Image Similarity Assessment Based on Coefficients of Spatial Association

Abstract

This paper focuses on the construction of image similarity indices that consider the hidden spatial association between two images. The proposal is a variant of a structural similarity (SSIM) coefficient and introduces a codispersion coefficient to capture the hidden spatial association between two images in a particular direction on a plane. The novel contribution of this article is the inclusion of the codispersion coefficient instead of the sample correlation coefficient. The difference between the codispersion and correlation coefficients is illustrated through two examples. We then show that this modified measure is a valid pseudo metric and has several useful properties, including quasi-convexity, which is established under very precise conditions. The quasi-convexity property is an attractive property that allows one to use this type of measure as a cost function in optimization problems. In addition, we introduce another variant of the SSIM with a contrast function that depends on the spatial lag on the space. This proposal trivially recovers the optimization properties of the SSIM. Various computational experiments with real datasets support our proposals and findings, and we characterize the practical advantages and drawbacks of the SSIM variants.

Appendix: Proofs

1.1 Derivation of (6)

Assuming that the parameters of the models lie in the stationary region, the following representations hold:

$$\begin{aligned} X(i,j)&=\sum _{k=0}^{\infty }\sum _{l=0}^{\infty }\sum _{m=0}^{\infty }\frac{(k+l+m)!}{k!l!m!}\\&\phi _1^k\phi _2^l\phi _3^m\varepsilon _1(i-k-m,j-l-m),\\ Y(i,j)&=\sum _{n=0}^{\infty }\sum _{p=0}^{\infty }\sum _{q=0}^{\infty }\frac{(p+q+r)!}{p!q!r!}\\&\psi _1^p\psi _2^q\psi _3^r\varepsilon _2(i-q-r,j-q-r). \end{aligned}$$

Then, we compute the term ${\mathbb {E}}[X(i,j)Y(i,j)]$, thereby obtaining

$$\begin{aligned}&\sum _{k,l,m,r}\frac{(k+l+m)!}{k!l!m!}\frac{(k+l+2m+r)!}{r!(k+m-r)!(l+m-r)!}\\&\quad \qquad \times \,(\phi _1\psi _1)^k(\phi _2\psi _2)^l(\phi _3\psi _1\psi _2)^m\left( \frac{\psi _3}{\psi _1\psi _2}\right) ^r. \end{aligned}$$

We recall that this expression is valid when $\phi _i\ne 0$, $\psi _i\ne 0$, $i=1,2,3$ and when the summation is over all non-negative k, l, m, r such that $(k+m-r)\ge 0$ and $(l+m-r)\ge 0$. Now, we define

$$\begin{aligned}&C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,h_1,h_2)\nonumber \\&\quad =\sum _{k,l,m,r}\frac{(k+l+m)!}{k!l!m!} \frac{(k+l+2m+r+h_1+h_2)!}{r!(k+m-r+h_1)!(l+m-r+h_2)!}\nonumber \\&\qquad \times (\phi _1\psi _1)^k(\phi _2\psi _2)^l(\phi _3\psi _1\psi _2)^m\left( \frac{\psi _3}{\psi _1\psi _2}\right) ^r. \end{aligned}$$

(50)

