If $$ T= \left[ \begin{array}{cccc} e^{\beta J} & e^{-\beta J} \\ e^{-\beta J} & e^{\beta J} \\ \end{array} \right] $$ and $$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \exp{\beta J(\vec{S_1}\vec{S_2}+\vec{S_2}\vec{S_3}+...+\vec{S_{N-1}}\vec{S_N}+\vec{S_N}\vec{S_1})} $$
Then why can we say that $$Z = \sum_{S_i=\pm 1} ... \sum_{S_N=\pm 1} \langle S_1|T|S_2\rangle\langle S_2|T|S_3\rangle...\langle S_N|T|S_1\rangle ? $$
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Consider two complex $n\m$vectors expressed also as kets \begin{equation} \mathbf x\e \begin{bmatrix} x_1 \vphantom{\dfrac{a}{b}}\\ x_2 \vphantom{\dfrac{a}{b}}\\ \vdots \vphantom{\dfrac{a}{b}}\\ x_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix}\e \vra{\mathbf x}\qquad \texttt{and} \qquad \mathbf y\e \begin{bmatrix} y_1 \vphantom{\dfrac{a}{b}}\\ y_2 \vphantom{\dfrac{a}{b}}\\ \vdots \vphantom{\dfrac{a}{b}}\\ y_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix}\e \vra{\mathbf y}\quad \in \mathbb C^n \tag{01}\label{01} \end{equation} Complex conjugating and transposing these one-column matrices we obtain the bras \begin{equation} \mathbf x^{\boldsymbol*}\e \begin{bmatrix} \overline x_1 & \overline x_2 & \cdots & \overline x_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix}\e \lav{\mathbf x}\quad \texttt{and} \quad \mathbf y^{\boldsymbol*}\e \begin{bmatrix} \overline y_1 & \overline y_2 & \cdots & \overline y_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix}\e \lav{\mathbf y} \tag{02}\label{02} \end{equation} Their usual inner product in $\,\mathbb C^n\,$ is \begin{equation} \overline x_1\,y_1\p\overline x_2\,y_2\p\cdots\overline x_n\,y_n\e \begin{bmatrix} \overline x_1 & \overline x_2 & \cdots & \overline x_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \begin{bmatrix} y_1 \vphantom{\dfrac{a}{b}}\\ y_2 \vphantom{\dfrac{a}{b}}\\ \vdots \vphantom{\dfrac{a}{b}}\\ y_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix}\e \lavra{\mathbf x}{\mathbf y} \tag{03}\label{03} \end{equation} Given a $\,n\times n\,$ complex matrix $\,\mathrm A\,$ \begin{equation} \mathrm A\e \begin{bmatrix} a_{11} & a_{11} & \cdots & a_{1n} \vphantom{\dfrac{a}{b}}\\ a_{21} & a_{22} & \cdots & a_{2n} \vphantom{\dfrac{a}{b}}\\ \vdots & \vdots & \vdots & \vdots\vphantom{\dfrac{a}{b}}\\ a_{n1} & a_{n2} & \cdots & a_{nn}\vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \tag{04}\label{04} \end{equation} the notation $\,\lavvra{\mathbf x}{\mathrm A}{\mathbf y}\,$ is the inner product of the vectors $\,\mathbf x\,$ and $\,\mathrm A\mathbf y\,$ expressed by matrices as \begin{equation} \lavvra{\mathbf x}{\mathrm A}{\mathbf y}\e \begin{bmatrix} \overline x_1 & \overline x_2 & \cdots & \overline x_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \begin{bmatrix} a_{11} & a_{11} & \cdots & a_{1n} \vphantom{\dfrac{a}{b}}\\ a_{21} & a_{22} & \cdots & a_{2n} \vphantom{\dfrac{a}{b}}\\ \vdots & \vdots & \vdots & \vdots\vphantom{\dfrac{a}{b}}\\ a_{n1} & a_{n2} & \cdots & a_{nn}\vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \begin{bmatrix} y_1 \vphantom{\dfrac{a}{b}}\\ y_2 \vphantom{\dfrac{a}{b}}\\ \vdots \vphantom{\dfrac{a}{b}}\\ y_n \vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \tag{05}\label{05} \end{equation} Under this spirit you could look at the $\,S_k\,$ as $2\times 1$ matrices and more precisely \begin{equation} S_k\e \left. \begin{cases} \begin{bmatrix} 1 \vphantom{\dfrac{a}{b}}\\ 0 \vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \texttt{for} \p 1\\ \\ \begin{bmatrix} 0 \vphantom{\dfrac{a}{b}}\\ 1 \vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \texttt{for} \m 1 \end{cases}\right\} \tag{06}\label{06} \end{equation} (note : this reminds us the up and down states of a spin-1/2 particle or the up and down quarks of isospin-1/2 particle).
So if for example $\,S_3\e\m 1\,$ and $\,S_8\e\p 1\,$ then \begin{equation} \lavvra{S_3}{\mathrm T}{S_8}\e \begin{bmatrix} 0 & 1 \vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \begin{bmatrix} t_{11} & t_{11} \vphantom{\dfrac{a}{b}}\\ t_{21} & t_{22} \vphantom{\dfrac{a}{b}} \end{bmatrix} \begin{bmatrix} 1 \vphantom{\dfrac{a}{b}}\\ 0 \vphantom{\dfrac{a}{b}} \end{bmatrix}\e \begin{bmatrix} 0 & 1 \vphantom{\dfrac{a}{b}}\\ \end{bmatrix} \begin{bmatrix} t_{11} \vphantom{\dfrac{a}{b}}\\ t_{21} \vphantom{\dfrac{a}{b}} \end{bmatrix}\e t_{21} \tag{07}\label{07} \end{equation} For the rest look in the other till now two answers.
Then why can we say that:
Because, each $S_i$ can only take on two values: +1 or -1
For example, if $S_1 = +1$ and $S_2 = +1$ then the $e^{\beta J S1S2}$ is ${e^{\beta J}}$, which is exactly what the ++ matrix element of $T$ says.
As another example, If $S_1 = +1$ and $S_2 = -1$ then the $e^{\beta J S1S2}$ is ${e^{-\beta J}}$, which is exactly what the +- matrix element of $T$ says.
And so on.
Update:
In the Dirac bra/ket notation, $\left < S | T | S' \right >$ is a number that depends on S and S'. (Note: We could use a different symbol to denote $T$ as a stand-alone matrix vs $T$ when it is in the Dirac bra/key notation, but in this case we dont.)
For example, when S=+ and S'=+: $$ \left < + | T | + \right > = e^{\beta J} $$
For example, when S=+ and S'=-: $$ \left < + | T | - \right > = e^{-\beta J} $$
For example, when S=- and S'=+: $$ \left < - | T | + \right > = e^{-\beta J} $$
For example, when S=- and S'=-: $$ \left < - | T | - \right > = e^{-\beta J} $$
And we write these four possible values together in a matrix, whose indices span + and - in both directions.
Because, with $S_i$ taking values $\pm 1$, we have $$ \langle S_1|e^{\beta J {S}_i {S}_{i+1}}| S_2\rangle= \left[\matrix{e^{\beta J}& e^{-\beta J} \cr e^{-\beta J} &e^{\beta J}}\right]_{S_1,S_2} $$ where the subscript on the matrix means the appropriate matrix entry.
