1 Introduction

Many problems in science and engineering lead to eigenvalue and singular value problems for matrices. Perturbation bounds of eigenvalues and singular values play an important role in matrix computations. Let \(S_{n}\) be the set of all n! permutations of \(\{1, 2, \ldots, n\} \). If \(x = (x_{1}, x_{2}, \ldots,x_{n})\) and \(\pi\in S_{n}\), then the vector \(x_{\pi}\) is defined as \((x_{\pi(1)}, x_{\pi(2)}, \ldots, x_{\pi(n)})\). A square matrix is called doubly stochastic if its elements are real nonnegative numbers and if the sum of the elements in each row and in each column is equal to 1. Let \(\mathbb{C}^{n\times n}\) be the set of \(n\times n\) complex matrices. Let \(A=(a_{ij})\in\mathbb{C}^{n\times n}\), we use the notation (see [1, 2])

$$\begin{aligned}& \Vert A\Vert _{p}= \Biggl(\sum _{i,j=1}^{n}\vert a_{ij}\vert ^{p} \Biggr)^{\frac{1}{p}} \quad\mbox{for } p \geq0, \end{aligned}$$
(1.1)
$$\begin{aligned}& \Vert A\Vert _{q,p}= \Biggl(\sum _{j=1}^{n} \Biggl(\sum_{k=1}^{n} \vert a_{kj}\vert ^{q} \Biggr)^{\frac{p}{q}} \Biggr)^{\frac{1}{p}} \quad\mbox{for } p>0, q>0, \frac{1}{p}+ \frac{1}{q}=1. \end{aligned}$$
(1.2)

Let \(T\in\mathbb{C}^{n\times n}\) and assume that

$$\Lambda^{(k)}= \operatorname{diag}\bigl(\lambda_{1}^{(k)}, \lambda_{2}^{(k)}, \ldots, \lambda_{n}^{(k)} \bigr) \in\mathbb{C}^{n\times n},\quad k=1, \ldots, 4, $$

are diagonal matrices. In [3], the following classical result is given:

$$ \bigl\Vert \Lambda^{(1)} T \Lambda^{(2)}- \Lambda^{(3)} T \Lambda^{(4)}\bigr\Vert _{2}^{2} \geq s_{n}^{2}(T)\sum_{i=1}^{n} \bigl\vert \lambda_{i}^{(1)}\lambda_{\pi(i)}^{(2)}- \lambda_{i}^{(3)}\lambda_{\pi(i)}^{(4)}\bigr\vert ^{2} $$
(1.3)

for some \(\pi\in S_{n}\), where \(s_{n}(T)\) is the smallest singular value of T. The inequality has many applications in bounding the (relative) perturbation for eigenvalues and singular values, such as [4–6] and the references therein. We generalize (1.3) in Section 2.

Let \(\lambda(A)\) denote the spectrum of matrix A. In 1970, Ikramov [7] defined the ‘Hölder distance \(d_{p}(\lambda(A), \lambda(B) )\) between the spectra’ of the matrices A and B, which have the eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\) and \(\mu _{1}, \mu_{2}, \ldots, \mu_{n}\), respectively, by the equation:

$$ d_{p} \bigl(\lambda(A), \lambda(B) \bigr) = \min\limits _{\pi\in S_{n}} \Biggl(\sum_{i=1}^{n}\vert \lambda_{i} - \mu_{\pi(i)}\vert ^{p} \Biggr)^{\frac{1}{p}}. $$
(1.4)

If A and B are Hermitian matrices and \(1\leq p < 2\), [7] obtained

$$ d_{p} \bigl(\lambda(A), \lambda(B) \bigr)\leq \Vert A-B \Vert _{p}, $$
(1.5)

which partially generalizes the Hoffman-Wielandt theorem [8]. However, for normal matrices (1.5) can no longer be valid. The purpose of this paper is to obtain several inequalities similar to (1.5) for diagonalizable matrices. We exhibit some upper bounds and lower bounds for \(d_{p}(\lambda(A), \lambda(B) )\) of diagonalizable matrices A and B in Section 2.

Majorization is one of the most powerful techniques for deriving inequalities. We use majorization to get some perturbation bounds for singular values. For simplicity of the notations, in most cases in this paper the vectors in \(\mathbb{R}^{n}\) are regarded as row vectors, but when they are multiplied by matrices we regard them as column vectors. Given a real vector \(x = (x_{1}, x_{2}, \ldots, x_{n}) \in\mathbb{R}^{n}\), we rearrange its components as \(x_{[1]} \geq x_{[2]} \geq\cdots\geq x_{[n]}\).

Definition 1.1

([9], p.14)

For \(x = (x_{1}, x_{2}, \ldots, x_{n})\), \(y = (y_{1}, y_{2}, \ldots, y_{n}) \in \mathbb{R}^{n}\), if

$$\sum_{i=1}^{k} x_{[i]}\leq\sum _{i=1}^{k}y_{[i]},\quad k=1, 2, \ldots,n, $$

then we say that x is weakly majorized by y and denote \(x \prec_{w} y\). If \(x \prec_{w} y\) and \(\sum_{i=1}^{n} x_{i} = \sum_{i=1}^{n}y_{i}\), then we say that x is majorized by y and denote \(x \prec y\).