Using (50), we obtain

$$\begin{aligned}&{\mathbb {E}}[X(i,j)Y(i,j)]=\rho \sigma \tau C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,0,0),\\&{\mathbb {E}}[X(i+h_1,j+h_2)Y(i+h_1,j+h_2)]\\&\quad =\rho \sigma \tau C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,0,0),\\&{\mathbb {E}}[X(i,j)Y(i+h_1,j+h_2)]\\&\quad =\rho \sigma \tau \psi _1^{h_1}\psi _2^{h_2}C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,h_1,h_2),\\&{\mathbb {E}}[X(i+h_1,j+h_2)Y(i,j)]\\&\quad =\rho \sigma \tau \phi _1^{h_1}\phi _2^{h_2}C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,h_1,h_2),\\&\displaystyle {\mathbb {E}}[(X(i,j)-X(i+h_1,j+h_2)) (Y(i,j) -Y(i+h_1,j+h_2))] \\&\quad =\rho \sigma \tau (2C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,0,0)\\&\qquad - [\psi _1^{h_1}\psi _2^{h_2}+\phi _1^{h_1}\phi _2^{h_2}]C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,h_1,h_2)),\\&\mathbb {V}[X(i,j)]=\sigma ^2C(\phi _1,\phi _2,\phi _3,\phi _1,\phi _2,\phi _3,0,0),\\&\mathbb {V}[Y(i,j)]=\tau ^2C(\psi _1,\psi _2,\psi _3,\psi _1,\psi _2,\psi _3,0,0),\\&{{\mathrm{cov}}}(X(i+h_1,j+h_2)Y(i,j))\\&\quad =\sigma ^2\phi _1^{h_1}\phi _2^{h_2}C(\phi _1,\phi _2,\phi _3,\phi _1,\phi _2,\phi _3,h_1,h_2)\\&{{\mathrm{cov}}}(X(i,j)Y(i+h_1,j+h_2))\\&\quad =\tau ^2\psi _1^{h_1}\psi _2^{h_2}C(\psi _1,\psi _2,\psi _3,\psi _1,\psi _2,\psi _3,h_1,h_2),\\&{\mathbb {E}}\left[ (X(i+h_1,j+h_2)-X(i,j))^2\right] \\&\quad =2\sigma ^2[C(\phi _1,\phi _2,\phi _3,\phi _1,\phi _2,\phi _3,0,0)-\phi _1^{h_1}\phi _2^{h_2}\\&\qquad \times C(\phi _1,\phi _2,\phi _3,\phi _1,\phi _2,\phi _3,h_1,h_2)],\\&{\mathbb {E}}\left[ (Y(i+h_1,j+h_2)-Y(i,j))^2\right] \\&\quad =2\tau ^2[C(\psi _1,\psi _2,\psi _3,\psi _1,\psi _2,\psi _3,0,0)-\psi _1^{h_1}\psi _2^{h_2}\\&\qquad \times C(\psi _1,\psi _2,\psi _3,\psi _1,\psi _2,\psi _3,h_1,h_2)]. \end{aligned}$$

Replacing the above quantities in (2), one obtains Equation (6), where $K=\frac{K_1}{2R_1R_2}$ and

$$\begin{aligned} K_1= & {} C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,0,0)\\&-\,[\psi _1^{h_1}\psi _2^{h_2}+\phi _1^{h_1}\phi _2^{h_2}]C(\phi _1,\phi _2,\phi _3,\psi _1,\psi _2,\psi _3,h_1,h_2),\\ R_1= & {} \sqrt{C(\phi _1,\phi _2,\phi _3,\phi _1,\phi _2,\phi _3,0,0)-\phi _1^{h_1}\phi _2^{h_2}C(\phi _1,\phi _2,\phi _3,\phi _1,\phi _2,\phi _3,h_1,h_2)},\\ R_2= & {} \sqrt{C(\psi _1,\psi _2,\psi _3,\psi _1,\psi _2,\psi _3,0,0)-\psi _1^{h_1}\psi _2^{h_2}C(\psi _1,\psi _2,\psi _3,\psi _1,\psi _2,\psi _3,h_1,h_2)}. \end{aligned}$$

Proof of Proposition 1

For the proof of Proposition 1, we need to establish the following lemma. $\square$

Lemma 2

A spatial lag ${\varvec{h}}=(h_1,h_2)$ that belongs to a rectangular grid of $\mathbb {Z}^{2}_{+}$ induces a lag $h^{*}>0$ in the space of images $\mathbb {R}_+^N.$

Proof

Let n, m $\in \mathbb {Z}_{+}$ be the number of rows and columns, respectively, of a rectangular grid of $\mathbb {Z}^{2}_{+}$. Then, the spatial lag ${\varvec{h}}=(h_1,h_2)\in \mathbb {Z}^2_{+}$ corresponds to a translation of $h_1$ units to the right and $h_2$ units down (depending on the convention used). Thus, for $\mathrm {{\varvec{X}}}$, fixing the point ${\varvec{s}}=(i,j)\in S_{{\varvec{h}}},$ we have that if $(i,j)\mapsto k$, then $(i+h_1,j+h_2)\mapsto k+h^*$, where $h^*=h_1+nh_2$.

