Review : Eigenvalues & Eigenvectors — Question 6

Question 6 – Solution

Strategy. Commutation forces preservation of this matrix’s two distinct eigendirections, allowing scalar tests in a common basis.

Step 1: Equate the products. Subtracting the products gives $$AB-BA=\begin{pmatrix}r-q&s-p\\p-s&q-r\end{pmatrix}.$$ Thus $r=q$ and $s=p$, with no further restrictions. Thus the entire commuting family is $\boxed{B=\begin{pmatrix}p&q\\q&p\end{pmatrix}}$.

Step 2: Use the common directions. For $u=(1,1)^T$ and $v=(1,-1)^T$, the eigenvalues of $A$ are $3,1$, and those of $B$ are $p+q,p-q$, respectively. These independent vectors form a common eigenvector basis, even if the two eigenvalues of $B$ coincide. Conversely, if $A=PDP^{-1}$ and $B=PEP^{-1}$ with $D,E$ diagonal, then $AB=PDE P^{-1}=PED P^{-1}=BA$.

Step 3: Analyze the concrete sum and product. For $p=1,q=2$, the eigenvalues of $B$ on $u,v$ are $3,-1$. Therefore the eigenvalues of $AB$ are $9,-1$ and those of $A+B$ are $6,0$. Both $A$ and $B$ are invertible, but their actions on $v$ cancel in the sum. The sum has kernel $\operatorname{span}\{v\}$.

Step 4: Classify all cancellation cases. Invertibility of $B$ requires $(p+q)(p-q)\ne 0$. Singularity of the sum requires $(p+q+3)(p-q+1)=0$. Equivalently the full allowed set is $$\boxed{\{p+q=-3,\ p-q\ne 0\}\ \cup\ \{p-q=-1,\ p+q\ne 0\}.}$$ On the first line only, the kernel is the line through $u$; on the second only, it is the line through $v$. Their intersection is $(p,q)=(-2,-1)$, where $B=-A$ and $A+B=0$, so the kernel is the entire plane. The excluded points are $(-3/2,-3/2)$ and $(-1/2,1/2)$, where $B$ itself is singular. The diagram marks these exclusions with open circles.