Some remarks on the monotonicity of primary matrix functions on the set of symmetric matrices

Abstract

This note contains some observations on primary matrix functions and different notions of monotonicity with relevance toward constitutive relations in nonlinear elasticity. Focusing on primary matrix functions on the set of symmetric matrices, we discuss and compare different criteria for monotonicity. The demonstrated results are particularly applicable to computations involving the true-stress–true-strain monotonicity condition, a constitutive inequality recently introduced in an Arch. Appl. Mech. article by C.S. Jog and K.D. Patil. We also clarify a statement by Jog and Patil from the same article which could be misinterpreted.

Appendix

1.1 On the derivative of the determinant function

Consider the first-order approximation

$$\begin{aligned} \det (A+H) = \det A + D\det [A].H + \underbrace{\cdots }_{{\begin{array}{c} \text {higher order}\\ \text {terms} \end{array}}} \end{aligned}$$

of the determinant function at a diagonal matrix $A={{\mathrm{diag}}}(a_1,\ldots ,a_n)$. First we assume that $H\in {{\mathrm{Sym}}}(n)$ is an off-diagonal matrix of the form

$$\begin{aligned} H = H^{\mathrm {off}} = \left( {\begin{matrix} 0&{}&{}h\\ {} &{}\ddots &{}\\ h&{}&{}0 \end{matrix}}\right) \end{aligned}$$

(17)

with $h\in \mathbb {R}$. We compute

$$\begin{aligned}&\det \begin{pmatrix} \vec {a}_1 + \left( {\begin{matrix} 0\\ \vdots \\ h \end{matrix}}\right) , \, \vec {a}_2\,\,,\,\, \cdots \,\,, \,\, \vec {a}_n + \left( {\begin{matrix} h\\ \vdots \\ 0 \end{matrix}}\right) \end{pmatrix} \nonumber \\&\quad = \det \left( \vec {a}_1, \, \vec {a}_2, \, \cdots , \, \vec {a}_n + \left( {\begin{matrix} h\\ \vdots \\ 0 \end{matrix}}\right) \right) \,+\, \det \left( \left( {\begin{matrix} 0\\ \vdots \\ h \end{matrix}}\right) , \, \vec {a}_2, \, \cdots , \, \vec {a}_n + \left( {\begin{matrix} h\\ \vdots \\ 0 \end{matrix}}\right) \right) \nonumber \\&\quad = \det \left( \vec {a}_1, \, \vec {a}_2, \, \cdots , \, \vec {a}_n \right) \,+\, \det \left( \vec {a}_1, \, \vec {a}_2, \, \cdots , \, \vec {a}_n + \left( {\begin{matrix} h\\ \vdots \\ 0 \end{matrix}}\right) \right) \nonumber \\&\quad \;\;+\, \det \left( \left( {\begin{matrix} 0\\ \vdots \\ h \end{matrix}}\right) , \, \vec {a}_2, \, \cdots , \, \vec {a}_n \right) \,+\, \det \left( \left( {\begin{matrix} 0\\ \vdots \\ h \end{matrix}}\right) , \, \vec {a}_2, \, \cdots , \, \left( {\begin{matrix} h\\ \vdots \\ 0 \end{matrix}}\right) \right) . \end{aligned}$$

(18)

Since $A$ is diagonal by assumption, the column vector $\vec {a}_1$ has the form $\vec {a}_1 = \begin{pmatrix} a_{1}&\cdots&0 \end{pmatrix}^T$ and $\vec {a}_n$ has the form $\vec {a}_n = \begin{pmatrix} 0&\cdots&a_{n} \end{pmatrix}^T$. Therefore the vectors $\vec {a}_1$ and $\begin{pmatrix} h&\cdots&0 \end{pmatrix}^T$ as well as $\vec {a}_n$ and $\begin{pmatrix} 0&\cdots&h \end{pmatrix}^T$ are linearly dependent. Thus (18) reduces to