Note that Lemma 2 is valid if the spatial lag ${{\varvec{h}}}$ is a valid position on the image, i.e., $(i+h_1,j+h_2)$ is a point belonging to the rectangular grid.

Proof of Proposition 1

The proof is developed by construction. Let ${{\varvec{h}}}=(h_1,h_2)\in \mathbb {Z}^{2}_{+}$ be a fixed spatial lag; let $h^{*}$ be the lag induced by ${{\varvec{h}}}$ according to Lemma 2; let us denote by ${\varvec{I}}_{k}$ the identity matrix of order $k\times k$ and by ${\varvec{0}}_{k,l}$ the null matrix of order $k\times l$; and let ${\varvec{A}}_{{\varvec{h}}}^{(1)}$ be a matrix of order $n-h_1\times N$ such that

$$\begin{aligned} {\varvec{A}}_{{\varvec{h}}}^{(1)} = \left( \begin{matrix} {\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,2h_1}&-{\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,N-2n} \end{matrix}\right) . \end{aligned}$$

Similarly, we define the following quantities:

$$\begin{aligned} {\varvec{A}}_{{\varvec{h}}}^{(2)}&= \left( \begin{matrix} {\varvec{0}}_{n-h_1,n}&{\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,2h_1}&-{\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,N-3n} \end{matrix}\right) ,\\ {\varvec{A}}_{{\varvec{h}}}^{(3)}&= \left( \begin{matrix} {\varvec{0}}_{n-h_1,2n}&{\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,2h_1}&-{\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,N-4n} \end{matrix}\right) ,\\ \vdots&\\ {\varvec{A}}_{{\varvec{h}}}^{(k)}&= \left( \begin{matrix} {\varvec{0}}_{n-h_1,(k-1)n}&{\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,2h_1}&-{\varvec{I}}_{n-h_1}&{\varvec{0}}_{n-h_1,N-(k+1)n} \end{matrix}\right) , \end{aligned}$$

where $k=1,\ldots ,m-h_2$.

In the previous calculations, for a given matrix ${{\varvec{X}}}$ of order $n\times m$ and a spatial lag ${{\varvec{h}}}=(h_1,h_2)$, $|S_{{\varvec{h}}}|=N_{{{\varvec{h}}}}=(n-h_1)(m-h_2)$ is implicit. Hence, defining

$$\begin{aligned} {\varvec{A}}_{{\varvec{h}}}= \left( \begin{matrix} {\varvec{A}}_{{\varvec{h}}}^{(1)}\\ {\varvec{A}}_{{\varvec{h}}}^{(2)}\\ \vdots \\ {\varvec{A}}_{{\varvec{h}}}^{(m-h_2)}\\ \end{matrix}\right) , \end{aligned}$$

it is straightforward to obtain

$$\begin{aligned} {\varvec{A}}_{{\varvec{h}}}{\varvec{x}}= (x_k-x_{k+h^*})_{k\in S_{{\varvec{h}}}^{im}}, \end{aligned}$$

where $S_{{\varvec{h}}}^{im} =\{k: 1\le k\le nm \wedge F^{-1}(k)\in S_{{\varvec{h}}}\},$ and F is the natural map associated with the indices given by

$$\begin{aligned} \begin{array}{lccl} F:&{}\mathbb {Z}_{+}^2&{}\longrightarrow &{}\mathbb {Z}_+\\ &{}(i,j)&{}\longmapsto &{} k=j+(n-1)i. \end{array} \end{aligned}$$

Clearly, F is a bijective map between $S_{{\varvec{h}}}$ and $S_{{\varvec{h}}}^{im}$. Because $N_{{\varvec{h}}}=(n-h_1)(m-h_2)$, the result follows after simple calculations of the norms and inner products.

