[proofplan] Since $A$ is real symmetric, the [Spectral Theorem for Hermitian Matrices](/theorems/925) provides an orthonormal eigenbasis. We express the quadratic form $x^\top A x$ in this eigenbasis, reducing it to a weighted sum of squares $\sum_k \lambda_k c_k^2$. The sign behaviour of this sum is then controlled entirely by the eigenvalues: positivity for all nonzero $x$ is equivalent to all weights being positive (i.e., $\lambda_1 > 0$), and the remaining cases follow by analogous reasoning. [/proofplan] [step:Diagonalise $A$ in an orthonormal eigenbasis] Since $A \in \mathbb{R}^{n \times n}$ is symmetric, it is Hermitian. By the [Spectral Theorem for Hermitian Matrices](/theorems/925), there exists an orthogonal matrix $P \in \mathbb{R}^{n \times n}$ (satisfying $P^\top P = I_n$) such that \begin{align*} P^\top A P = \operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n), \end{align*} where $\lambda_1 \le \lambda_2 \le \cdots \le \lambda_n$ are the eigenvalues of $A$. The columns of $P$ form an orthonormal basis $\{v_1, \ldots, v_n\}$ of $\mathbb{R}^n$ consisting of eigenvectors of $A$ with $A v_k = \lambda_k v_k$ for each $k = 1, \ldots, n$. [guided] The proof hinges on reducing the quadratic form $x^\top A x$ to a diagonal expression. How do we achieve this? The key tool is the [Spectral Theorem for Hermitian Matrices](/theorems/925), which applies because every real symmetric matrix is in particular Hermitian. The theorem requires that $A$ is Hermitian (i.e., $A^\dagger = A$); since $A$ is real and symmetric, $A^\dagger = \overline{A}^\top = A^\top = A$, so the hypothesis is satisfied. Moreover, since $A$ is real symmetric, the theorem guarantees that $P$ can be chosen to be real orthogonal (not merely unitary). The spectral theorem produces an orthogonal matrix $P \in \mathbb{R}^{n \times n}$ with $P^\top P = I_n$ such that \begin{align*} P^\top A P = \operatorname{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n). \end{align*} Writing $P = [v_1 \mid v_2 \mid \cdots \mid v_n]$, the columns $\{v_1, \ldots, v_n\}$ form an orthonormal basis of $\mathbb{R}^n$ with $A v_k = \lambda_k v_k$ for each $k$. That the eigenvalues are real follows from the [Eigenvalues of Symmetric Matrices are Real](/theorems/3279) theorem, so the ordering $\lambda_1 \le \cdots \le \lambda_n$ is well-defined. [/guided] [/step] [step:Express the quadratic form as a weighted sum of squares] Let $x \in \mathbb{R}^n$ be arbitrary. Since $\{v_1, \ldots, v_n\}$ is an orthonormal basis, we may write \begin{align*} x = \sum_{k=1}^{n} c_k v_k, \quad \text{where } c_k = v_k^\top x \in \mathbb{R}. \end{align*} Substituting into the quadratic form and using $A v_k = \lambda_k v_k$ together with orthonormality ($v_j^\top v_k = \delta_{jk}$): \begin{align*} x^\top A x = \left(\sum_{j=1}^{n} c_j v_j\right)^\top A \left(\sum_{k=1}^{n} c_k v_k\right) = \sum_{j=1}^{n} \sum_{k=1}^{n} c_j c_k \, v_j^\top (\lambda_k v_k) = \sum_{j=1}^{n} \sum_{k=1}^{n} c_j c_k \lambda_k \, \delta_{jk} = \sum_{k=1}^{n} \lambda_k c_k^2. \end{align*} Equivalently, defining $y = P^\top x \in \mathbb{R}^n$ (so $y_k = c_k$), we have $x^\top A x = y^\top \operatorname{diag}(\lambda_1, \ldots, \lambda_n) y = \sum_{k=1}^n \lambda_k y_k^2$. [guided] The change of coordinates $y = P^\top x$ is the core of the argument. Since $P$ is orthogonal, the map $x \mapsto y = P^\top x$ is a bijection $\mathbb{R}^n \to \mathbb{R}^n$ that preserves the Euclidean norm: $|y|^2 = y^\top y = x^\top P P^\top x = x^\top x = |x|^2$. In particular, $x \neq 0$ if and only if $y \neq 0$. Expanding $x$ in the eigenbasis $\{v_1, \ldots, v_n\}$: \begin{align*} x = \sum_{k=1}^{n} c_k v_k, \quad c_k = v_k^\top x. \end{align*} The coefficients $c_k$ are exactly the components of $y = P^\top x$, since $(P^\top x)_k = v_k^\top x = c_k$. Now compute the quadratic form. Using linearity of $A$ and orthonormality: \begin{align*} x^\top A x &= \left(\sum_{j=1}^{n} c_j v_j\right)^\top \left(\sum_{k=1}^{n} c_k A v_k\right) = \left(\sum_{j=1}^{n} c_j v_j\right)^\top \left(\sum_{k=1}^{n} c_k \lambda_k v_k\right) \\ &= \sum_{j=1}^{n} \sum_{k=1}^{n} c_j c_k \lambda_k \, v_j^\top v_k = \sum_{j=1}^{n} \sum_{k=1}^{n} c_j c_k \lambda_k \, \delta_{jk} = \sum_{k=1}^{n} \lambda_k c_k^2. \end{align*} This is the key identity: the quadratic form $x^\top A x$ decomposes as a weighted sum of squares, with the eigenvalues as weights. The sign of $x^\top A x$ for arbitrary nonzero $x$ is therefore governed entirely by the signs of $\lambda_1, \ldots, \lambda_n$ and the values of $c_1^2, \ldots, c_n^2$. Why is this decomposition useful? Because each $c_k^2 \ge 0$, and for any nonzero $x$, the vector $y = (c_1, \ldots, c_n)$ is also nonzero (since $P$ is invertible), so at least one $c_k^2 > 0$. Moreover, for any specific index $k$, we can choose $x = v_k$ to obtain $c_j = \delta_{jk}$, so $x^\top A x = \lambda_k$. This gives us both directions of the equivalences. [/guided] [/step] [step:Prove the positive definite characterisation] **($\Rightarrow$)** Suppose $A$ is positive definite, i.e., $x^\top A x > 0$ for all $x \in \mathbb{R}^n \setminus \{0\}$. For each eigenvalue $\lambda_k$, choose $x = v_k \neq 0$. Then $c_j = v_j^\top v_k = \delta_{jk}$, so \begin{align*} 0 < v_k^\top A v_k = \sum_{j=1}^{n} \lambda_j \delta_{jk}^2 = \lambda_k. \end{align*} In particular $\lambda_1 > 0$. **($\Leftarrow$)** Suppose $\lambda_1 > 0$. Since $\lambda_1 \le \lambda_2 \le \cdots \le \lambda_n$, all eigenvalues are positive. For any $x \neq 0$, the vector $y = P^\top x$ satisfies $y \neq 0$ (since $P$ is invertible), so at least one $c_k^2 > 0$. Therefore \begin{align*} x^\top A x = \sum_{k=1}^{n} \lambda_k c_k^2 > 0, \end{align*} since each summand $\lambda_k c_k^2 \ge 0$ and at least one is strictly positive. Hence $A$ is positive definite. [guided] We prove both directions of the equivalence: $A$ is positive definite if and only if $\lambda_1 > 0$. **Forward direction.** Assume $A$ is positive definite: $x^\top A x > 0$ for every nonzero $x \in \mathbb{R}^n$. We must show $\lambda_1 > 0$, which (since $\lambda_1$ is the smallest eigenvalue) implies all eigenvalues are positive. For any eigenvalue $\lambda_k$, the corresponding eigenvector $v_k$ is nonzero (eigenvectors are nonzero by definition). Choosing $x = v_k$, the expansion in the eigenbasis gives $c_j = v_j^\top v_k = \delta_{jk}$ (by orthonormality), so \begin{align*} v_k^\top A v_k = \sum_{j=1}^{n} \lambda_j \delta_{jk}^2 = \lambda_k. \end{align*} Positive definiteness requires $v_k^\top A v_k > 0$, hence $\lambda_k > 0$. Since this holds for every $k$, and $\lambda_1$ is the smallest, $\lambda_1 > 0$. **Backward direction.