I'm currently looking at achieving some proof that an $n \times n$ matrix (let's say $Y$) is idempotent if and only if $I_n-Y$ is an idempotent matrix, (with $I_n$ being the identity matrix).
I have so far proved that $I_n-Y$ is idempotent if $Y$ is idempotent as:
$(I_n-Y)^2 = I_n – 2Y + Y = I_n – Y$ hence idempotent.
But I don't know if this helps me at all and how to flip it so that if $I_n – Y$ is idempotent then $Y$ must be also idempotent.
Any help would be great!
Best Answer
Let $n \in \mathbb{N}$ and let $Y \in \cal{M}_{n \times n}\mathbb{(K)}$ where $\mathbb{K}$ is some field (with characteristic zero).
$\implies.$ Suppose that $Y$ is idempotent. Then, we know that $Y^2=Y.$ Hence
$$\begin{align}(I_n - Y)^2 & = (I_n - Y)(I_n - Y)\\ & = I_n^2 - I_nY -YI_n + Y^2\\ & = I_n -Y - Y + Y\\ & = I_n - Y \end{align}$$
Therefore, $Y - I_n$ is also idempotent.
$\Longleftarrow.$ Supposee that $I_n - Y$ is idempotent. Then, we know that $(I_n - Y)^2 = I_n - Y.$ Hence
$$\begin{align} (I_n - Y)^2 = I_n - Y & \implies I_n^2 - I_nY -YI_n + Y^2 = I_n - Y\\ & \implies I_n -2Y + Y^2 = I_n - Y\\ & \implies Y^2 = Y \end{align}$$
Therefore, $Y$ is also idempotent. $\square$