Proof of Proposition 5

Without loss of generality, we assume that $h_1,h_2\ge 0$. $h_1$ corresponds to a column translation of the matrix ${{\varvec{X}}}$, and $h_2$ corresponds to a row translation of ${{\varvec{X}}}.$ Now, ${\varvec{A}}_{{\varvec{h}}}{\varvec{x}}$ is equal to the vector ${\varvec{x}}_{{\varvec{h}}}= ({\varvec{X}}(j,i)-{\varvec{X}}(j+h_2,i+h_1)),$ for all $(j,i)\ \text {such that} \ (j+h_2,i+h_1)\in [1,2,\ldots ,n]\times [1,2,\ldots ,m].$ Specifically,

$$\begin{aligned} {\varvec{x}}_{{\varvec{h}}}=\left( \begin{matrix} {\varvec{X}}(1,1)-{\varvec{X}}(1+h_2,1+h_1)\\ {\varvec{X}}(1,2)-{\varvec{X}}(1+h_2,2+h_1)\\ \vdots \\ {\varvec{X}}(1,m-h_1)-{\varvec{X}}(1+h_2,m)\\ {\varvec{X}}(2,1)-{\varvec{X}}(2+h_2,2+h_1)\\ {\varvec{X}}(2,2)-{\varvec{X}}(2+h_2,2+h_1)\\ \vdots \\ {\varvec{X}}(2,m-h_1)-{\varvec{X}}(2+h_2,m)\\ \vdots \\ {\varvec{X}}(n-h_2, m-h_1)-{\varvec{X}}(n,m) \end{matrix}\right) . \end{aligned}$$

Clearly, every pixel belonging to the subimage

$$\begin{aligned} I= & {} \{ {{\varvec{X}}}(j,i):\quad (j,i)\in [h_2+1,\ldots ,n-h_2]\\&\times \, [h_1+1,\ldots ,m-h_1]\} \end{aligned}$$

appears once with a positive sign and once with a negative sign in ${\varvec{x}}_{{\varvec{h}}}$; thus, they do not contribute to the sum of the entries of ${\varvec{x}}_{{\varvec{h}}}$. The terms that remain are those that are located outside of the central subimage. Among them, the terms that are in $L_+=\{{\varvec{X}}(j,i):\quad (j,i)\in [1,\ldots ,h_2]\times [1,\ldots ,m-h_1]\cup [h_2+1,\ldots ,n-h_2]\times [1,\ldots ,h_1]\}$ have a positive sign, while those that are in $L_{-}=\{{{\varvec{X}}}(j,i):\quad (j,i)\in [n-h_2,\ldots ,n]\times [h_1,\ldots ,m]\cup [n-h_1+1,\ldots ,n]\times [h_2+1,n-h_2]\}$ have a negative sign. Given that the intensities of the central part of the image is zero, we want the components of ${\varvec{x}}_{{\varvec{h}}}$ that have the same property. Then, we require that

$$\begin{aligned} \sum _{{{\varvec{X}}}(j,i)\in L_{+}}{{\varvec{X}}}(j,i) - \sum _{{{\varvec{X}}}(j,i)\in L_{-}} {{\varvec{X}}}(j,i)= 0, \end{aligned}$$

(51)

Using the definition of $L_{+}$ and $L_{-}$, it is straightforward to prove that Equation (51) is equivalent to

$$\begin{aligned}&\sum _{i=1}^{m-h_1}\sum _{j=1}^{h_2}{\varvec{X}}(j,i)+\sum _{i=1}^{h_1}\sum _{j=h_2+1}^{n-h_2} {\varvec{X}}(j,i)\\&\quad = \sum _{i=h_1+1}^{m}\sum _{j=n-h_2+1}^{n} {\varvec{X}}(j,i)+\sum _{i=m-h_1+1}^{m}\sum _{j=h_2+1}^{n-h_2} {\varvec{X}}(j,i). \end{aligned}$$

This completes the proof.

