Find scalars that result in an idempotent matrix

Question

For what scalars, $b$ and $c$ will the following matrix be idempotent:

$bI_m + c1_m1_m^T$

I know that idempotency implies that each eigenvalue is 0 or 1, but I am still having trouble finding the b and c.

Answer 1

Assuming that by $1_m$ you mean a column matrix of $1$s, then call $A = 1_m1_m^T$ and $B = bI_m + cA$ is the idempotent matrix: $B^n = B$ for some $n$. Now for any $k > 0$, it is easy to check that $A^k = m^{k-1}A$. So $$\begin{align}B^n = (bI_m + cA)^n &= \sum_{k=0}^n \binom n k b^{n-k}c^kA^k\\ &= b^nI_m + \left(\sum_{k=1}^n \binom n k b^{n-k}c^km^{k-1}\right)A\\ &= bI_m + cA = B \end{align}$$

By looking at the off-diagonal elements, we see that $$\sum_{k=1}^n \binom n k b^{n-k}c^km^{k-1} = c$$ And subtracting that from the diagonal elements shows that $$b^n = b$$ Thus, $$(b + cm)^n = \sum_{k=0}^n \binom n k b^{n-k}c^km^k = b^n + m\sum_{k=1}^n \binom n k b^{n-k}c^km^{k-1} = b + mc$$ Thus both $b$ and $b + cm$ are either $0$ or $(n-1)^\text{th}$-roots of unity.

Answer 2

An idempotent matrix satisfies $A^2=A$.

Let $J$ be the matrix of all ones. Then squaring your matrix $$\eqalign{ A &= bI + cJ \cr A^2 &= b^2I + (2bc+mc^2)J \cr }$$ and equating coefficients yields $$\eqalign{ b &= b^2 \cr c &= (2b+mc)c \cr }$$ These equations are satisfied by 4 sets of coefficients $$\eqalign{ (b,c) &= (0,0) \cr (b,c) &= (1,0) \cr (b,c) &= (0,\,1/m) \cr (b,c) &= (1,\,-1/m) \cr }$$ The key insight for getting to this solution is that $J^2=mJ$.