Let \(s_{1}\geq s_{2} \geq\cdots\geq s_{n}\) and \(\delta_{1}\geq\delta_{2} \geq \cdots\geq\delta_{n}\) be the singular values of the complex matrices \(A=(a_{ij}) \in\mathbb{C}^{n\times n}\) and \(B=(b_{ij}) \in\mathbb{C}^{n\times n}\), respectively. In [10], p.215, and [11], p.199, the following classical result is given:

$$ \sum_{i=1}^{n} \vert s_{i}-\delta_{i}\vert ^{2}\leq\sum _{i,j=1}^{n}\vert a_{ij}-b_{ij} \vert ^{2}. $$
(1.6)

We generalize the inequality (1.6) in Section 3.

2 Perturbation bounds for eigenvalues of diagonalizable matrices

Let \(A \circ B\) denote the Hadamard product of matrices A and B. \(\Vert A\Vert \) denotes the spectral norm of matrix A. \(A^{T}\) denotes the transpose of matrix A. For two n-square real matrices A, B, we write \(A \leq_{e} B\) to mean that \(B-A\) is (entrywise) nonnegative. For \(A=(a_{ij}) \in\mathbb{C}^{n\times n}\) and a real number \(t > 0\), we denote \(A^{\vert \circ \vert t}\equiv(\vert a_{ij}\vert ^{t})\in\mathbb{C}^{n\times n}\). Let \(s_{n}(A)\) and \(s_{1}(A)\) be the smallest and the largest singular values of A, respectively. The following entrywise inequalities involve the smallest and the largest singular values.

Lemma 2.1

([9], p.52)

Let \(A \in\mathbb{C}^{n\times n}\) and let p, q be real numbers with \(0 < p \leq2\) and \(q \geq2\). Then there exist two doubly stochastic matrices \(B,C \in\mathbb{C}^{n\times n}\) such that

$$ s_{n}(A)^{p}B \leq_{e} A^{\vert \circ \vert p} $$
(2.1)

and

$$ A^{\vert \circ \vert q} \leq_{e} s_{1}(A)^{q}C. $$
(2.2)

Theorem 2.2

Let \(T\in\mathbb{C}^{n\times n}\) and let p, q be real numbers with \(0 < p \leq2\) and \(q \geq2\). Assume that \(\Lambda^{(k)}= \operatorname{diag}(\lambda _{1}^{(k)}, \lambda_{2}^{(k)}, \ldots, \lambda_{n}^{(k)}) \in\mathbb {C}^{n\times n}\), \(k=1, \ldots, 4\), are diagonal matrices. Then there are permutations π and ν of \(S_{n}\) such that

$$ \bigl\Vert \Lambda^{(1)} T \Lambda^{(2)}- \Lambda^{(3)} T \Lambda^{(4)}\bigr\Vert _{p}^{p} \geq s_{n}^{p}(T) \sum_{i=1}^{n} \bigl\vert \lambda_{i}^{(1)} \lambda_{\pi(i)}^{(2)}- \lambda_{i}^{(3)} \lambda_{\pi(i)}^{(4)}\bigr\vert ^{p} $$
(2.3)

and

$$ \bigl\Vert \Lambda^{(1)} T \Lambda^{(2)}- \Lambda^{(3)} T \Lambda^{(4)}\bigr\Vert _{q}^{q} \leq s_{1}^{q}(T) \sum_{i=1}^{n} \bigl\vert \lambda_{i}^{(1)} \lambda_{\nu(i)}^{(2)}- \lambda_{i}^{(3)} \lambda_{\nu(i)}^{(4)}\bigr\vert ^{q}. $$
(2.4)

Proof

Set \(T=(t_{ij})\in\mathbb{C}^{n\times n}\). Then

$$\begin{aligned} \bigl\Vert \Lambda^{(1)} T \Lambda^{(2)}- \Lambda^{(3)} T \Lambda^{(4)}\bigr\Vert _{p}^{p} =& \sum_{i,j=1}^{n} \vert t_{ij}\vert ^{p} \bigl\vert \lambda_{i}^{(1)} \lambda_{j}^{(2)}- \lambda_{i}^{(3)} \lambda_{j}^{(4)}\bigr\vert ^{p} \\ =&e^{T} \bigl(T^{\vert \circ \vert p}\circ M \bigr)e, \end{aligned}$$
(2.5)

where \(M=(\vert \lambda_{i}^{(1)}\lambda_{j}^{(2)}-\lambda_{i}^{(3)}\lambda _{j}^{(4)}\vert ^{p})\in\mathbb{C}^{n\times n}\), \(e=(1, 1, \ldots, 1)^{T} \in \mathbb{C}^{n}\). Applying inequality (2.1), we have

$$T^{\vert \circ \vert p}\geq s_{n}^{p}(T)B, $$

where \(B=(b_{ij})\) is a doubly stochastic matrix. Then

$$ \bigl\Vert \Lambda^{(1)} T \Lambda^{(2)}- \Lambda^{(3)} T \Lambda^{(4)}\bigr\Vert _{p}^{p} \geq e^{T} \bigl(s_{n}^{p}(T)B\circ M \bigr)e=s_{n}^{p}(T)e^{T}(B \circ M)e. $$

Since B is doubly stochastic, by Birkhoff’s theorem ([12, 13], p.527) B is a convex combination of permutation matrices:

$$ B=\sum_{i=1}^{n!}\tau_{i}P_{i},\quad \tau_{i}\geq0, \sum_{i=1}^{n!} \tau_{i}=1, P_{i} \mbox{ are permutation matrices.} $$

