Review : Eigenvalues & Eigenvectors — Question 5

Question 5 – Solution

Strategy. Independent eigenvectors form an input basis, so their prescribed images determine every column of the transformation.

Step 1: Build the spectral representation. The basis matrix and inverse are $$P=\begin{pmatrix}1&2\\1&-1\end{pmatrix},\qquad P^{-1}=\frac 13\begin{pmatrix}1&2\\1&-1\end{pmatrix}.$$ Therefore $$\boxed{A=P\begin{pmatrix}2&0\\0&-1\end{pmatrix}P^{-1} =\begin{pmatrix}0&2\\1&1\end{pmatrix}.}$$ Its products with $u,v$ are $(2,2)^T=2u$ and $(-2,1)^T=-v$. Since $u,v$ span the plane, these images determine $A$ uniquely.

Step 2: Separate notation changes from changed assignments. Multiplying either basis column by a nonzero scalar leaves its eigenvalue relation unchanged, so the reconstructed matrix stays the same. Reordering columns and the matching diagonal entries also leaves the map unchanged. Assigning $-1$ to $u$ and $2$ to $v$ instead produces $\boxed{I-A=\begin{pmatrix}1&-2\\-1&0\end{pmatrix}}$, a different map. The correspondence between each direction and its scalar matters.

Step 3: Test a proposed third direction. Directly, $A(1,0)^T=(0,1)^T$, not a multiple of $(1,0)^T$. Thus no choice of eigenvalue makes the additional request possible. Equivalently, $(1,0)^T=(u+v)/3$ combines both distinct eigendirections.

Step 4: Recover the general two-scalar family. The same basis computation yields $$\boxed{A_{r,s}=\frac 13\begin{pmatrix}r+2s&2r-2s\\r-s&2r+s\end{pmatrix}.}$$ For $r\ne s$, the eigenspaces are exactly the two specified lines. For $r=s$, the matrix becomes $rI_2$ and every nonzero vector is an eigenvector; the eigenspace is the whole plane. Repeating an eigenvalue here enlarges the eigenspace, unlike a defective repeated-root example.