Real Eigenvalues — Question 6

Question 6 – Solution

Strategy. An eigenbasis fixes the linear map, while eigenvalues alone do not fix its directions.

Step 1: Reconstruct the operator. Let $P=\begin{pmatrix}1&2\\2&1\end{pmatrix}$ and $D=\operatorname{diag}(-1,2)$. Since $\det P=-3\ne 0$, $$\boxed{A=PDP^{-1}=\begin{pmatrix}3&-2\\2&-2\end{pmatrix}.}$$ A linear map is uniquely determined by its values on a basis, so there is no other matrix satisfying both given observations.

Step 2: Solve and check the IVP. The initial vector is $(1,2)^T+(2,1)^T$, giving $$\boxed{X=e^{-t}\binom 12+e^{2t}\binom 21.}$$ At zero this is $(3,3)^T$, and its derivative is $(-1,-2)^T+(4,2)^T =(3,0)^T=A(3,3)^T$. More generally each mode differentiates to its eigenvalue times itself, verifying the ODE for every time.

Step 3: Separate stable data from generic growth. For initial $(p,q)$, the unstable coefficient is $(2p-q)/3$. Decay therefore occurs exactly on $q=2p$, the line spanned by $(1,2)^T$. For the stated IVP the unstable coefficient is one, so $X/e^{2t}\to(2,1)^T$ and $X/\|X\|\to(2,1)^T/\sqrt 5$. The drawn eigenlines are invariant; the off-line curve cannot cross them.

Step 4: Distinguish rates from directions. The matrix $B=\operatorname{diag}(-1,2)$ has the same trace and determinant, hence the same eigenvalues, but its decaying line is the horizontal axis. Trace and determinant alone do not determine the stable line. Multiplying either supplied eigenvector by any nonzero scalar changes its coordinate coefficient but leaves $PDP^{-1}$ unchanged: the diagonal rescaling commutes with $D$. Zero rescaling is not allowed.