Suppose \(e^{T}(B\circ P_{k})e=\min\{e^{T}(B\circ P_{i})e\mid 1\leq i \leq n!\}\) and \(P_{k}\) corresponds to \(\pi\in S_{n}\). Then

$$ \begin{aligned} \bigl\Vert \Lambda^{(1)} T \Lambda^{(2)}- \Lambda^{(3)} T \Lambda^{(4)}\bigr\Vert _{p}^{p}&\geq s_{n}^{p}(T)e^{T}(B \circ M)e \\ &=s_{n}^{p}(T)e^{T} \Biggl(\sum _{i=1}^{n!}\tau_{i}P_{i}\circ M \Biggr)e \\ &\geq s_{n}^{p}(T)\sum_{i=1}^{n!} \tau_{i}e^{T}(P_{k}\circ M)e \\ &=s_{n}^{p}(T)e^{T}(P_{k}\circ M)e \\ &=s_{n}^{p}(T)\sum_{i=1}^{n} \bigl\vert \lambda_{i}^{(1)}\lambda_{\pi(i)}^{(2)}- \lambda_{i}^{(3)}\lambda_{\pi(i)}^{(4)}\bigr\vert ^{p}. \end{aligned} $$

Proving (2.3).

Use (2.2), (2.5), and the Birkhoff theorem, we can deduce the inequality (2.4). □

Remark 2.3

If we take \(p=2\), we get Theorem 3.2 in [3]. So, the bound in inequality (2.3) generalizes the bound of Theorem 3.2 in [3].

Next, we apply Theorem 2.2 to get some perturbation bounds for the eigenvalues of diagonalizable matrices. Let \(A=(a_{ij}) \in\mathbb{C}^{n\times n}\) and \(B=(b_{ij}) \in\mathbb {C}^{n\times n}\). When \(p>0\), \(q>0\), \(\frac{1}{p}+\frac{1}{q}=1\), then (see [2])

$$\begin{aligned}& \Vert AB\Vert _{p}\leq \Vert A\Vert _{p}\Vert B \Vert _{q,p}, \end{aligned}$$
(2.6)
$$\begin{aligned}& \Vert AB\Vert _{p}\leq\bigl\Vert A^{T}\bigr\Vert _{q,p}\Vert B \Vert _{p}. \end{aligned}$$
(2.7)

If B is nonsingular, then we have

$$ \begin{aligned} \frac{\Vert A\Vert _{p}}{\Vert B^{-1}\Vert _{q,p}}\leq \Vert AB\Vert _{p}. \end{aligned} $$
(2.8)

If A is nonsingular, then we have

$$ \begin{aligned} \frac{\Vert B\Vert _{p}}{\Vert (A^{-1})^{T}\Vert _{q,p}}\leq \Vert AB\Vert _{p}. \end{aligned} $$
(2.9)

For normal matrices the statement of Theorem 3 in [7] (inequality (1.5)) can no longer be valid. However, we have the following theorem.

Theorem 2.4

Assume that both \(A \in\mathbb{C}^{n\times n}\) and \(B \in\mathbb {C}^{n\times n}\) are diagonalizable and admit the following decompositions:

$$ \begin{aligned} A=D_{1}\Lambda_{1}D_{1}^{-1} \quad\textit{and}\quad B=D_{2}\Lambda_{2}D_{2}^{-1}, \end{aligned} $$
(2.10)

where \(D_{1}\) and \(D_{2}\) are nonsingular, and \(\Lambda_{1}=\operatorname{diag}(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n})\) and \(\Lambda_{2}=\operatorname{diag}(\mu_{1}, \mu_{2}, \ldots, \mu_{n})\). Then there are permutations π and ν of \(S_{n}\) such that

$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{p} \Biggr)^{\frac{1}{p}} \leq\bigl\Vert \bigl(D_{1}^{-1} \bigr)^{T}\bigr\Vert _{q,p} \Vert D_{2}\Vert _{q,p}\bigl\Vert D_{2}^{-1}\bigr\Vert \Vert D_{1}\Vert \Vert A-B\Vert _{p}, \end{aligned}$$
(2.11)
$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\nu(i)}\vert ^{p} \Biggr)^{\frac{1}{p}} \leq\bigl\Vert \bigl(D_{2}^{-1} \bigr)^{T}\bigr\Vert _{q,p} \Vert D_{1}\Vert _{q,p}\bigl\Vert D_{1}^{-1}\bigr\Vert \Vert D_{2}\Vert \Vert A-B\Vert _{p}, \end{aligned}$$
(2.12)

where \(1 < p \leq2\) and \(\frac{1}{p}+\frac{1}{q}=1\).

