What's the smallest eigenvalue of correlation matrix with all non-diagonal elements equal to alpha, where alpha: -1<alpha<1?

(1 - alpha) ?

Note: alpha has a lower-bounds that is higher than -1 and is an increasing function of the size of the matrix

and at that lower bound the min eigenvalue will be zero.ps. by the way that lower bound for alpha is -1/(n-1) where n is the size of the matrix.

if it's the smallest eigenvalues for all possible alpha, one could just use the fact that all eigenvalues of a semi-positive definite matrix are non-negative, so the minimum is zero (with either alpha=1 or -1/(N-1), where N is the matrix dimension)if you want the minimum for particular alpha and N, first consider when the determinant of the correlation matrix is zero, by arranging N unit vector in N dimensional space, and consider the special case that they degenerate into one single vector, or they form a regular (N-1)-dimensional polyhedron. so alpha = 1 or -1/(N-1), which also gives the possible range for alpha.(e.g., when N=3, either the end of these three vectors become one point or form a regular triangle in two dimension)now for the eigenvalue equation, it's simply replacing alpha with alpha/(1-lambda), thus lambda=1-alpha or 1+(N-1)*alpha. with the range given above, one gets the minimum is 1+(N-1)*alpha when -1/(N-1)<=alpha<=0, and 1-alpha when 0<=alpha<=1.----- ----- ----- ----- -----scooped...

Quoteone gets the minimum is 1+(N-1)*alpha when -1/(N-1)<=alpha<=0, and 1-alpha when 0<=alpha<=1.That's correct!

Matrix has two eigenvalues: non degenerate lambda1= 1+(N-1)*a, and (n-1) fold degenerate lambda2=1-a. All properties of eigenvalues, including the above follow from this.