[proofplan] We prove the four characterisations are equivalent by establishing the cycle $(1) \Rightarrow (2) \Rightarrow (1)$, $(1) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$. For $(1) \Leftrightarrow (2)$, the spectral theorem reduces the quadratic form to a diagonal one whose sign is controlled by the eigenvalues. For $(1) \Rightarrow (3)$, positive definiteness passes to principal submatrices, forcing each leading minor to be a product of positive eigenvalues. For $(3) \Rightarrow (4)$, we construct the Cholesky factor $L$ column by column via an inductive block factorisation, using the positivity of the leading minors to guarantee that each diagonal entry of $L$ is real and positive. Finally, $(4) \Rightarrow (1)$ follows by writing the quadratic form as $\|L^\top x\|^2$. [/proofplan]

[step:Show that positive definiteness is equivalent to all eigenvalues being strictly positive]Since $A \in \mathbb{R}^{n \times n}$ is symmetric, the [Spectral Theorem for Hermitian Matrices](/theorems/925) guarantees an orthogonal matrix $Q \in \mathbb{R}^{n \times n}$ (i.e., $Q^\top Q = I_n$) and a diagonal matrix $\Lambda = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$ with $\lambda_1, \dots, \lambda_n \in \mathbb{R}$ such that $A = Q \Lambda Q^\top$. $(1) \Rightarrow (2)$: Suppose $A$ is positive definite. Let $\lambda_j$ be an eigenvalue with corresponding unit eigenvector $q_j$ (the $j$-th column of $Q$). Then \begin{align*} 0 < q_j^\top A q_j = q_j^\top (\lambda_j q_j) = \lambda_j |q_j|^2 = \lambda_j, \end{align*} so every eigenvalue is strictly positive. $(2) \Rightarrow (1)$: Suppose $\lambda_j > 0$ for all $j = 1, \dots, n$. For any $x \in \mathbb{R}^n \setminus \{0\}$, define $y = Q^\top x$. Since $Q$ is orthogonal, $y \neq 0$, so at least one component $y_j \neq 0$. Then \begin{align*} x^\top A x = x^\top Q \Lambda Q^\top x = y^\top \Lambda y = \sum_{j=1}^{n} \lambda_j y_j^2 > 0, \end{align*} where the strict inequality holds because each $\lambda_j > 0$ and at least one $y_j^2 > 0$.[/step]

[guided]The spectral theorem is the key structural tool: it tells us that every real symmetric matrix is orthogonally diagonalisable. This means we can study the quadratic form $x^\top A x$ by changing to the eigenbasis, where $A$ acts as a diagonal scaling. Since $A$ is symmetric, the [Spectral Theorem for Hermitian Matrices](/theorems/925) provides an orthogonal matrix $Q \in \mathbb{R}^{n \times n}$ (satisfying $Q^\top Q = Q Q^\top = I_n$) and real eigenvalues $\lambda_1, \dots, \lambda_n$ such that $A = Q \Lambda Q^\top$, where $\Lambda = \operatorname{diag}(\lambda_1, \dots, \lambda_n)$. $(1) \Rightarrow (2)$: Suppose $A$ is positive definite, meaning $x^\top A x > 0$ for every $x \in \mathbb{R}^n \setminus \{0\}$. To show each eigenvalue is positive, we test the quadratic form on the eigenvectors. Let $q_j$ be the $j$-th column of $Q$, so $A q_j = \lambda_j q_j$ and $|q_j| = 1$. Then \begin{align*} 0 < q_j^\top A q_j = q_j^\top (\lambda_j q_j) = \lambda_j |q_j|^2 = \lambda_j. \end{align*} Since $q_j \neq 0$, positive definiteness forces $\lambda_j > 0$. $(2) \Rightarrow (1)$: Suppose all eigenvalues $\lambda_1, \dots, \lambda_n$ are strictly positive. For any $x \in \mathbb{R}^n \setminus \{0\}$, define $y := Q^\top x \in \mathbb{R}^n$. Since $Q$ is orthogonal, $|y| = |Q^\top x| = |x| > 0$, so $y \neq 0$. The orthogonal change of basis transforms the quadratic form into a sum of squares weighted by the eigenvalues: \begin{align*} x^\top A x = x^\top Q \Lambda Q^\top x = y^\top \Lambda y = \sum_{j=1}^{n} \lambda_j y_j^2. \end{align*} Each term $\lambda_j y_j^2 \ge 0$ (since $\lambda_j > 0$), and at least one $y_j \neq 0$, so the sum is strictly positive. This is the geometric content: in the eigenbasis, the quadratic form is a weighted sum of squares, and the weights are exactly the eigenvalues.[/guided]