Proof

Using (2.10), we have

$$\begin{aligned} \Vert A-B\Vert _{p}^{p} = &\bigl\Vert D_{1} \Lambda_{1}D_{1}^{-1}-D_{2} \Lambda_{2}D_{2}^{-1}\bigr\Vert _{p}^{p} \\ =&\bigl\Vert D_{1} \bigl(\Lambda_{1}D_{1}^{-1}D_{2}-D_{1}^{-1}D_{2} \Lambda_{2} \bigr)D_{2}^{-1}\bigr\Vert _{p}^{p} \end{aligned}$$
(2.13)

and

$$ \begin{aligned}[b] \Vert A-B\Vert _{p}^{p}&=\bigl\Vert D_{1} \Lambda_{1}D_{1}^{-1}-D_{2} \Lambda_{2}D_{2}^{-1}\bigr\Vert _{p}^{p} \\ &=\bigl\Vert D_{2} \bigl(D_{2}^{-1}D_{1} \Lambda_{1}-\Lambda_{2}D_{2}^{-1}D_{1} \bigr)D_{1}^{-1}\bigr\Vert _{p}^{p}. \end{aligned} $$
(2.14)

We give a proof of (2.11) with the help of (2.13). Similarly one can prove (2.12) using (2.14). Applying (2.8) and (2.9) to (2.13) we obtain

$$ \begin{aligned} \Vert A-B\Vert _{p}^{p}\geq \frac{\Vert \Lambda_{1}D_{1}^{-1}D_{2}-D_{1}^{-1}D_{2}\Lambda_{2}\Vert _{p}^{p}}{\Vert (D_{1}^{-1})^{T}\Vert _{q,p}^{p}\Vert D_{2}\Vert _{q,p}^{p}}. \end{aligned} $$

Using inequality (2.3), there is a permutation π of \(S_{n}\) such that

$$ \begin{aligned} \bigl\Vert \Lambda_{1}D_{1}^{-1}D_{2}-D_{1}^{-1}D_{2} \Lambda_{2}\bigr\Vert _{p}^{p}\geq s_{n}^{p} \bigl(D_{1}^{-1}D_{2} \bigr)\sum_{i=1}^{n}\vert \lambda_{i} - \mu_{\pi(i)}\vert ^{p}. \end{aligned} $$

So we have

$$ \begin{aligned} \sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{p} \leq \frac{\Vert (D_{1}^{-1})^{T}\Vert _{q,p}^{p}\Vert D_{2}\Vert _{q,p}^{p}}{s_{n}^{p}(D_{1}^{-1}D_{2})} \Vert A-B\Vert _{p}^{p}. \end{aligned} $$

We use the relations

$$s_{n}^{-1}\bigl(D_{1}^{-1}D_{2} \bigr)=\bigl\Vert \bigl(D_{1}^{-1}D_{2} \bigr)^{-1}\bigr\Vert \leq \Vert D_{1}\Vert \bigl\Vert D_{2}^{-1}\bigr\Vert $$

to get the inequality in (2.11). □

Theorem 2.5

Under the hypotheses of Theorem  2.4, there are permutations π and ν of \(S_{n}\) such that

$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{q} \Biggr)^{\frac{1}{q}} \geq\frac{\Vert A-B\Vert _{q}}{\Vert (D_{1})^{T}\Vert _{p,q}\Vert D_{2}^{-1}\Vert _{p,q}\Vert D_{1}^{-1}\Vert \Vert D_{2}\Vert }, \end{aligned}$$
(2.15)
$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\nu(i)}\vert ^{q} \Biggr)^{\frac{1}{q}} \geq\frac{\Vert A-B\Vert _{q}}{\Vert (D_{2})^{T}\Vert _{p,q}\Vert D_{1}^{-1}\Vert _{p,q}\Vert D_{1}\Vert \Vert D_{2}^{-1}\Vert }, \end{aligned}$$
(2.16)

where \(q \geq2\) and \(\frac{1}{p}+\frac{1}{q}=1\).

Proof

Using (2.10), we have

$$ \begin{aligned}[b] \Vert A-B\Vert _{q}^{q}&=\bigl\Vert D_{1} \Lambda_{1}D_{1}^{-1}-D_{2} \Lambda_{2}D_{2}^{-1}\bigr\Vert _{q}^{q} \\ &=\bigl\Vert D_{1} \bigl(\Lambda_{1}D_{1}^{-1}D_{2}-D_{1}^{-1}D_{2} \Lambda_{2} \bigr)D_{2}^{-1}\bigr\Vert _{q}^{q} \end{aligned} $$
(2.17)

and

$$ \begin{aligned}[b] \Vert A-B\Vert _{q}^{q}&=\bigl\Vert D_{1} \Lambda_{1}D_{1}^{-1}-D_{2} \Lambda_{2}D_{2}^{-1}\bigr\Vert _{q}^{q} \\ &=\bigl\Vert D_{2} \bigl(D_{2}^{-1}D_{1} \Lambda_{1}-\Lambda_{2}D_{2}^{-1}D_{1} \bigr)D_{1}^{-1}\bigr\Vert _{q}^{q}. \end{aligned} $$
(2.18)

Applying (2.6) and (2.7) to (2.17) we obtain

$$ \begin{aligned} \Vert A-B\Vert _{q}^{q}\leq \bigl\Vert (D_{1})^{T}\bigr\Vert _{p,q}^{q} \bigl\Vert \Lambda_{1}D_{1}^{-1}D_{2}-D_{1}^{-1}D_{2} \Lambda_{2}\bigr\Vert _{q}^{q}\bigl\Vert D_{2}^{-1} \bigr\Vert _{p,q}^{q}. \end{aligned} $$

