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A cyclic vector gives a companion matrix, whose characteristic relation annihilates its generator and hence every iterate. If there is no cyclic vector spanning the whole space, a nonzero proper cyclic subspace is invariant. Induction annihilates the restriction and the quotient; the quotient polynomial first sends the whole space into the invariant subspace, and the restriction polynomial then kills it. The block-triangular characteristic polynomial is the product of these two polynomials.

The $n$ spanning vectors form a basis. WriteIn this cyclic basis the matrix is the companion matrixExpanding $\det (tI-C)$ along its last column givesFor example, the term containing $a_j$ has minor $t^j$ and the signs from the subdiagonal cancel the cofactor sign; the diagonal term supplies $t^n$. The defining relation says ${\chi }_T(T)x=0$. Every polynomial in $T$ commutes with $T$, soIt therefore vanishes on a basis, provingThis direct argument for a cyclic vector does not assume the Cayley-Hamilton theorem.

Induct on $n$. The one-dimensional assertion is immediate, and the zero-dimensional case is vacuous. If $T$ has a cyclic vector spanning all of $V$, the direct proof in part (ii) applies. Otherwise choose $x\ne 0$ and let $W$ be its cyclic subspace. Part (i) makes $W$ invariant; the absence of a cyclic vector makes it proper and nonzero.

By induction, the restriction to $W$ is annihilated by ${\chi }_A$, and the induced map on $V/W$ is annihilated by ${\chi }_D$. The quotient assertion means ${\chi }_D(T)V\subset W$. Applying the restriction assertion next givesBy the characteristic factorization in part (iii), this is precisely ${\chi }_T(T)V=0$. HenceThis is the Cayley-Hamilton theorem from cyclic subspaces; using the quotient map is what makes the argument valid even when the block $B$ is nonzero.

For the standard basis vector $e_j$, matrix multiplication selects the $j$th column, soConsequentlyand the columns of $PM$ are $P{c}_1,\dots ,P{c}_n$.

Suppose first that $v,Av,\dots ,A^{n-1}v$ are linearly independent, and defineThen $S$ is invertible. The first $n-1$ columns of $AS$ are the last $n-1$ columns of $S$. By the Cayley-Hamilton theorem,This is exactly the last-column rule for the companion matrix $C$, so

Conversely, suppose $S^{-1}AS=C$, or $AS=SC$, and write the columns of $S$ as $s_1,\dots ,s_n$. Comparing the first $n-1$ columns givesHenceSince $S$ is invertible, its columns are linearly independent. Taking $v=s_1$ therefore makeslinearly independent. Thus $A$ is similar to $C$ exactly when it has the stated cyclic vector.

If $V=V_1\oplus V_2$ were a nontrivial invariant decomposition, each restricted minimal polynomial would be a power of $x-\alpha$ of degree at most $\dim V_i<n$. Their least common multiple is the minimal polynomial of $\varphi$, contradicting its degree $n$.

Since $m_{\varphi }$ has degree $n$, $\varphi$ has a cyclic vector $b_1$. Put $b_k={\varphi }^{k-1}{b}_1$. Then $\varphi (b_k)=b_{k+1}$ for $k<n$. In this basis the matrix is the companion matrix with ones immediately below the diagonal and last columnobtained by expanding $(x-\alpha {)}^n$.