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I knew the setup immediately but fumbled the justification.
First, recognize that the matrix has a special structure: it can be written as (1-a)I + aJ, where J is the all-ones matrix. Then, find the eigenvalues of this matrix and determine the condition on a for all eigenvalues to be nonnegative. Alternatively, use Sylvester's criterion to derive the same condition.
Pro tip: Mention that the matrix is a common example in covariance matrices and that the condition a >= -1/2 ensures positive semidefiniteness, which is crucial for applications like correlation matrices.
Write the 3x3 matrix explicitly and observe that it can be expressed as (1-a)I + aJ, where J is the all-ones matrix.
Find the eigenvalues of the matrix. Since J has eigenvalues 3 (multiplicity 1) and 0 (multiplicity 2), the eigenvalues of the matrix are (1-a) + 3a = 1+2a and (1-a) (with multiplicity 2).
For the matrix to be positive semidefinite, all eigenvalues must be nonnegative. Thus, require 1+2a >= 0 and 1-a >= 0, which gives a >= -1/2 and a <= 1.
The smallest value of a that satisfies both inequalities is a = -1/2. Verify that at a = -1/2, the matrix is positive semidefinite (one eigenvalue zero, others positive).
Optionally, use Sylvester's criterion: all principal minors must be nonnegative. Compute the 1x1 minors (1), 2x2 minors (1 - a^2), and the determinant (1 - a)^2(1+2a). This yields the same condition a >= -1/2.
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