Using inequality (2.4), there exists a permutation π of \(S_{n}\) such that

$$ \begin{aligned} \bigl\Vert \Lambda_{1}D_{1}^{-1}D_{2}-D_{1}^{-1}D_{2} \Lambda_{2}\bigr\Vert _{q}^{q}\leq s_{1}^{q} \bigl(D_{1}^{-1}D_{2} \bigr)\sum_{i=1}^{n}\vert \lambda_{i} - \mu_{\pi(i)}\vert ^{q}, \end{aligned} $$

so we have

$$ \begin{aligned} \sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{q} \geq \frac{\Vert A-B\Vert _{q}^{q}}{\Vert (D_{1})^{T}\Vert _{p,q}^{q}\Vert D_{2}^{-1}\Vert _{p,q}^{q}s_{1}^{q}(D_{1}^{-1}D_{2})}. \end{aligned} $$

We use the relations

$$s_{1}\bigl(D_{1}^{-1}D_{2}\bigr)=\bigl\Vert D_{1}^{-1}D_{2}\bigr\Vert \leq\bigl\Vert D_{1}^{-1}\bigr\Vert \Vert D_{2}\Vert $$

to get the inequality in (2.15).

The proof of inequality (2.16) is similar to the proof of (2.15) and is omitted here. □

For \(1 \leq p \leq2\), it is well known [2] that the scalar function (1.1) of a matrix A is a submultiplicative matrix norm. However, it is true for \(0 < p \leq2\). Actually, according to the Cauchy-Schwarz inequality, we have

$$ \begin{aligned} \Vert AB\Vert _{p}^{p}&= \sum _{i=1}^{n}\sum_{j=1}^{n} \Biggl\vert \sum_{k=1}^{n} a_{ik}b_{kj}\Biggr\vert ^{p} \\ &\leq\sum_{i=1}^{n}\sum _{j=1}^{n} \Biggl( \Biggl(\sum _{k=1}^{n} \vert a_{ik}\vert ^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum_{k=1}^{n} \vert b_{kj}\vert ^{2} \Biggr)^{\frac{1}{2}} \Biggr)^{p}. \end{aligned} $$

Since for fixed vector \(x=(x_{1}, x_{2}, \ldots, x_{n})\), the function \(p \rightarrow(\vert x_{1}\vert ^{p}+\vert x_{2}\vert ^{p}+\cdots+\vert x_{n}\vert ^{p})^{\frac{1}{p}}\) is decreasing on \((0,\infty)\),

$$ \begin{aligned} \Vert AB\Vert _{p}^{p}&\leq\sum _{i=1}^{n}\sum_{j=1}^{n} \Biggl(\Biggl(\sum_{k=1}^{n} \vert a_{ik}\vert ^{p} \Biggr) \Biggl(\sum _{k=1}^{n}\vert b_{kj}\vert ^{p} \Biggr)\Biggr) \\ &= \Biggl(\sum_{i=1}^{n}\sum _{k=1}^{n} \vert a_{ik}\vert ^{p} \Biggr) \Biggl(\sum_{j=1}^{n} \sum_{k=1}^{n}\vert b_{kj}\vert ^{p} \Biggr) \\ &=\Vert A\Vert _{p}^{p}\Vert B\Vert _{p}^{p}. \end{aligned} $$

That is,

$$ \Vert AB\Vert _{p}\leq \Vert A\Vert _{p}\Vert B \Vert _{p}. $$
(2.19)

If B is nonsingular, then

$$ \Vert A\Vert _{p}=\bigl\Vert ABB^{-1}\bigr\Vert _{p}\leq \Vert AB \Vert _{p}\bigl\Vert B^{-1} \bigr\Vert _{p}. $$

So we have

$$ \frac{\Vert A\Vert _{p}}{\Vert B^{-1}\Vert _{p}}\leq \Vert AB\Vert _{p}. $$
(2.20)

Similarly, when A is nonsingular, then

$$ \Vert B\Vert _{p}=\bigl\Vert A^{-1}AB\bigr\Vert _{p}\leq\bigl\Vert A^{-1} \bigr\Vert _{p} \Vert AB\Vert _{p}. $$

That is,

$$ \frac{\Vert B\Vert _{p}}{\Vert A^{-1}\Vert _{p}}\leq \Vert AB\Vert _{p}. $$
(2.21)

Theorem 2.6

Under the assumptions of Theorem  2.4, there are permutations π and ν of \(S_{n}\) such that

$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{p} \Biggr)^{\frac{1}{p}} \leq\bigl\Vert D_{1}^{-1}\bigr\Vert _{p}\Vert D_{2} \Vert _{p}\bigl\Vert D_{2}^{-1}\bigr\Vert \Vert D_{1}\Vert \Vert A-B \Vert _{p}, \end{aligned}$$
(2.22)
$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\nu(i)}\vert ^{p} \Biggr)^{\frac{1}{p}} \leq\bigl\Vert D_{2}^{-1}\bigr\Vert _{p}\Vert D_{1} \Vert _{p}\bigl\Vert D_{1}^{-1}\bigr\Vert \Vert D_{2}\Vert \Vert A-B \Vert _{p}, \end{aligned}$$
(2.23)

where \(0 < p \leq2\).