[step:Show that positive definiteness implies all leading principal minors are positive]$(1) \Rightarrow (3)$: Suppose $A$ is positive definite. For each $k \in \{1, \dots, n\}$, let $A_k \in \mathbb{R}^{k \times k}$ denote the top-left $k \times k$ submatrix of $A$, called the $k$-th leading principal submatrix. We claim $A_k$ is positive definite. For any $y \in \mathbb{R}^k \setminus \{0\}$, define $x = (y_1, \dots, y_k, 0, \dots, 0)^\top \in \mathbb{R}^n$. Then $x \neq 0$ and \begin{align*} x^\top A x = \sum_{i=1}^{k} \sum_{j=1}^{k} A_{ij} y_i y_j = y^\top A_k y. \end{align*} Since $A$ is positive definite, $x^\top A x > 0$, hence $y^\top A_k y > 0$. Since $A_k$ is a $k \times k$ real symmetric positive definite matrix, by the equivalence $(1) \Leftrightarrow (2)$ established in the previous step, all eigenvalues of $A_k$ are strictly positive. The determinant equals the product of the eigenvalues, so $\det A_k > 0$.[/step]

[guided]$(1) \Rightarrow (3)$: Suppose $A$ is positive definite. We need to show that every leading principal minor $\det A_k$ is positive for $k = 1, \dots, n$. The idea is that positive definiteness is inherited by principal submatrices: restricting the quadratic form $x^\top A x$ to vectors supported on the first $k$ coordinates yields $y^\top A_k y$. For each $k \in \{1, \dots, n\}$, let $A_k \in \mathbb{R}^{k \times k}$ be the leading principal submatrix (the top-left $k \times k$ block of $A$). $A_k$ is symmetric because $A$ is symmetric and taking a principal submatrix preserves symmetry. For any $y \in \mathbb{R}^k \setminus \{0\}$, embed $y$ into $\mathbb{R}^n$ by defining $x = (y_1, \dots, y_k, 0, \dots, 0)^\top \in \mathbb{R}^n$. Then $x \neq 0$, and by the block structure of the quadratic form: \begin{align*} x^\top A x = \sum_{i=1}^{n} \sum_{j=1}^{n} A_{ij} x_i x_j = \sum_{i=1}^{k} \sum_{j=1}^{k} A_{ij} y_i y_j = y^\top A_k y. \end{align*} The cross terms with indices $i > k$ or $j > k$ vanish because $x_i = 0$ for $i > k$. Since $A$ is positive definite and $x \neq 0$, we have $y^\top A_k y = x^\top A x > 0$. This holds for all $y \neq 0$, so $A_k$ is positive definite. Now, by the equivalence $(1) \Leftrightarrow (2)$ proved in the previous step, all eigenvalues $\mu_1, \dots, \mu_k$ of $A_k$ are strictly positive. Since the determinant of a matrix equals the product of its eigenvalues: \begin{align*} \det A_k = \prod_{j=1}^{k} \mu_j > 0. \end{align*}[/guided]

