434

b=1.3455

4 Acknowledgments

I would like to express my sincere gratitude to professor Xiao-Wu Chen from University of Science and Technology of China.

References

[1] L. Zou, A lower bound for the smallest singular value, J. Math. Inequal 6 (4) (2012) 625–629.
[2] M. Lin, M. Xie, On some lower bounds for smallest singular value of matrices, Applied Mathematics Letters 121 (2021) 107411.
[3] R. A. Horn, C. R. Johnson, Matrix analysis, New York (1985).
[4] R. A. Horn, R. A. Horn, C. R. Johnson, Topics in matrix analysis, Cambridge University Press, 1994.
[5] Y. Yisheng, G. Dunhe, A note on a lower bound for the smallest singular value, Linear algebra and its Applications 253 (1-3) (1997) 25–38.
[6] G. Piazza, T. Politi, An upper bound for the condition number of a matrix in spectral norm, Journal of Computational and Applied Mathematics 143 (1) (2002) 141–144.

	$\displaystyle\sigma$	$\displaystyle\geqslant\|{\det}A\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-\sigma_{n}^{2}}\right)^{(n-1)/2}$
		$\displaystyle\geqslant\|{\det}A\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-l_{0}^{2}}\right)^{(n-1)/2}$
		$\displaystyle>\|{\det}A\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-l^{2}}\right)^{(n-1)/2}=l_{0}$

	$\displaystyle\sigma_{n}$	$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-\sigma_{n}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}$
		$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-(n-1)l_{0}^{2}-b_{1}^{2}}\right)^{n-1}\right)^{1/2}=b_{2}$
		$\displaystyle>\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{1}>0$

	$\displaystyle\sigma_{n}$	$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-\sigma_{n}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}$
		$\displaystyle\geqslant\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-b_{m+1}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{m+2}$
		$\displaystyle>\left(l_{0}^{2}+\|\det(l_{0}^{2}I_{n}-A^{H}A)\|\left(\frac{n-1}{\\|A\\|_{F}^{2}-b_{m}^{2}-(n-1)l_{0}^{2}}\right)^{n-1}\right)^{1/2}=b_{m+1}>0$