Proof

The proof is similar to the proof of Theorem 2.4 and is omitted here. □

Remark 2.7

Since

$$ \Vert A\Vert _{q,p}= \Biggl(\sum_{j=1}^{n} \Biggl(\sum_{k=1}^{n}\vert a_{kj}\vert ^{q} \Biggr)^{\frac{p}{q}} \Biggr)^{\frac{1}{p}}\leq \Biggl(\sum_{j=1}^{n} \Biggl(\sum_{k=1}^{n}\vert a_{kj}\vert ^{p} \Biggr)^{\frac{p}{p}} \Biggr)^{\frac{1}{p}}=\Vert A\Vert _{p} $$

and

$$ \bigl\Vert A^{T}\bigr\Vert _{q,p}\leq\bigl\Vert A^{T}\bigr\Vert _{p}=\Vert A \Vert _{p} $$

for \(1< p \leq2\) and \(\frac{1}{p}+\frac{1}{q}=1\), the bounds in (2.11) and (2.12) are always sharper than those in (2.22) and (2.23), respectively.

When \(p = q = 2\), then \(\Vert AB\Vert _{2} \leq \min\{\Vert A\Vert _{2}\Vert B\Vert , \Vert A\Vert \Vert B\Vert _{2}\} \). We obtain

$$\begin{aligned}& \begin{aligned} \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{2} \Biggr)^{\frac{1}{2}}& \leq\bigl\Vert D_{1}^{-1}\bigr\Vert \Vert D_{2}\Vert \bigl\Vert D_{2}^{-1}D_{1} \bigr\Vert \Vert A-B\Vert _{2} \\ &\leq\bigl\Vert D_{1}^{-1}\bigr\Vert \Vert D_{2}\Vert \bigl\Vert D_{2}^{-1}\bigr\Vert \Vert D_{1}\Vert \Vert A-B\Vert _{2}, \end{aligned} \\& \begin{aligned} \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\nu(i)}\vert ^{2} \Biggr)^{\frac{1}{2}}& \leq\bigl\Vert D_{2}^{-1}\bigr\Vert \Vert D_{1}\Vert \bigl\Vert D_{1}^{-1}D_{2} \bigr\Vert \Vert A-B\Vert _{2} \\ &\leq\bigl\Vert D_{2}^{-1}\bigr\Vert \Vert D_{1}\Vert \bigl\Vert D_{1}^{-1}\bigr\Vert \Vert D_{2}\Vert \Vert A-B\Vert _{2}. \end{aligned} \end{aligned}$$

Since

$$ \Vert A\Vert _{p,q}= \Biggl(\sum_{j=1}^{n} \Biggl(\sum_{k=1}^{n}\vert a_{kj}\vert ^{p} \Biggr)^{\frac{q}{p}} \Biggr)^{\frac{1}{q}}\leq \Biggl(\sum_{j=1}^{n} \Biggl(\sum_{k=1}^{n}\vert a_{kj}\vert ^{p} \Biggr)^{\frac{p}{p}} \Biggr)^{\frac{1}{p}}=\Vert A\Vert _{p} $$

and

$$ \bigl\Vert A^{T}\bigr\Vert _{p,q}\leq\bigl\Vert A^{T}\bigr\Vert _{p}=\Vert A \Vert _{p} $$

for \(q\geq2\) and \(\frac{1}{p}+\frac{1}{q}=1\), we have the following corollary.

Corollary 2.8

Under the same conditions as in Theorem  2.4, there are permutations π and ν of \(S_{n}\) such that

$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{q} \Biggr)^{\frac{1}{q}} \geq\frac{\Vert A-B\Vert _{q}}{\Vert D_{1}\Vert _{p}\Vert D_{2}^{-1}\Vert _{p}\Vert D_{1}^{-1}\Vert \Vert D_{2}\Vert },\\& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\nu(i)}\vert ^{q} \Biggr)^{\frac{1}{q}} \geq\frac{\Vert A-B\Vert _{q}}{\Vert D_{1}^{-1}\Vert _{p}\Vert D_{2}\Vert _{p}\Vert D_{1}\Vert \Vert D_{2}^{-1}\Vert }, \end{aligned}$$

where \(q \geq2\) and \(\frac{1}{p}+\frac{1}{q}=1\).

Remark 2.9

When \(p = q = 2\), we obtain

$$\begin{aligned}& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\pi(i)}\vert ^{2} \Biggr)^{\frac{1}{2}} \geq\frac{\Vert A-B\Vert _{2}}{\Vert D_{1}\Vert \Vert D_{2}^{-1}\Vert \Vert D_{1}^{-1}D_{2}\Vert } \geq\frac{\Vert A-B\Vert _{2}}{\Vert D_{1}\Vert \Vert D_{2}^{-1}\Vert \Vert D_{1}^{-1}\Vert \Vert D_{2}\Vert }, \\& \Biggl(\sum_{i=1}^{n} \vert \lambda_{i} - \mu_{\nu(i)}\vert ^{2} \Biggr)^{\frac{1}{2}} \geq\frac{\Vert A-B\Vert _{2}}{\Vert D_{1}^{-1}\Vert \Vert D_{2}\Vert \Vert D_{2}^{-1}D_{1}\Vert } \geq\frac{\Vert A-B\Vert _{2}}{\Vert D_{1}^{-1}\Vert \Vert D_{2}\Vert \Vert D_{1}\Vert \Vert D_{2}^{-1}\Vert }. \end{aligned}$$

3 Perturbation bounds for singular values

For brevity we only consider square matrices. The generalizations from square matrices to rectangular matrices are obvious, and usually problems on singular values of rectangular matrices can be converted to the case of square matrices by adding zero rows or zero columns.

