# How to decompose a 2 x 2 matrix into projection matrices from its eigenvalues, eigenvectors

Consider the matrix $$B = \begin{bmatrix} 2 & 2 \\ 1 & 3 \end{bmatrix}$$. Find projection matrices $$P_1, P_2$$ such that (1) $$B = \lambda_1 P_1 + \lambda_2 P_2$$ where $$\lambda_1, \lambda_2$$ are the eigenvalues of $$B$$, (2) $$P_1 P_2 = 0$$, and (3) $$P_1 + P_2 = I_2$$, the $$2 \times 2$$ identity. (Note: a projection matrix $$P$$ satisfies $$P^2 = P$$.

The eigenvalues are $$\lambda_1 = 1, \lambda_2 = 4$$ and eigenvectors $$\begin{bmatrix} 2 \\ -1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$. The problem comes from this past QR exam - https://lsa.umich.edu/content/dam/math-assets/math-document/AIM/DELA/DELA_Sep18%20-%20Differential%20Eqns%20%26%20Linear%20Algebra%20Fall%202018.pdf - and I thought I could figure it out for practice, but I haven't been able to solve it. In particular, I'm not familiar with how to decompose a matrix into projection matrices using its eigenvalues. Any help or hints?

As I explain in the first case of this answer, $$P_1={A-\lambda_2 I\over\lambda_1-\lambda_2} \\ P_2 = {A-\lambda_1I\over\lambda_2-\lambda_1}.$$ One way to obtain this is to note that when expressed relative to the eigenbasis, the two projectors are simply $$\operatorname{diag}(1,0)$$ and $$\operatorname{diag}(0,1)$$. Perform a change of basis to the standard basis. Another way to derive these is to note, for instance, that if $$\mathbf v_1$$ and $$\mathbf v_4$$ are eigenvectors with eigenvalues $$1$$ and $$4$$, then $$(A-4I)\mathbf v_1 = (1-4)\mathbf v_1$$ and by definition $$(A-4I)\mathbf v_4=0$$. We want $$P_1\mathbf v_1=\mathbf v_1$$ and $$P_1\mathbf v_2=0$$, which we almost have with $$A-4I$$: we just have to divide by $$3$$ to make this the identity map on the span of $$\mathbf v_1$$.
Another approach that one doesn’t see as often comes up in this question: if $$x$$ and $$y^*$$ are right and left eigenvectors, respectively, of $$A$$ corresponding to the same simple eigenvalue, then the projector onto the right eigenspace (the span of $$x$$) is $${xy^*\over y^*x}$$. (This looks a lot like the formula for orthogonal projection onto a vector.) You can find a derivation of this in the answer to that question. For example, a left eigenvector of $$1$$ for your matrix is $$(-1,1)$$ and a right eigenvector is $$(-2,1)^T$$, yielding $$P_1 = \frac13\begin{bmatrix}-2\\1\end{bmatrix}\begin{bmatrix}-1&1\end{bmatrix} = \frac13\begin{bmatrix}2&-2\\-1&1\end{bmatrix},$$ which matches the result of applying the formula at the top of this answer.
• Thanks! I appreciate this detailed explanation. I played around with scaled versions of the idempotent matrices $(B-\lambda I)$ but to no avail, and I also tried orthogonal projection matrices onto each eigenvectors, but that wasn't quite the answer. So, abstractly, I very much understand this answer. Sep 2 '19 at 23:03
$$B=\lambda_1P_1+\lambda_2P_2=P_1+4P_2=P_1+4(I-P_1)=4I-3P_1$$ so $$P_1=(1/3)(4I-B)$$.