$$\begin{aligned} \det \left( \vec {a}_1, \, \vec {a}_2, \, \cdots , \, \vec {a}_n\right) \,+\, \det \left( \left( {\begin{matrix} 0\\ \vdots \\ h \end{matrix}}\right) , \, \vec {a}_2, \, \cdots , \, \left( {\begin{matrix} h\\ \vdots \\ 0 \end{matrix}}\right) \right) \,=\, \det A \,+\, \underbrace{\det \left( \left( {\begin{matrix} 0\\ \vdots \\ h \end{matrix}}\right) , \, \vec {a}_2, \, \cdots , \, \left( {\begin{matrix} h\\ \vdots \\ 0 \end{matrix}}\right) \right) }_{=:R(h)}. \end{aligned}$$

The term $R(h)$ is quadratic in $h$, thus the linear approximation is simply

$$\begin{aligned} D\det [A].H = 0 \end{aligned}$$

for a matrix $H\in {{\mathrm{Sym}}}(n)$ of the form (17). Through similar computations, it is easy to show that $D\det [A].H = 0$ for any off-diagonal $H\in {{\mathrm{Sym}}}(n)$.

To find the derivative $D\det [A].1\!\!1$ we compute

$$\begin{aligned} \det (A+h1\!\!1)&= \det \left( {\begin{matrix} a_1+h &{} &{}0\\ {} &{}\ddots &{}\\ 0&{}&{}a_n+h \end{matrix}}\right) = \prod _{i=1}^n (a_i+h) = \prod _{i=1}^n a_i + \left( \sum _{i=1}^n\prod _{\begin{array}{c} j=1\\ j\ne i \end{array}}^n a_j\right) \cdot h + h^2\cdot [\dots ], \end{aligned}$$

thus

$$\begin{aligned} D\det [A].1\!\!1 = \frac{\mathrm {d}}{\mathrm {dh}}\det (A+h1\!\!1) = \sum _{i=1}^n \prod _{\begin{array}{c} j=1\\ j\ne i \end{array}}^n a_j. \end{aligned}$$

(19)

For $n=3$ we obtain

$$\begin{aligned} D\det [A].1\!\!1 = a_1 a_2 + a_2 a_3 + a_1 a_3. \end{aligned}$$

Furthermore, if we assume that 0 is a simple eigenvalue of $A$ [which is the case for matrices of the form $D-\lambda _i(D) 1\!\!1$ where $D$ is a diagonal matrix with simple eigenvalues; such matrices appear in Eq. (22)], then (19) can be written as

$$\begin{aligned} D\det [A].1\!\!1 = \prod _{\begin{array}{c} j=1\\ j\ne k \end{array}}^n a_j \ne 0, \end{aligned}$$

where 0 is the $k$th eigenvalue of $A$.

1.2 On the derivative of isotropic functions

Lemma 7.1

Let $W:{{\mathrm{Sym}}}(n)\rightarrow \mathbb {R}$ be an isotropic real valued function, i.e.

Then

$$\begin{aligned} DW[Q^TXQ] = Q^TDW[X]Q. \end{aligned}$$

Proof

We directly compute:

$$\begin{aligned}&W(Q^T(X+H)Q) = W(X+H)\\&\quad \Rightarrow W(Q^TXQ + Q^THQ) = W(X) + \left\langle DW[X],\,H \right\rangle + \cdots \\&\quad \Rightarrow \mathop {\overbrace{W(Q^TXQ)}}\limits ^{{=W(X)}} + \left\langle DW[Q^TXQ],\,Q^THQ \right\rangle + \cdots = W(X) + \left\langle DW[X],\,H \right\rangle + \cdots \\&\quad \Rightarrow \left\langle DW[Q^TXQ],\,Q^THQ \right\rangle = \left\langle DW[X],\,H \right\rangle \\&\quad \Rightarrow \left\langle QDW[Q^TXQ]Q^T,\,H \right\rangle = \left\langle DW[X],\,H \right\rangle . \end{aligned}$$