** Assume $\lambda_1 > 0$. Then $\lambda_k \ge \lambda_1 > 0$ for all $k = 1, \ldots, n$. Let $x \in \mathbb{R}^n \setminus \{0\}$. We need to show $x^\top A x > 0$. From the diagonal decomposition, $x^\top A x = \sum_{k=1}^n \lambda_k c_k^2$. Each term satisfies $\lambda_k c_k^2 \ge 0$ since $\lambda_k > 0$ and $c_k^2 \ge 0$. At least one $c_k \neq 0$: if all $c_k = 0$, then $x = \sum_k c_k v_k = 0$, contradicting $x \neq 0$. For that index, $\lambda_k c_k^2 > 0$. Therefore \begin{align*} x^\top A x = \sum_{k=1}^{n} \lambda_k c_k^2 \ge \lambda_k c_k^2 > 0. \end{align*} We can also give a quantitative lower bound: since $\lambda_k \ge \lambda_1$ for all $k$, \begin{align*} x^\top A x = \sum_{k=1}^{n} \lambda_k c_k^2 \ge \lambda_1 \sum_{k=1}^{n} c_k^2 = \lambda_1 |y|^2 = \lambda_1 |x|^2 > 0, \end{align*} where we used $|y|^2 = |x|^2$ (since $P$ is orthogonal) and $|x|^2 > 0$ (since $x \neq 0$). [/guided] [/step] [step:Prove the positive semidefinite characterisation] **($\Rightarrow$)** Suppose $A$ is positive semidefinite: $x^\top A x \ge 0$ for all $x \in \mathbb{R}^n$. Choosing $x = v_k$ gives $\lambda_k = v_k^\top A v_k \ge 0$ for each $k$, so $\lambda_1 \ge 0$. **($\Leftarrow$)** Suppose $\lambda_1 \ge 0$. Then $\lambda_k \ge 0$ for all $k$. For any $x \in \mathbb{R}^n$, \begin{align*} x^\top A x = \sum_{k=1}^{n} \lambda_k c_k^2 \ge 0, \end{align*} since each summand is non-negative. Hence $A$ is positive semidefinite. [/step] [step:Prove the negative definite characterisation] **($\Rightarrow$)** Suppose $A$ is negative definite: $x^\top A x < 0$ for all $x \neq 0$. Choosing $x = v_n$ gives $\lambda_n = v_n^\top A v_n < 0$. **($\Leftarrow$)** Suppose $\lambda_n < 0$. Then $\lambda_k \le \lambda_n < 0$ for all $k$. For any $x \neq 0$, at least one $c_k \neq 0$, so \begin{align*} x^\top A x = \sum_{k=1}^{n} \lambda_k c_k^2 < 0, \end{align*} since each nonzero summand $\lambda_k c_k^2 < 0$ (as $\lambda_k < 0$ and $c_k^2 > 0$) and the remaining summands are zero. Hence $A$ is negative definite. [/step] [step:Prove the indefinite characterisation] **($\Rightarrow$)** Suppose $A$ is indefinite: there exist $x, y \in \mathbb{R}^n \setminus \{0\}$ with $x^\top A x > 0$ and $y^\top A y < 0$. From the diagonal representation, $x^\top A x = \sum_k \lambda_k c_k^2$. If all eigenvalues were non-negative, every such sum would be non-negative, contradicting $y^\top A y < 0$. Hence some $\lambda_k < 0$, which forces $\lambda_1 < 0$. Similarly, if all eigenvalues were non-positive, every quadratic form value would be non-positive, contradicting $x^\top A x > 0$. Hence some $\lambda_k > 0$, which forces $\lambda_n > 0$. Therefore $\lambda_1 < 0 < \lambda_n$. **($\Leftarrow$)** Suppose $\lambda_1 < 0 < \lambda_n$. Choosing $x = v_1$ gives $x^\top A x = \lambda_1 < 0$. Choosing $x = v_n$ gives $x^\top A x = \lambda_n > 0$. Since the quadratic form takes both positive and negative values, $A$ is indefinite. [/step]