= Solution
The <characteristic polynomials>, in the convention $\det(\lambda I-T)$, are
$$
\chi_A(\lambda)=(\lambda-1)(\lambda-2)^2,\qquad\chi_B(\lambda)=(\lambda-2)(\lambda-4)(\lambda-6).
$$
Solving the <eigenvalue equations> gives the complete <eigenspaces>
$$
\begin{aligned}
E_1(A)&=\operatorname{span}\{(1,1,1)^T\},\\
E_2(A)&=\{(x,y,z)^T:y=x+z\}=\operatorname{span}\{(1,0,-1)^T,(1,2,1)^T\},\\
E_2(B)&=\operatorname{span}\{(1,1,1)^T\},\\
E_4(B)&=\operatorname{span}\{(1,0,-1)^T\},\\
E_6(B)&=\operatorname{span}\{(1,2,1)^T\}.
\end{aligned}
$$
Every nonzero vector in a listed <eigenspace> is an <eigenvector>, and these are all the <eigenvectors>. For $A$, the repeated <eigenvalue> $2$ has a two-dimensional <eigenspace>; for $B$, all three <eigenvalues> are simple. The three displayed generating vectors form a <basis> because their column matrix
$$
S=\begin{pmatrix}1&1&1\\1&0&2\\1&-1&1\end{pmatrix}
$$
has <determinant> $2$. Thus \b[both matrices are diagonalizable], with diagonal entries $(1,2,2)$ for $A$ and $(2,4,6)$ for $B$ in this <basis>.
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