[step:Construct the Cholesky factor from positive leading minors by induction on dimension]$(3) \Rightarrow (4)$: Suppose all leading principal minors $\det A_k > 0$ for $k = 1, \dots, n$. We construct a lower triangular matrix $L \in \mathbb{R}^{n \times n}$ with strictly positive diagonal entries such that $A = L L^\top$ by induction on $n$. **Base case** ($n = 1$): $A = (a_{11})$ with $\det A_1 = a_{11} > 0$. Set $L = (\sqrt{a_{11}})$. Then $L L^\top = a_{11} = A$ and the diagonal entry $\sqrt{a_{11}} > 0$. **Inductive step**: Assume the result holds for $(n-1) \times (n-1)$ symmetric matrices with positive leading principal minors. Write $A$ in block form: \begin{align*} A = \begin{pmatrix} A_{n-1} & b \\ b^\top & a_{nn} \end{pmatrix}, \end{align*} where $A_{n-1} \in \mathbb{R}^{(n-1) \times (n-1)}$ is the leading principal submatrix, $b \in \mathbb{R}^{n-1}$, and $a_{nn} \in \mathbb{R}$. The leading principal minors of $A_{n-1}$ are $\det A_1, \dots, \det A_{n-1}$, all positive by hypothesis. By the inductive hypothesis, there exists a unique lower triangular $L_{n-1} \in \mathbb{R}^{(n-1) \times (n-1)}$ with strictly positive diagonal entries such that $A_{n-1} = L_{n-1} L_{n-1}^\top$. We seek $L$ in the form \begin{align*} L = \begin{pmatrix} L_{n-1} & 0 \\ w^\top & \ell_{nn} \end{pmatrix}, \end{align*} where $w \in \mathbb{R}^{n-1}$ and $\ell_{nn} > 0$. The equation $A = L L^\top$ yields the system: \begin{align*} A_{n-1} &= L_{n-1} L_{n-1}^\top, \\ b &= L_{n-1} w, \\ a_{nn} &= |w|^2 + \ell_{nn}^2. \end{align*} The first equation is satisfied by construction. Since $L_{n-1}$ is lower triangular with positive diagonal entries, it is invertible, so $w = L_{n-1}^{-1} b$ is uniquely determined. For the third equation, we need $a_{nn} - |w|^2 > 0$. Observe that \begin{align*} \det A = \det \begin{pmatrix} A_{n-1} & b \\ b^\top & a_{nn} \end{pmatrix} = \det(A_{n-1}) \cdot (a_{nn} - b^\top A_{n-1}^{-1} b), \end{align*} where the second equality follows from expanding the determinant along the last row: subtracting $b^\top A_{n-1}^{-1}$ times the first $n-1$ rows from the last row eliminates the off-diagonal entry $b^\top$, leaving $\det(A_{n-1}) \cdot (a_{nn} - b^\top A_{n-1}^{-1} b)$. Since $A_{n-1}^{-1} = (L_{n-1} L_{n-1}^\top)^{-1} = L_{n-1}^{-\top} L_{n-1}^{-1}$, we have \begin{align*} b^\top A_{n-1}^{-1} b = b^\top L_{n-1}^{-\top} L_{n-1}^{-1} b = |L_{n-1}^{-1} b|^2 = |w|^2. \end{align*} Therefore $a_{nn} - |w|^2 = \det A / \det A_{n-1} > 0$, since both $\det A > 0$ and $\det A_{n-1} > 0$ by hypothesis. We set $\ell_{nn} = \sqrt{a_{nn} - |w|^2} > 0$, completing the construction. **Uniqueness**: Suppose $A = L L^\top = \tilde{L} \tilde{L}^\top$ with both $L, \tilde{L}$ lower triangular with positive diagonal entries. The block structure of the inductive construction shows that $L_{n-1}$ is uniquely determined by the inductive hypothesis, $w = L_{n-1}^{-1} b$ is then uniquely determined, and $\ell_{nn} = \sqrt{a_{nn} - |w|^2}$ is uniquely determined (taking the positive root). By induction, $L$ is unique.[/step]

[guided]$(3) \Rightarrow (4)$: We construct the Cholesky factor by peeling off one row and column at a time. The key insight is that the block factorisation of a symmetric matrix reduces the problem from dimension $n$ to dimension $n-1$, and the positivity of the leading principal minors ensures that the diagonal entries of $L$ are real and positive at each stage. Suppose all leading principal minors $\det A_k > 0$ for $k = 1, \dots, n$. We construct a lower triangular $L$ with strictly positive diagonal entries satisfying $A = L L^\top$ by induction on $n$. **Base case** ($n = 1$): $A = (a_{11})$ with $a_{11} = \det A_1 > 0$. Set $L = (\sqrt{a_{11}})$. Then $L L^\top = a_{11} = A$. **Inductive step**: Assume the result holds for dimension $n - 1$. Partition $A$ as \begin{align*} A = \begin{pmatrix} A_{n-1} & b \\ b^\top & a_{nn} \end{pmatrix}, \end{align*} where $A_{n-1} \in \mathbb{R}^{(n-1) \times (n-1)}$ is the $(n-1)$-th leading principal submatrix, $b \in \mathbb{R}^{n-1}$ is the last column of $A$ restricted to the first $n-1$ rows, and $a_{nn} \in \mathbb{R}$. The leading principal minors of $A_{n-1}$ are $\det A_1, \dots, \det A_{n-1}$, all positive. By the inductive hypothesis, there exists a unique lower triangular $L_{n-1} \in \mathbb{R}^{(n-1) \times (n-1)}$ with positive diagonal entries such that $A_{n-1} = L_{n-1} L_{n-1}^\top$. We look for $L$ in the block lower triangular form \begin{align*} L = \begin{pmatrix} L_{n-1} & 0 \\ w^\top & \ell_{nn} \end{pmatrix}, \end{align*} where $w \in \mathbb{R}^{n-1}$ and $\ell_{nn} > 0$ are to be determined. Expanding $L L^\top$: \begin{align*} L L^\top = \begin{pmatrix} L_{n-1} L_{n-1}^\top & L_{n-1} w \\ w^\top L_{n-1}^\top & |w|^2 + \ell_{nn}^2 \end{pmatrix}. \end{align*} Matching blocks with $A$: - Top-left: $L_{n-1} L_{n-1}^\top = A_{n-1}$ (satisfied by construction). - Off-diagonal: $L_{n-1} w = b$. Since $L_{n-1}$ is lower triangular with positive diagonal entries, $\det L_{n-1} = \prod_{i=1}^{n-1} (L_{n-1})_{ii} > 0$, so $L_{n-1}$ is invertible. Thus $w = L_{n-1}^{-1} b$ is uniquely determined. - Bottom-right: $|w|^2 + \ell_{nn}^2 = a_{nn}$, i.e., $\ell_{nn}^2 = a_{nn} - |w|^2$. The critical question: is $a_{nn} - |w|^2 > 0$? We compute the determinant of the block matrix directly. Since $A_{n-1}$ is invertible, we may subtract $b^\top A_{n-1}^{-1}$ times the first $n-1$ rows from the last row, eliminating the off-diagonal block $b^\top$ without changing the determinant. This block row reduction gives: \begin{align*} \det A = \det(A_{n-1}) \cdot \det(a_{nn} - b^\top A_{n-1}^{-1} b) = \det(A_{n-1}) \cdot (a_{nn} - b^\top A_{n-1}^{-1} b). \end{align*} Now we connect $b^\top A_{n-1}^{-1} b$ to $|w|^2$. Since $A_{n-1} = L_{n-1} L_{n-1}^\top$: \begin{align*} A_{n-1}^{-1} = (L_{n-1} L_{n-1}^\top)^{-1} = L_{n-1}^{-\top} L_{n-1}^{-1}. \end{align*} Therefore: \begin{align*} b^\top A_{n-1}^{-1} b = b^\top L_{n-1}^{-\top} L_{n-1}^{-1} b = (L_{n-1}^{-1} b)^\top (L_{n-1}^{-1} b) = |w|^2. \end{align*} Substituting: $\det A = \det(A_{n-1}) \cdot (a_{nn} - |w|^2)$. Both $\det A > 0$ (hypothesis, the $n$-th leading minor) and $\det A_{n-1} > 0$ (hypothesis), so \begin{align*} a_{nn} - |w|^2 = \frac{\det A}{\det A_{n-1}} > 0. \end{align*} We set $\ell_{nn} = \sqrt{a_{nn} - |w|^2} > 0$. This completes the construction of $L$. **Uniqueness**: The inductive hypothesis gives uniqueness of $L_{n-1}$. Once $L_{n-1}$ is fixed, $w = L_{n-1}^{-1} b$ is uniquely determined. Since $\ell_{nn}$ must be strictly positive, $\ell_{nn} = \sqrt{a_{nn} - |w|^2}$ is the only choice. By induction, $L$ is unique.[/guided]

