MHB If matrix lambda is diagonal with entries 0,1 then lambda squared is lambda

[Jameson]

This is part of a proof I am working on involving idempotent matrices. I believe it is true that for any real symmetric matrix A ($n \times n$ for this), even with repeat eigenvalues, it can be decomposed into the form $A=Q \Lambda Q^{T}$. For the matrix I'm working on, we assume that all eigenvalues of A are 0 or 1.

What I need to show now is that $\Lambda^2=\Lambda$ (which in turn helps me show that $A^2=A$, thus is idempotent). It makes sense intuitively since when doing the row-column multiplications, the only time you would get a non-zero answer is when you have two non-zero elements being multiplied together in the sum, which only occurs when $i=j$. I'm trying to formulate a more rigorous argument for this though.

Just to make sure I'm clear, $\Lambda$ is an $n \times n$ matrix with all non-diagonal entries equal to 0, and diagonal entries equal to 0 or 1.

I know that $ \displaystyle \Lambda^2_{ij} = \sum_{k=1}^{n} \Lambda_{ik}\Lambda_{kj}$

Can I start here to make my argument you think?

 

[Nono713]

I'd use the dot product definition of matrix multiplication for this one. First get the $i \ne j$ case out of the way by showing that if $i \ne j$ then the $i$th row and $j$th column are either zero or orthogonal (since $\Lambda$ is diagonal so they can't have nonzero entries in the same dimension) so their dot product is zero. For $i = j$, use the fact that $\Lambda$ is diagonal and contains only entries 0, 1 to argue that the $i$th row and the $j$th column are equal, so their dot product is either 0 if they are zero (if the diagonal entry contains zero) or 1 otherwise (if the diagonal entry contains a 1, i.e. the two vectors are of unit length).

 

[Jameson]

Thanks, Bacterius. That was the logic I was thinking of but I didn't state it fully as you did. I think I can use your argument to sufficiently show what I need to. :)