As an illustration of the terms used in the proof of Proposition 5, Fig. 7 shows an example with an image of size $9\times 9$ and a spatial lag ${\varvec{h}}=(1,2).$ The sign displayed in each cell shows whether the term associated with that cell can be positive, negative, or both when applying the transformation ${\varvec{A}}_{{\varvec{h}}}$ to ${\varvec{x}}=\mathrm {vec}({\varvec{X}}).$ We observe from Fig. 7 that the subimage I corresponds to the area colored in green. $L_+$ corresponds to the area colored in light blue, while $L_+$ corresponds to the red area. The gray cells are those pixels that are not used in the construction of ${\varvec{x}}_{{\varvec{h}}}$ and are located at the upper-right corner and at the bottom-left corner of the matrix shown in Fig. 7 .

Fig. 7

Example of the terms used in the proof of Proposition 5 for an image of size $9 \times 9$ and ${{\varvec{h}}}=(1,2)$

Proof of Proposition 9

We have to prove that ${\varvec{H}}_g{\varvec{v}}^{\lambda }=\lambda {\varvec{v}}^{\lambda },$ where $\lambda$ and ${\varvec{v}}^{\lambda }$ are as in (36) and (37).

Let $v_k^{\lambda }$, $k=1,\ldots ,N$ be the components of ${\varvec{v}}^{\lambda }$, and let ${\varvec{H}}^{k}_{g}$ be the vector associated with the k-th row of ${\varvec{H}}_{g}$. We will prove that

$$\begin{aligned} \langle {\varvec{H}}^{k}_{g}, {\varvec{v}}^{\lambda }\rangle = \lambda v^{\lambda }_{k}\qquad k=1,\ldots ,N. \end{aligned}$$

(52)

Let $k>1$. Using the partial derivatives given by

$$\begin{aligned}&\frac{\partial ^2g}{\partial x_1^2}=\frac{3x_1(||{\varvec{x}}||^2-x_1^2)}{||{\varvec{x}}||^5}, \end{aligned}$$

(53)

$$\begin{aligned}&\frac{\partial ^2g}{\partial x_i^2}=\frac{x_1(||{\varvec{x}}||^2-3x_i^2)}{||{\varvec{x}}||^5},\qquad i\ne 1, \end{aligned}$$

(54)

$$\begin{aligned}&\frac{\partial ^2g}{\partial x_1\partial x_i}=\frac{x_i(||{\varvec{x}}||^2-3x_1^2)}{||{\varvec{x}}||^5}, \qquad i\ne 1,\end{aligned}$$

(55)

$$\begin{aligned}&\frac{\partial ^2g}{\partial x_j\partial x_i}=-\frac{3x_1x_ix_j}{||{\varvec{x}}||^5}, \qquad i\ne 1, i>j. \end{aligned}$$

(56)

we have that

$$\begin{aligned}&\langle {\varvec{H}}^{k}_{g}, {\varvec{v}}^{\lambda }\rangle \nonumber \\&\quad = \frac{x_k(||{\varvec{x}}||^2-3x_1^2)}{||{\varvec{x}}||^5}\frac{5x_1||{\varvec{x}}||^2-6x_1^3-||{\varvec{x}}||^2\sqrt{4||{\varvec{x}}||^2 -3x_1^2}}{2(||{\varvec{x}}||^2-3x_1^2)}\\&\quad \qquad -\,3{\mathop {\mathop {\sum }_{i=2}}\limits _{i\ne k}}^{N}\frac{x_1x_kx_i^2}{x_N||{\varvec{x}}||^5}+\frac{x_1x_k(||{\varvec{x}}||^2-3x_k^2)}{x_N||{\varvec{x}}||^5}\\&\quad =\frac{5x_1x_k||{\varvec{x}}||^2-6{\varvec{x}}_1^3x_k-||{\varvec{x}}||^2\sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2x_N||{\varvec{x}}||^5}\\&\quad \qquad -\,3x_1x_k\frac{||{\varvec{x}}||^2-x_1^2-x_k^2}{x_N||{\varvec{x}}||^5} +\frac{x_1x_k||{\varvec{x}}||^2-3x_1x_k^3}{x_N||{\varvec{x}}||}\\&\quad =x_k\frac{x_1-\sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2||{\varvec{x}}||^3} =x_k\cdot \lambda =v_{k}^{\lambda } \cdot \lambda . \end{aligned}$$

