Eigenvalues of $I - [Y(Y'Y)^{-1}Y']$ Let $Y$ be a $(r \times s)$ matrix with linearly independent columns. Let 
$$T= (I_r - Y(Y'Y)^{-1}Y'$$
I'm looking for advice on finding the eigenvalues of $T$. $T$ is idempotent so I believe the only possibilities are $0,1$, and I believe they also must be of the form $1- \lambda$ where $\lambda$ is an eigenvalue of $Y(Y'Y)^{-1}Y'$. $T$ is also symmetric, if that helps.
Is it possible to be more specific than that? If so, can I get a hint or two? Perhaps the fact that $(r\times s)$ has linearly independent columns can be used?
I prefer hints over being given the answer here.
Thanks.
 A: Using the identity $\text{trace}[AB] = \text{trace}[BA]$ for compatible matrices $A$,$B$, we get that $\text{trace}[Y(Y'Y)^{-1}Y'] = \text{trace}[Y'Y(Y'Y)^{-1}] = \text{trace}[I_s] = s$. 
Hence, the sum of the eigenvalues of $Y(Y'Y)^{-1}Y'$ is $s$. 
Since $Y(Y'Y)^{-1}Y'$ is idempotent, all of its eigenvalues are $0$ or $1$ (as you mentioned in your question). Can you figure out how many of the eigenvalues are $0$ and how many of them are $1$?
After you do that, use the relationship between the eigenvalues of $Y(Y'Y)^{-1}Y'$ and the eigenvalues of $T = I_r -Y(Y'Y)^{-1}Y'$ which you mentioned.

EDIT: A more intuitive explanation is that $Y(Y'Y)^{-1}Y'$ is the matrix which projects a vector onto the span of the columns of $Y$, which has dimension $s$ since $Y$ has $s$ linearly independent columns. So $T = I-Y(Y'Y)^{-1}Y'$ projects vectors onto the orthogonal complement of the span of the columns of $Y$. What is the dimension of this space? That will also tell you how many eigenvalues are $1$ and how many are $0$.