For a Hermitian matrix \(G \in\mathbb{C}^{n \times n}\), we always denote \(\lambda(G)=(\lambda_{1}(G), \lambda_{2}(G), \ldots, \lambda _{n}(G))\), where \(\lambda_{1}(G)\geq\lambda_{2}(G)\geq\cdots\geq\lambda _{n}(G)\) are the eigenvalues of G in decreasing order.

Lemma 3.1

(Lidskii [14], Lemma 3.18 [9])

If \(G,H \in\mathbb {C}^{n\times n}\) are Hermitian matrices, then

$$\lambda(G)-\lambda(H) \prec\lambda(G-H). $$

Lemma 3.2

([9], p.18)

Let \(f(t)\) be a convex function, \(x = (x_{1}, x_{2}, \ldots, x_{n})\), \(y = (y_{1}, y_{2}, \ldots, y_{n}) \in\mathbb {R}^{n}\). Then

$$ \begin{aligned} x \prec y\quad\implies\quad \bigl(f(x_{1}), f(x_{2}), \ldots, f(x_{n}) \bigr) \prec_{w} \bigl(f(y_{1}), f(y_{2}), \ldots, f(y_{n}) \bigr). \end{aligned} $$

Theorem 3.3

Let \(\sigma_{1}(A)\geq\sigma_{2}(A) \geq\cdots\geq\sigma_{n}(A)\), \(\sigma _{1}(B) \geq\sigma_{2}(B) \geq\cdots\geq\sigma_{n}(B)\) and \(\sigma_{1}(A-B) \geq\sigma_{2}(A-B) \geq\cdots\geq\sigma_{n}(A-B)\) be the singular values of the complex matrices \(A=(a_{ij})\), \(B=(b_{ij})\) and \(A-B\), respectively. Then

$$\begin{aligned}& \sum _{i=1}^{n} \bigl\vert \sigma_{i}(A)- \sigma_{i}(B)\bigr\vert ^{p}\leq\sum _{i,j=1}^{n}\vert a_{ij}-b_{ij} \vert ^{p}, \end{aligned}$$
(3.1)
$$\begin{aligned}& \sum _{i=1}^{n} \bigl\vert \sigma_{i}(A)- \sigma_{i}(B)\bigr\vert ^{q}\geq\sum _{i=1}^{n}\sigma_{i}^{q}(A-B), \end{aligned}$$
(3.2)

where \(1\leq p \leq2\), \(0< q \leq1\).

Proof

Let \(\varphi(A)\triangleq \left({\scriptsize\begin{matrix}{} 0 & A^{*} \cr A & 0\end{matrix}}\right)\), \(\varphi(B)\triangleq \left({\scriptsize\begin{matrix}{} 0 & B^{*} \cr B & 0\end{matrix}}\right)\). Then

$$\varphi(A-B)= \left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} 0 & (A-B)^{*} \\ A-B & 0 \end{array}\displaystyle \right ). $$

\(\varphi(A)\), \(\varphi(B)\), \(\varphi(A-B)\) are three Hermitian matrices. Assume that \(U_{1}^{*}AV_{1}= \operatorname{diag}(\sigma_{1}(A), \sigma_{2}(A), \ldots, \sigma_{n}(A))\), \(U_{2}^{*}BV_{2}= \operatorname{diag}(\sigma_{1}(B), \sigma_{2}(B), \ldots, \sigma_{n}(B))\) and \(U_{3}^{*}(A-B)V_{3}= \operatorname{diag}(\sigma_{1}(A-B), \sigma_{2}(A-B), \ldots, \sigma _{n}(A-B))\) are singular value decompositions with \(U_{1}\), \(U_{2}\), \(U_{3}\), \(V_{1}\), \(V_{2}\), \(V_{3}\) unitary. Then

$$Q_{1}\triangleq\frac{1}{\sqrt{2}}\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} V_{1} & V_{1} \\ U_{1} & -U_{1} \end{array}\displaystyle \right ),\qquad Q_{2}\triangleq\frac{1}{\sqrt{2}}\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} V_{2} & V_{2} \\ U_{2} & -U_{2} \end{array}\displaystyle \right ) \quad\mbox{and}\quad Q_{3} \triangleq\frac{1}{\sqrt{2}}\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} V_{3} & V_{3} \\ U_{3} & -U_{3} \end{array}\displaystyle \right ) $$

are unitary matrices and

$$\begin{aligned}& Q_{1}^{*}\varphi(A)Q_{1}=\operatorname{diag} \bigl( \sigma_{1}(A), \sigma_{2}(A), \ldots, \sigma_{n}(A), -\sigma_{1}(A), -\sigma_{2}(A), \ldots, - \sigma_{n}(A) \bigr), \\& Q_{2}^{*}\varphi(B)Q_{2}=\operatorname{diag}\bigl( \sigma_{1}(B), \sigma_{2}(B), \ldots, \sigma_{n}(B), -\sigma_{1}(B), -\sigma_{2}(B), \ldots, - \sigma_{n}(B)\bigr) \end{aligned}$$

and

$$\begin{aligned}& \begin{aligned} Q_{3}^{*}\varphi(A-B)Q_{3} =& \operatorname{diag} \bigl(\sigma_{1}(A-B), \sigma_{2}(A-B), \ldots, \sigma_{n}(A-B), \\ &{}-\sigma_{1}(A-B), {-}\sigma_{2}(A-B), \ldots, - \sigma_{n}(A-B) \bigr). \end{aligned} \end{aligned}$$

