Positive Semidefiniteness of a 3×3 Correlation Matrix
We are given the symmetric matrix:
\[
\Sigma =
\begin{pmatrix}
1 & 0.6 & -0.3
0.6 & 1 & \rho
-0.3 & \rho & 1
\end{pmatrix}
\]
We are to find all real values of \( \rho \) for which this matrix is a valid correlation matrix, i.e., positive semidefinite (PSD).
Step 1: Symmetry and Unit Diagonal
The matrix is symmetric and has 1’s on the diagonal, satisfying basic properties of correlation matrices.
Step 2: Leading Principal Minors
To be PSD, all leading principal minors must be nonnegative.
First minor (top-left entry):
\[ 1 \geq 0 \quad \text{(True)} \]
Second-order minor:
\[
\begin{vmatrix}
1 & 0.6
0.6 & 1
\end{vmatrix}
= 1 - 0.6^2 = 1 - 0.36 = 0.64 > 0
\]
Full determinant:
\[ \det(\Sigma) = 1(1 \cdot 1 - \rho^2)
- 0.6(0.6 \cdot 1 - \rho \cdot (-0.3))
- (-0.3)(0.6 \cdot \rho - 1 \cdot (-0.3)) \]
Expanding and simplifying:
\[ = 1(1 - \rho^2) - 0.6(0.6 + 0.3\rho) - 0.3(0.6\rho + 0.3) \] \[ = 1 - \rho^2 - 0.36 - 0.18\rho - 0.18\rho - 0.09 \] \[ = (1 - 0.36 - 0.09) - \rho^2 - 0.36\rho = 0.55 - \rho^2 - 0.36\rho \]
So the condition is:
\[ \det(\Sigma) = 0.55 - \rho^2 - 0.36\rho \geq 0 \]
Step 3: Solve the Inequality
Solve:
\[ -\rho^2 - 0.36\rho + 0.55 \geq 0 \]
Multiply by -1 (flip the inequality):
\[ \rho^2 + 0.36\rho - 0.55 \leq 0 \]
Solve the quadratic:
\[ \rho = \frac{-0.36 \pm \sqrt{0.36^2 + 4 \cdot 0.55}}{2} = \frac{-0.36 \pm \sqrt{0.1296 + 2.2}}{2} = \frac{-0.36 \pm \sqrt{2.3296}}{2} \]
\[ \sqrt{2.3296} \approx 1.525 \Rightarrow \rho \in \left[ \frac{-0.36 - 1.525}{2}, \frac{-0.36 + 1.525}{2} \right] = [-0.9425, 0.5825] \]
Conclusion
The matrix is a valid correlation matrix if and only if:
\[ \rho \in [-0.9425, 0.5825] \]
This is the full set of real values for \( \rho \) making the matrix positive semidefinite.