0
$\begingroup$

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?

$\endgroup$
1
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – CyCeez
    Sep 2 '19 at 23:03
  • $\begingroup$ @CyrilCeez Orthogonal projection doesn’t usually work because the eigenbasis isn’t usually orthogonal. You need to project parallel to the other eigenvector(s). $\endgroup$
    – amd
    Sep 3 '19 at 1:36
0
$\begingroup$

$$B=\lambda_1P_1+\lambda_2P_2=P_1+4P_2=P_1+4(I-P_1)=4I-3P_1$$ so $P_1=(1/3)(4I-B)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.