Review : Eigenvalues & Eigenvectors — Question 9

Question 9 – Solution

Strategy. A polynomial changes eigenvalues without changing each original eigenvector, but different eigenvalues can acquire the same image.

Step 1: Compare the two spectra. For $A$, the eigenspaces are $$E_1=\operatorname{span}\{(1,1,0)^T\},\quad E_{-1}=\operatorname{span}\{(1,-1,0)^T\},\quad E_2=\operatorname{span}\{(0,0,1)^T\}.$$ These vectors form a basis. Since $B=\operatorname{diag}(1,1,4)$, its eigenspace for $1$ is the entire plane $x_3=0$, and its eigenspace for $4$ is the third coordinate axis. Squaring sends both $1$ and $-1$ to $1$.

Step 2: Prove the forward rule and refute the converse. The vector $e_1=(1,0,0)^T$ satisfies $Be_1=e_1$ but $Ae_1=(0,1,0)^T$, not a multiple of $e_1$. In general, $Av=\lambda v$ implies $A^kv=\lambda^kv$ by induction; linearity then gives $\boxed{p(A)v=p(\lambda)v}$. The merged eigenspace permits new combinations that are eigenvectors of $p(A)$ without being eigenvectors of $A$.

Step 3: Find the least-degree annihilating polynomial. Any polynomial with $p(A)=0$ must vanish at $1,-1,2$, by applying the identity to the three nonzero eigenvectors. A nonzero polynomial therefore needs degree at least three. The monic candidate is $$\boxed{p(z)=(z-1)(z+1)(z-2)=z^3-2z^2-z+2.}$$ It annihilates all three basis eigenvectors, hence every vector, proving $p(A)=0$ and minimality. It is the unique monic cubic with those roots.

Step 4: Recover the inverse from the identity. Rearrange $A^3-2A^2-A+2I=0$ to get $A(-A^2+2A+I)/2=I$. The factors commute, so the reverse product is also $I$. Thus $$\boxed{A^{-1}=\frac{-A^2+2A+I}{2} =\begin{pmatrix}0&1&0\\1&0&0\\0&0&1/2\end{pmatrix}.}$$ Direct multiplication confirms the inverse without division by a matrix expression.