Find a Projection that has the Eigenvalues 0 and 1

Question

$P \in \mathbb{R}^{NxN}$ is a Projection Matrix, so $P^2=P$.

So the first task was to calculate all possible Eigenvalues.

I got $\lambda_1=1$ and $\lambda_2=0$.

Now the task is: Find a Projection $P: \mathbb{R}^2 \Rightarrow \mathbb{R}^2$, that has the calculated Eigenvalues

So in the solutions it's written:

$P: \mathbb{R}^2 \Rightarrow \mathbb{R}^2$

$P(\begin{matrix} v_1\\ v_2 \end{matrix})=(\begin{matrix}v_1\\0\end{matrix})$

$P(\begin{matrix} a\\ 0 \end{matrix})=(\begin{matrix}a\\0\end{matrix})= 1*(\begin{matrix}a\\0\end{matrix})$

$P(\begin{matrix} 0\\ b \end{matrix})=(\begin{matrix}0\\0\end{matrix})= 0*(\begin{matrix}0\\b\end{matrix})$

How did they come to this?

Is it because $P* x = \lambda * x$?

Answer 1

As you say, projections can only have eigenvalues $0$ or $1$. Thus the task here is simply to present an example which has both, right?

Guess-and-check would be a natural first thing to try. If I ask you to think of a projection $P:\mathbb R^2 \rightarrow \mathbb R ^2$, chances are the first one you'll come up with would be $$ P\begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}x\\0\end{bmatrix} $$ Computing the eigenvalues of $P$ shows that it does, in fact, have both $0$ and $1$ as eigenvalues.