Intuition behind finding projection matrices of transformation

Question

Supposedly a matrix $A$ is diagonalisable iff it can be written $A=\sum \lambda_i \pi_i$ where $\lambda_i$ are the distinct eigenvalues of $A$, and $\pi_i$ are projections satisfying $i \neq j \implies \pi_i \pi_j=0$ and $\sum \pi_i=I$. I'm willing to take this fact for granted at the moment.

Given $A=\begin{bmatrix} -6 & 6 \\ -12 & 11 \end{bmatrix}$, I'm not following how my course's notes find the projection matrices $P_1$ and $P_2$ satisfying these properties. It finds the eigenvectors, $\begin{bmatrix} 3 \\ 4 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$. Then claims that the projection matrices will have their columns being the scalar multiples of these, letting $P_1=\begin{bmatrix}3a & 3b \\ 4a & 4b \end{bmatrix}$ and $P_2=\begin{bmatrix}2c & 2d \\ 3c & 3d \end{bmatrix}$.

Then it uses the aforementioned properties of these projections to form a system of many equations, eventually arriving at $a=3,b=-2,c=-4,d=3$, and $P_1=\begin{bmatrix}9 & -6 \\ 12 & -8 \end{bmatrix}$, $P_2=\begin{bmatrix} -8 & 6 \\ -12 & 9 \end{bmatrix}$.

But why does this process produce a projection matrix? It's not intuitive to me why we start by assuming they have their columns as scalar multiples of the eigenvectors. We've shown that $P_1$ and $P_2$ are matrices satisfying the required properties (or rather defined them to be that way), but a priori I wouldn't know that they're projections.

Answer 1

I don't think it is the easiest way to find the projection matrices, but I can say why it works. You naturally want the projection on the eigenspace generated by eigenvector $3\choose 4$ to be that eigenspace, which means the column of the projection matrix should be multiples of that vector. Similarly for the other projection matrix.

However, rather than setting up and solving lots of (non linear) equations, you could go a bit further and use that the projection on the eigenspace generated one eigenvector must send the other eigenvector to $0$. Thus $P_1$ must have rows that are multiples of $(3~~{-}2)$ so $b=-\frac23a$, and $P_1$ must be a multiple of $$ \pmatrix{9&-6\\12&-8}. $$ Now using that the trace of a projection matrix onto a line is $1$ (namely, te trace is the sum of the eigenvalues with their multiplicities, which for a projection is the dimension of the image subspace), you can see that the multiple is by$~1$, in other words that is already the projection matrix. For $P_2$ just use $P_2=I_2-P_1$.