By Lemma 3.1, we have

$$\begin{aligned} & \bigl(\sigma_{1}(A)- \sigma_{1}(B), \sigma_{2}(A)-\sigma_{2}(B), \ldots, \sigma_{n}(A)-\sigma_{n}(B), \\ &\qquad \sigma_{n}(B)-\sigma_{n}(A), \ldots, \sigma_{2}(B)- \sigma_{2}(A), \sigma_{1}(B)-\sigma_{1}(A) \bigr) \\ & \quad \prec \bigl(\sigma_{1}(A-B), \sigma_{2}(A-B), \ldots, \sigma_{n}(A-B), \\ &\qquad{} {-}\sigma_{n}(A-B), \ldots, -\sigma_{2}(A-B), - \sigma_{1}(A-B) \bigr). \end{aligned}$$
(3.3)

First consider the case \(1\leq p\leq2\). Since the function \(f(t)=\vert t\vert ^{p}\) is convex on \((-\infty, +\infty)\), applying Lemma 3.2 with \(f(t)\) to the majorization (3.3) yields

$$ \begin{aligned} & \bigl(\bigl\vert \sigma_{1}(A)- \sigma_{1}(B)\bigr\vert ^{p}, \bigl\vert \sigma_{2}(A)-\sigma_{2}(B)\bigr\vert ^{p}, \ldots, \bigl\vert \sigma_{n}(A)-\sigma_{n}(B)\bigr\vert ^{p}, \\ &\qquad\bigl\vert \sigma_{n}(B)-\sigma_{n}(A)\bigr\vert ^{p}, \ldots, \bigl\vert \sigma_{2}(B)- \sigma_{2}(A)\bigr\vert ^{p}, \bigl\vert \sigma_{1}(B)-\sigma_{1}(A)\bigr\vert ^{p} \bigr) \\ &\quad \prec_{w} \bigl(\sigma_{1}^{p}(A-B), \sigma_{2}^{p}(A-B), \ldots, \sigma_{n}^{p}(A-B), \sigma_{n}^{p}(A-B), \ldots, \sigma_{2}^{p}(A-B), \sigma_{1}^{p}(A-B) \bigr). \end{aligned} $$

In particular,

$$ \begin{aligned}[b] &\bigl(\bigl\vert \sigma_{1}(A)- \sigma_{1}(B)\bigr\vert ^{p}, \bigl\vert \sigma_{2}(A)- \sigma_{2}(B)\bigr\vert ^{p}, \ldots, \bigl\vert \sigma_{n}(A)- \sigma_{n}(B)\bigr\vert ^{p} \bigr) \\ &\quad\prec_{w} \bigl(\sigma_{1}^{p}(A-B), \sigma_{2}^{p}(A-B), \ldots, \sigma_{n}^{p}(A-B) \bigr). \end{aligned} $$
(3.4)

According to Theorem 1 of [15] or Theorem 3.32 of [9], we have

$$ \begin{aligned} \sum _{i=1}^{n} \sigma_{i}^{p}(A-B) \leq\sum_{i,j=1}^{n}\vert a_{ij}-b_{ij}\vert ^{p} \end{aligned} $$
(3.5)

for \(1\leq p\leq2\). Combining (3.4) and (3.5), we obtain (3.1).

When \(0< q\leq1\), by considering the convex function \(g(t)=-\vert t\vert ^{q}\), on \((-\infty, +\infty)\), applying Lemma 3.2 with \(g(t)\) to the majorization (3.3) yields

$$ \begin{aligned} & \bigl(-\bigl\vert \sigma_{1}(A)- \sigma_{1}(B)\bigr\vert ^{q}, \ldots, -\bigl\vert \sigma_{n}(A)-\sigma_{n}(B)\bigr\vert ^{q}, -\bigl\vert \sigma_{n}(B)-\sigma_{n}(A)\bigr\vert ^{q}, \ldots, -\bigl\vert \sigma_{1}(B)- \sigma_{1}(A)\bigr\vert ^{q} \bigr) \\ &\quad \prec_{w} \bigl(-\sigma_{1}^{q}(A-B), \ldots, - \sigma_{n}^{q}(A-B), -\sigma_{n}^{q}(A-B), \ldots, -\sigma_{1}^{q}(A-B) \bigr). \end{aligned} $$

In particular,

$$ \begin{aligned}[b] &\bigl(-\bigl\vert \sigma_{1}(A)- \sigma_{1}(B)\bigr\vert ^{q}, - \bigl\vert \sigma_{2}(A)- \sigma_{2}(B)\bigr\vert ^{q}, \ldots, -\bigl\vert \sigma_{n}(A)- \sigma_{n}(B)\bigr\vert ^{q} \bigr) \\ &\quad\prec_{w} \bigl(-\sigma_{1}^{q}(A-B), - \sigma_{2}^{q}(A-B), \ldots, -\sigma_{n}^{q}(A-B) \bigr). \end{aligned} $$
(3.6)

From (3.6), we get (3.2). □

Remark 3.4

From inequality (3.1), if \(p=2\), we get the inequality (1.6). So the inequality (3.1) generalizes the inequality (3.5.33) of [10], p.215, and Theorem 3.12 of [11], p.199.