Since this holds for all $H\in {{\mathrm{Sym}}}(n)$, we obtain

$$\begin{aligned} QDW[Q^TXQ]Q^T = DW[X] \end{aligned}$$

and thus

$$\begin{aligned} DW[Q^TXQ] = Q^TDW[X]Q. \end{aligned}$$

$\square$

1.3 The eigenvalue function

We could also try to prove Proposition 2.5 for the more general case of non-analytic functions by directly computing the derivative of the function

$$\begin{aligned} W: {{\mathrm{Sym}}}(n)\rightarrow \mathbb {R},\quad W(A) = \sum _{i=1}^n F(\lambda _i(A)). \end{aligned}$$

Unfortunately, while the derivative of $W$ at a point $A\in {{\mathrm{Sym}}}(n)$ in directions $H$ can be explicitly computed if $A$ and $H$ commute, it is difficult to do so for arbitrary choices of $H\in {{\mathrm{Sym}}}(n)$.

One possible approach is to assume that the function $\lambda : {{\mathrm{Sym}}}(n)\rightarrow \mathbb {R}^n$ mapping a matrix $M\in {{\mathrm{Sym}}}(n)$ to its (ordered) eigenvalues $\lambda (M)$ is differentiable in a neighborhood of $A\in {{\mathrm{Sym}}}(n)$. For example, this is the case if all eigenvalues of $A$ are simple [16]. The basic idea is to write $W(A) = \Psi (\lambda (A))$ with $\Psi (\lambda _1,\ldots ,\lambda _n)=\sum _{i=1}^n F(\lambda _i)$. Then

$$\begin{aligned} DW[A] = D\Psi [\lambda (A)] \cdot D\lambda [A]. \end{aligned}$$

(20)

It is therefore useful to compute the derivative $D\lambda [A]$ of the eigenvalue function. Since Lemma 7.1 implies

$$\begin{aligned} \lambda (Q^TAQ) = \lambda (A) \,\Longrightarrow \, D\lambda [Q^TAQ] = Q^T\,D\lambda [A]\,Q, \end{aligned}$$

the derivative of $\lambda$ at $A$ is determined by the derivative at the diagonal matrix corresponding to $A$. We will therefore assume w.l.o.g. that $A$ is already a diagonal matrix.

The eigenvalues $\lambda _i$ of $A$ are characterized by

$$\begin{aligned} \det (A-\lambda _i 1\!\!1)=0. \end{aligned}$$

(21)

Let $H\in {{\mathrm{Sym}}}(n)$. We compute the first-order approximation of (21):

$$\begin{aligned}&\det (A+H - \lambda _i(A+H)\cdot 1\!\!1) = 0 \nonumber \\&\quad \Rightarrow \, \det (A+H-[\lambda _i(A)1\!\!1 + [D\lambda _i(A).H]\cdot 1\!\!1 + \cdots ]) = 0 \nonumber \\&\quad \Rightarrow \, \det ([A-\lambda _i(A)1\!\!1]+H-[D\lambda _i(A).H]\cdot 1\!\!1 + \cdots ) = 0 \nonumber \\&\quad \Rightarrow \, \underbrace{\det [A-\lambda _i(A)1\!\!1]}_{=0} + \left\langle {{\mathrm{Cof}}}[A-\lambda _i(A)1\!\!1],\, H-[D\lambda _i(A).H]\cdot 1\!\!1 + \cdots \right\rangle + \cdots = 0. \end{aligned}$$

(22)

By ignoring higher-order terms we obtain

$$\begin{aligned} \left\langle {{\mathrm{Cof}}}[A-\lambda _i(A)1\!\!1],\, H-[D\lambda _i(A).H]\cdot 1\!\!1 \right\rangle = 0 \end{aligned}$$

(23)

