Real Eigenvalues — Question 9

Question 9 – Solution

Strategy. Polynomial projections isolate the two real eigenspaces and make the evolution law a scalar calculation on each one.

Step 1: Verify complementary projections. The characteristic polynomial is $(\lambda+1)(\lambda+2)$, and direct multiplication gives $A^2+3A+2I=0$. Here $P=\begin{pmatrix}-2&2\\-3&3\end{pmatrix}$ and $Q=\begin{pmatrix}3&-2\\3&-2\end{pmatrix}$. Thus $P+Q=I$, $P^2=P$, $Q^2=Q$, and $PQ=QP=0$. For example $PQ=-(A+2I)(A+I)=0$; the other identities follow by multiplication or from $Q=I-P$.

Step 2: Identify the oblique directions. The range of $P$ is $\operatorname{span}(2,3)^T$, the $-1$ eigenspace; its kernel is $\operatorname{span}(1,1)^T$, the $-2$ eigenspace. For $Q$ the range and kernel are interchanged. These two directions have nonzero dot product $5$, so neither projection is orthogonal. The matrices are also visibly nonsymmetric.

Step 3: Construct and verify evolution. The same polynomial identity gives $AP=-P$, $AQ=-2Q$. Consequently $F'=AF$ and $F(0)=P+Q=I$. For arbitrary initial vector $X_0$, $F(t)X_0$ solves the IVP; constant linear-system uniqueness gives the complete solution. In particular $$\boxed{X=e^{-t}\binom{-2}{-3}+e^{-2t}\binom 33.}$$ Its initial state is $(1,0)^T$ and its initial derivative $(-4,-3)^T=A(1,0)^T$.

Step 4: Multiply by modes. All mixed products vanish, so $$F(t)F(s)=e^{-(t+s)}P+e^{-2(t+s)}Q=F(t+s).$$ Taking $s=-t$ yields $\boxed{F(t)^{-1}=F(-t)=e^tP+e^{2t}Q}$. No finite real time makes either scalar exponential vanish. This is the normalized evolution of one constant matrix; the fixed complementary projections justify the multiplication rule for every $s,t\in\mathbb R$. No orthogonality or eigenvector normalization was used.