If $k=1$, the left-hand side of Eq. (52) is

$$\begin{aligned}&\langle {\varvec{H}}^{1}_{g}, {\varvec{v}}^{\lambda }\rangle \nonumber \\&\quad = \frac{3x_1(||{\varvec{x}}||^2-x_1^2)}{||{\varvec{x}}||^5}\cdot \frac{5x_1||{\varvec{x}}||^2-6x_1^3-||{\varvec{x}}||^2\sqrt{4||{\varvec{x}}||-3x_1^2}}{2x_N(||{\varvec{x}}||^2-3x_1^2)}\nonumber \\&\qquad +\sum _{i=2}^{N}\frac{x_i^2(||{\varvec{x}}||^2-3x_1^2)}{||{\varvec{x}}||^5}\nonumber \\&\quad =\frac{||{\varvec{x}}||^2-x_1^2}{x_N||{\varvec{x}}||^5}\nonumber \\&\qquad \cdot \left( \frac{15x_1^2||{\varvec{x}}||^2-18x_1^4-3x_1||{\varvec{x}}||^2 \sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2(||{\varvec{x}}||^2-3x_1^2)} +||{\varvec{x}}||^2-3x_1^2\right) \nonumber \\&\quad = \frac{||{\varvec{x}}||^2-x_1^2}{x_n||{\varvec{x}}||^3}\cdot \left( \frac{2||{\varvec{x}}||^2+3x_1^2-3x_1\sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2(||{\varvec{x}}||^2-3x_1^2)}\right) . \end{aligned}$$

(57)

Now, the right-hand side of Equation (52) is given by

$$\begin{aligned}&\lambda \cdot v^{\lambda }_1\nonumber \\&\quad =\frac{x_1-\sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2||{\varvec{x}}||^3}\nonumber \\&\qquad \cdot \, \frac{5x_1||{\varvec{x}}||^2-6x_1^3-||{\varvec{x}}||^2\sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2x_N(||{\varvec{x}}||^2-3x_1^2)}\nonumber \\&\quad =\frac{2||{\varvec{x}}||^2 +x_1^2||{\varvec{x}}||^2-3x_1^4-3x_1||{\varvec{x}}||^2 \sqrt{4||{\varvec{x}}||-3x_1^2}+3x_1^3\sqrt{4||{\varvec{x}}||^2 -3x_1^2}}{2x_N||{\varvec{x}}||^3(||{\varvec{x}}||-3x_1^2)}\nonumber \\&\quad =\frac{(||{\varvec{x}}||^2-x_1^2)(2||{\varvec{x}}||^2+3x_1^2)-3x_1(||{\varvec{x}}||^2-x_1^2)\sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2x_N||{\varvec{x}}||^3(||{\varvec{x}}||^2-3x_1^2)}\nonumber \\&\quad =\frac{||{\varvec{x}}||^2-x_1^2}{x_N||{\varvec{x}}||^3}\cdot \left( \frac{2||{\varvec{x}}||^2+3x_1^2-3x_1\sqrt{4||{\varvec{x}}||^2-3x_1^2}}{2(||{\varvec{x}}||^2-3x_1^2)}\right) . \end{aligned}$$

(58)

Because the expressions (57) and (58) are equal, the proof for $k=1$ is complete. This concludes the proof.