Recall that $A$ is diagonal by assumption. Since $A$ commutes with diagonal matrices $H$ (and thus the derivative $DW[A].H$ could be computed by more direct means), we are only interested in cases where the symmetric matrix $H$ is off-diagonal, i.e. $H_{i,i}=0$ for $i=1,\ldots ,n$. But then

$$\begin{aligned} \langle {\,\underbrace{{{\mathrm{Cof}}}[A-\lambda _i(A)1\!\!1]}_{\text {diagonal}},\, \underbrace{H}_{{\text {off-diagonal}}}}\rangle = 0, \end{aligned}$$

thus (23) reduces to

$$\begin{aligned} \left\langle {{\mathrm{Cof}}}[A-\lambda _i(A)1\!\!1],\, [D\lambda _i(A).H]\cdot 1\!\!1 \right\rangle = 0, \end{aligned}$$

which we can also write as

$$\begin{aligned} (D\lambda _i(A).H)\cdot {{\mathrm{tr}}}\left( {{\mathrm{Cof}}}[A-\lambda _i(A)1\!\!1]\right) = 0. \end{aligned}$$

To conclude that $D\lambda _i(A).H=0$ it remains to show that ${{\mathrm{tr}}}\left( {{\mathrm{Cof}}}[A-\lambda _i(A)1\!\!1]\right) \ne 0$. Assuming that the diagonal entries of $A$ are ordered we write $A={{\mathrm{diag}}}(\lambda _1, \ldots , \lambda _n)$ and find

$$\begin{aligned} A - \lambda _i(A)1\!\!1 = \begin{pmatrix} \lambda _1-\lambda _i &{} &{} 0\\ &{} \ddots &{} \\ 0 &{} &{} \lambda _n-\lambda _i \end{pmatrix} \end{aligned}$$

and thus

$$\begin{aligned} {{\mathrm{Cof}}}[A - \lambda _i(A)1\!\!1] = \begin{pmatrix} \prod \limits _{k\ne 1}(\lambda _k-\lambda _i) &{} &{} 0\\ &{} \ddots &{} \\ 0 &{} &{} \prod \limits _{k\ne n}(\lambda _k-\lambda _i) \end{pmatrix}. \end{aligned}$$

We compute the trace:

$$\begin{aligned} {{\mathrm{tr}}}\left( {{\mathrm{Cof}}}[A - \lambda _i(A)1\!\!1]\right)&= \sum _{j=1}^n \, \prod \limits _{\begin{array}{c} k=1\\ k\ne j \end{array}}^n (\lambda _j-\lambda _i) = \prod \limits _{\begin{array}{c} k=1\\ k\ne i \end{array}}^n (\lambda _j-\lambda _i), \end{aligned}$$

where the second equality holds due to the fact that the product is zero if it contains the factor $(\lambda _i-\lambda _i)$. Hence this term is nonzero if and only if all eigenvalues of $A$ are simple, in which case we can conclude that $D\lambda _i(A).H = 0$ for all off-diagonal $H\in {{\mathrm{Sym}}}(n)$.

Using these results, we can prove the following, which is a simple corollary to Proposition 2.5:

Corollary 7.2

Let $f\in C^1(\mathbb {R})$, $F\in C^2(\mathbb {R})$ with $F'=f$ and let $A\in {{\mathrm{Sym}}}(n)$ such that all eigenvalues of $A$ are simple. Then the function

$$\begin{aligned} W: {{\mathrm{Sym}}}(n)\rightarrow \mathbb {R},\quad W(M) = \sum _{i=1}^n F(\lambda _i(M)) \end{aligned}$$

is differentiable at $A$ with

$$\begin{aligned} DW[A] = f(A) = Q^T {{\mathrm{diag}}}(f(\lambda _1),\ldots ,f(\lambda _n)) \,Q, \end{aligned}$$

where $A=Q^T {{\mathrm{diag}}}(\lambda _1,\ldots ,\lambda _n) \,Q$ is the spectral decomposition of $A$.

