Correlation Matrices Are Positive Semidefinite
A correlation matrix $C \in \mathbb{R}^{d \times d}$ is the matrix of pairwise correlations of some random vector $Z = (Z_1, \ldots, Z_d)^\top$, where $C_{ij} = \text{Corr}(Z_i, Z_j)$, so the diagonal entries are all 1.
- Define what it means for a symmetric matrix $M \in \mathbb{R}^{d \times d}$ to be positive semidefinite (PSD).
- Prove that every correlation matrix is PSD.
- Address the converse: if $C$ is a symmetric matrix with ones on the diagonal and $C$ is PSD, must there exist a random vector $Z$ whose correlation matrix is exactly $C$? Prove your answer.
Hints
- Think about what it means to form the linear combination $v^\top Z$ for an arbitrary vector $v$ -- what property of variance can you exploit?
- For the PSD proof, standardize the components of $Z$ first so the covariance matrix equals the correlation matrix, then compute $\text{Var}(v^\top \tilde{Z})$ and expand.
- For the converse, the multivariate normal $\mathcal{N}(0, C)$ is well-defined whenever $C$ is symmetric PSD -- and if $C$ has ones on the diagonal, its components each have unit variance, so covariances equal correlations directly.
Worked Solution
How to Think About It: This is a classic linear algebra meets probability question. The PSD condition is all about variance -- and variance is always non-negative. That single fact is the engine of the entire proof. For part (1), write the definition. For part (2), construct a variance: take any vector $v$, form the linear combination $v^\top Z$, and use the fact that its variance must be non-negative. For part (3), the answer is yes, and the construction is direct -- use a multivariate normal.
Key Insight: Positive semidefiniteness of a covariance or correlation matrix is not a coincidence or a constraint you impose -- it is a logical necessity that falls out of $\text{Var}(v^\top Z) \geq 0$.
Part 1: Definition of Positive Semidefinite
A symmetric matrix $M \in \mathbb{R}^{d \times d}$ is positive semidefinite if for every vector $v \in \mathbb{R}^d$,
$v^\top M v \geq 0.$
Equivalently, $M$ is PSD if and only if all its eigenvalues are non-negative.
Part 2: Every Correlation Matrix Is PSD
Let $Z = (Z_1, \ldots, Z_d)^\top$ be a random vector with finite second moments, and assume each $Z_i$ has positive variance (so correlations are well-defined). Without loss of generality, standardize each component: let $\tilde{Z}_i = (Z_i - \mu_i) / \sigma_i$ where $\mu_i = E[Z_i]$ and $\sigma_i = \sqrt{\text{Var}(Z_i)}$. Then $C_{ij} = E[\tilde{Z}_i \tilde{Z}_j]$.
For any $v \in \mathbb{R}^d$, consider the scalar random variable $W = v^\top \tilde{Z} = \sum_{i=1}^d v_i \tilde{Z}_i$. Since $E[\tilde{Z}_i] = 0$ for all $i$, we have $E[W] = 0$, so
$\text{Var}(W) = E[W^2] = E\!\left[\left(\sum_i v_i \tilde{Z}_i\right)^2\right] = \sum_{i,j} v_i v_j E[\tilde{Z}_i \tilde{Z}_j] = \sum_{i,j} v_i v_j C_{ij} = v^\top C v.$
Since variance is always non-negative, $v^\top C v = \text{Var}(W) \geq 0$ for every $v$. Therefore $C$ is PSD. $\square$
Part 3: Converse -- Every PSD Matrix with Ones on the Diagonal Is a Correlation Matrix
The converse holds. The construction: let $Z \sim \mathcal{N}(0, C)$, i.e., a multivariate normal with mean zero and covariance matrix $C$.
First, this is a valid distribution: a multivariate normal $\mathcal{N}(0, \Sigma)$ exists if and only if $\Sigma$ is symmetric PSD -- and by assumption $C$ is symmetric PSD. (In the degenerate case where $C$ has zero eigenvalues, the distribution is supported on a lower-dimensional subspace, but it still exists.)
Next, compute the correlation matrix of $Z$. Since $C_{ii} = 1$ for all $i$, each $Z_i$ has variance 1, so $\text{Var}(Z_i) = 1$. The covariance between $Z_i$ and $Z_j$ is $C_{ij}$ by construction. Therefore
$\text{Corr}(Z_i, Z_j) = \frac{\text{Cov}(Z_i, Z_j)}{\sqrt{\text{Var}(Z_i) \, \text{Var}(Z_j)}} = \frac{C_{ij}}{1 \cdot 1} = C_{ij}.$
So the correlation matrix of $Z$ is exactly $C$. $\square$
Answer:
- $M$ is PSD if $v^\top M v \geq 0$ for all $v \in \mathbb{R}^d$.
- Every correlation matrix is PSD because $v^\top C v = \text{Var}(v^\top \tilde{Z}) \geq 0$.
- The converse holds: any symmetric PSD matrix with ones on the diagonal is the correlation matrix of $Z \sim \mathcal{N}(0, C)$.
Intuition
The PSD property of correlation matrices is not a constraint that statisticians decided to impose -- it is forced on you by probability itself. Variance cannot be negative. The moment you write down a linear combination of random variables, its variance must be $\geq 0$, and that variance is exactly $v^\top C v$. So any matrix of pairwise correlations, no matter how it was formed, inherits this property for free. This is also why, in practice, estimated correlation matrices can fail to be PSD: sample correlations computed from finite data can violate this because rounding, missing data, or pairwise-computed correlations (where each pair uses a different sample) break the structure. When a risk model or factor model produces a non-PSD correlation matrix, you need to project it back onto the PSD cone -- a common task in quant risk work.
The converse tells you the PSD-plus-unit-diagonal characterization is tight: every such matrix can be realized, and the multivariate normal is the canonical witness. In practice this means PSD with unit diagonal is not just a necessary condition for being a correlation matrix -- it is also sufficient. This matters whenever you are constructing or stress-testing correlation matrices for a portfolio model: as long as you maintain PSD and unit diagonal (e.g., via eigenvalue clipping and rescaling), you are guaranteed a valid probabilistic interpretation.