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(i) Find the general solution of [itex]\dot{\mathbf{x}}=B\mathbf{x}[/itex].

(ii) Show that 1 is the only eigenvalue of [itex]B[/itex].

It's clear that [itex]\mathbf{v_3},\mathbf{v_2}[/itex] and [itex]\mathbf{v_1}[/itex] form a chain of generalized eigenvectors associated with [itex]\lambda=1[/itex] and hence are linearly independent. From this I can find the general solution of [itex]\dot{\mathbf{x}}=A\mathbf{x}=SBS^{-1}\mathbf{x}[/itex] but how can I proceed from here to find the general solution of [itex]\dot{\mathbf{x}}=B\mathbf{x}[/itex]?

Any help is much appreciated, thank you!