Proof

According to Lemma 7.1, $DW[Q^TXQ] = Q^TDW[X]Q$, hence we find

$$\begin{aligned} DW[A] = Q^T DW[{{\mathrm{diag}}}(\lambda _1,\ldots ,\lambda _n)] \,Q. \end{aligned}$$

Therefore it remains to show that

$$\begin{aligned} DW[{{\mathrm{diag}}}(\lambda _1,\ldots ,\lambda _n)].H = \left\langle {{\mathrm{diag}}}(f(\lambda _1),\ldots ,f(\lambda _n)),\, H \right\rangle \end{aligned}$$

(24)

for all $H\in {{\mathrm{Sym}}}(n)$ and pairwise different $\lambda _1,\ldots ,\lambda _n$.

We first consider the case of diagonal matrices $H = H^{\mathrm {diag}}= {{\mathrm{diag}}}(h_1,\ldots ,h_n)$. Writing $A^{\mathrm {diag}}= {{\mathrm{diag}}}(\lambda _1,\ldots ,\lambda _n)$ we find

$$\begin{aligned} W(A^{\mathrm {diag}}+tH^{\mathrm {diag}}) = W ({{\mathrm{diag}}}(\lambda _1+th_1,\ldots ,\lambda _n+th_n)) = \sum _{i=1}^n F(\lambda _i+th_i), \end{aligned}$$

thus

$$\begin{aligned} DW[A^{\mathrm {diag}}].H^{\mathrm {diag}}&= \lim _{t\rightarrow 0} \,\frac{1}{t} \left( W(A^{\mathrm {diag}}+tH^{\mathrm {diag}})-W(A^{\mathrm {diag}}) \right) \nonumber \\&= \lim _{t\rightarrow 0} \,\frac{1}{t} \sum _{i=1}^n F(\lambda _i+th_i) - F(\lambda _i) = \sum _{i=1}^n F'(\lambda _i)\,h_i \nonumber \\&= \left\langle {{\mathrm{diag}}}(f(\lambda _1),\ldots ,f(\lambda _n)),\, {{\mathrm{diag}}}(h_1,\ldots ,h_n) \right\rangle = \left\langle f(A^{\mathrm {diag}}),\,H^{\mathrm {diag}} \right\rangle . \end{aligned}$$

(25)

Now let $H=H^{\mathrm {off}}$ be a symmetric off-diagonal matrix, i.e. $H^{\mathrm {off}}_{i,i}=0$ for $i=1,\ldots ,n$. Using Eq. (20):

$$\begin{aligned} DW[A] = D\Psi [\lambda (A)] \cdot D\lambda [A], \end{aligned}$$

as well as the result of the previous considerations for diagonal $A$ and off-diagonal $H^{\mathrm {off}}$:

$$\begin{aligned} D\lambda [{{\mathrm{diag}}}(\lambda _1,\ldots ,\lambda _n)].H^{\mathrm {off}}= 0, \end{aligned}$$

we conclude

$$\begin{aligned} DW[{{\mathrm{diag}}}(\lambda _1,\ldots ,\lambda _n)].H^{\mathrm {off}}= 0. \end{aligned}$$

Finally, for arbitrary $H\in {{\mathrm{Sym}}}(n)$, we can write $H=H^{\mathrm {diag}}+H^{\mathrm {off}}$ with a diagonal matrix $H^{\mathrm {diag}}$ and a symmetric off-diagonal matrix $H^{\mathrm {off}}$. Then

$$\begin{aligned} DW[A^{\mathrm {diag}}].H&= DW[A^{\mathrm {diag}}].H^{\mathrm {diag}}+ DW[A^{\mathrm {diag}}].H^{\mathrm {off}}= DW[A^{\mathrm {diag}}].H^{\mathrm {diag}}\nonumber \\&= \left\langle f(A^{\mathrm {diag}}),\,H^{\mathrm {diag}} \right\rangle = \left\langle f(A^{\mathrm {diag}}),\,H \right\rangle , \end{aligned}$$

(26)

showing (24) and concluding the proof. $\square$