Let Bn be the unit matrix and A be the identity matrix then C= Bn-A .What is the rank of C?
Jack
It is not clear what you mean by "the unit matrix". If you mean the n x n matrix all of whose entries are 1, then the fact that B^2 = nB means B is diagonalizable with spectrum {0,n} since the minimum polynomial of B must divide x^2 - nx = x(x - n). Indeed, the all ones vector v is the unique (up to scalar multiple) eigenvector for B for the eigenvalue n. Why unique? Two ways to see this: BIG Theorem approach - use the Perron-Frobenius Theorem. SMALL Theorem approach: B = v v^T, so B is clearly rank 1 with eigenvector v for the eigenvalue v^T v = n. In any case, given all of this information about B, the eigenvalues of diagonalizable matrix B - cI where I is the n x n identity matrix are n - c (multiplicity 1) and - c (multiplicity n - 1). So, B - cI has rank n except when either c = 0 or c = n. When c = 1 (your question), B - I has rank n except when n = 1.
If I have misunderstood your question, please ask again, but be much more explicit about what you mean by the matrix B.
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