[step:Close the cycle by showing the Cholesky factorisation implies positive definiteness]$(4) \Rightarrow (1)$: Suppose $A = L L^\top$ where $L \in \mathbb{R}^{n \times n}$ is lower triangular with strictly positive diagonal entries. For any $x \in \mathbb{R}^n \setminus \{0\}$, define $y = L^\top x \in \mathbb{R}^n$. Since $L^\top$ is upper triangular with strictly positive diagonal entries, $\det L^\top = \prod_{i=1}^n L_{ii} > 0$, so $L^\top$ is invertible. Therefore $y = L^\top x \neq 0$, and \begin{align*} x^\top A x = x^\top L L^\top x = (L^\top x)^\top (L^\top x) = |y|^2 > 0. \end{align*} Since $x^\top A x > 0$ for every $x \neq 0$, the matrix $A$ is positive definite. This completes the equivalence of all four characterisations.[/step]

[guided]$(4) \Rightarrow (1)$: This is the most direct implication. Suppose $A = L L^\top$ where $L$ is lower triangular with strictly positive diagonal entries. The idea is that $x^\top A x$ factors as the squared Euclidean norm of $L^\top x$. For any $x \in \mathbb{R}^n \setminus \{0\}$, define $y := L^\top x$. We need $y \neq 0$: since $L$ is lower triangular with each diagonal entry $L_{ii} > 0$, the transpose $L^\top$ is upper triangular with the same positive diagonal entries, so $\det L^\top = \prod_{i=1}^n L_{ii} > 0$. In particular, $L^\top$ is invertible, and since $x \neq 0$, we conclude $y \neq 0$. Now compute: \begin{align*} x^\top A x = x^\top L L^\top x = (L^\top x)^\top (L^\top x) = y^\top y = |y|^2 = \sum_{i=1}^{n} y_i^2 > 0, \end{align*} where the strict inequality holds because $y \neq 0$ implies at least one $y_i \neq 0$. Since $x^\top A x > 0$ for every nonzero $x$, the matrix $A$ is positive definite. Why does this direction feel so clean? Because the Cholesky factorisation literally encodes $A$ as a "squared" object: the quadratic form $x^\top A x$ becomes $|L^\top x|^2$, which is manifestly non-negative and zero only if $L^\top x = 0$. The invertibility of $L$ then forces $x = 0$. This is the same principle as completing the square, but systematised for all dimensions.[/guided]