Proof of (40) and (41)

The proof is similar to that of Proposition 9. First, note that

$$\begin{aligned}&\frac{\partial ^2f}{\partial x_1^2}=-\frac{2(||{\varvec{x}}||^2-x_1^2)(||{\varvec{x}}||^2-4x_1^2)}{||{\varvec{x}}||^6},\\&\frac{\partial ^2f}{\partial x_i^2}=\frac{2x^2_1(||{\varvec{x}}||^2-4x_i^2)}{||{\varvec{x}}||^6},\qquad i\ne 1,\\&\frac{\partial ^2f}{\partial x_1\partial x_i}=\frac{4x_1x_i(||{\varvec{x}}||^2-2x_1^2)}{||{\varvec{x}}||^6}, \qquad i\ne 1,\\&\frac{\partial ^2f}{\partial x_j\partial x_i}=-\frac{8x^2_1x_ix_j}{||{\varvec{x}}||^6}, \qquad i\ne 1, i>j. \end{aligned}$$

Then, if $k=1$, we obtain

$$\begin{aligned} \langle {\varvec{H}}_{f}^1,{\varvec{v}}^{\lambda } \rangle= & {} -2\frac{(||{\varvec{x}}||^2-x_1^2)(||{\varvec{x}}||^2-4x_1^2)}{||{\varvec{x}}||^6}\cdot \frac{(2x_1^2-||{\varvec{x}}||^2)}{2x_1x_N}\\&+\sum _{i=2}^{N}\frac{4x_ix_1(||{\varvec{x}}||^2-2x_1^2)}{x_N||{\varvec{x}}||^6}\\= & {} -\frac{2(||{\varvec{x}}||^2-x_1^2)}{||{\varvec{x}}||^4}\left[ \frac{(||{\varvec{x}}||^2-4x_1^2)(2x_1^2-||{\varvec{x}}||^2)}{2x_1x_N||{\varvec{x}}||^2}\right. \\&\left. -\frac{2x_1(||{\varvec{x}}||^2-2x_1^2)}{x_N||{\varvec{x}}||^2}\right] \\= & {} \lambda \left[ \frac{(2x_1^2-||{\varvec{x}}||^2)}{2x_1x_N}\left\{ \frac{||{\varvec{x}}||^2-4x_1^2}{||{\varvec{x}}||^2}+\frac{4x_1^2}{||{\varvec{x}}||^2}\right\} \right] \\= & {} \lambda \cdot \frac{(2x_1^2-||{\varvec{x}}||^2)}{2x_1x_N}\\= & {} \lambda \cdot v_1^{\lambda }. \end{aligned}$$

If $k>1$, one obtains

$$\begin{aligned}&\langle {\varvec{H}}_{f}^k,{\varvec{v}}^{\lambda } \rangle \\&\quad = \frac{4x_1x_k(||{\varvec{x}}||^2-2x_1^2)}{||{\varvec{x}}||^6}\cdot \frac{(2x_1^2-||{\varvec{x}}||)}{2x_1x_N}\\&\qquad -\,{\mathop {\mathop {\sum }\limits _{i=2}}\limits _{i\ne k}}^{N}\frac{8x_1^2x_i^2x_k}{x_N||{\varvec{x}}||^6}+\frac{2x_1x_k(||{\varvec{x}}||^2-4x_k^2)}{x_N||{\varvec{x}}||^6}\\&\quad =\frac{-2x_1x_k(||{\varvec{x}}||^4-4x_1^2 ||{\varvec{x}}||^2+4x_1^4)-6x_1^3x_k||{\varvec{x}}||^2 +8x_1^5x_k+8x_1^3x_k^3-8x_1^3x_k^3}{x_1x_N||{\varvec{x}}||^6}\\&\quad =\frac{-2x_1x_k(||{\varvec{x}}||^4-2x_1^2)}{x_1x_N||{\varvec{x}}||^6} =\frac{-2(||{\varvec{x}}||^2-2x_1^2)}{||{\varvec{x}}||^4}\cdot x_k\\&\quad =\lambda \cdot v_k^{\lambda }. \end{aligned}$$