Existence by induction. For $n = 1$: $A = (a_{11})$ with $a_{11} > 0$, so $L = (\sqrt{a_{11}})$.

For the inductive step, partition $A = \begin{pmatrix} A_{n-1} & c \\ c^T & a_{nn} \end{pmatrix}$ where $A_{n-1}$ is $(n-1) \times (n-1)$. Since any principal submatrix of an SPD matrix is SPD, $A_{n-1}$ is SPD. By the induction hypothesis, $A_{n-1} = L_{n-1} L_{n-1}^T$.

Seek $L = \begin{pmatrix} L_{n-1} & 0 \\ \ell^T & \ell_{nn} \end{pmatrix}$ so that $LL^T = \begin{pmatrix} L_{n-1}L_{n-1}^T & L_{n-1}\ell \\ \ell^T L_{n-1}^T & \ell^T\ell + \ell_{nn}^2 \end{pmatrix} = A$.

From $L_{n-1}\ell = c$, we get $\ell = L_{n-1}^{-1}c$ (unique, since $L_{n-1}$ is nonsingular).

From $\ell^T\ell + \ell_{nn}^2 = a_{nn}$, we need $\ell_{nn}^2 = a_{nn} - \ell^T\ell = a_{nn} - c^T(A_{n-1})^{-1}c$.

Positivity of $\ell_{nn}^2$. The quantity $a_{nn} - c^T A_{n-1}^{-1} c$ is the Schur complement of $A_{n-1}$ in $A$. For SPD $A$, all Schur complements are positive definite. Concretely: for any $\alpha \neq 0$, let $v = \begin{pmatrix} -A_{n-1}^{-1}c\alpha \\ \alpha \end{pmatrix}$. Then $v^T A v = \alpha^2(a_{nn} - c^T A_{n-1}^{-1}c) > 0$ since $A$ is positive definite and $v \neq 0$.

Therefore $\ell_{nn} = \sqrt{a_{nn} - c^T A_{n-1}^{-1}c} > 0$, completing the existence proof.

Uniqueness. Suppose $A = L_1 L_1^T = L_2 L_2^T$ with $L_1, L_2$ lower triangular with positive diagonals. Then $L_2^{-1}L_1 = L_2^T L_1^{-T}$. The left side is lower triangular and the right side is upper triangular, so both equal a diagonal matrix $D$. From $L_1 = L_2 D$: $A = L_2 D D^T L_2^T = L_2 D^2 L_2^T$. Comparing with $A = L_2 L_2^T$ gives $D^2 = I$, so $D = \pm I$ on each diagonal entry. Since $L_1$ and $L_2$ have positive diagonals, $D = I$, hence $L_1 = L_2$.

Explicit formula. The diagonal entries satisfy $\ell_{ii}^2 = a_{ii} - \sum_{k=1}^{i-1} \ell_{ik}^2$. Since $a_{ii} > 0$ (diagonal of SPD matrix) and the construction guarantees positivity at each step, $\ell_{ii} > 0$ for all $i$. $\square$