Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 1 / 9G (Linear Algebra)

If $x=0$, the generated subspace is zero and invariant. Otherwise, let $r$ be the first index for which $x,Tx,\dots ,T^{r}x$ are linearly dependent. Such an index exists with $1\le r\le n$, since there are $n+1$ vectors. The preceding vectors are independent, so the coefficient of $T^{r}x$ in this first relation cannot vanish. HenceThe linear operator $T$ sends each of the first $r-1$ generators to the next and sends the last into their span. This proves that their span is an invariant subspace. Every later iterate lies in the same subspace by induction, so it equals the span of the first $n$ iterates. Therefore the indicated subspace is $T$-invariant. This proof of a cyclic subspace uses only finite dimension and linear dependence.

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.

Since the first $k$ basis vectors span the invariant subspace $W$, their images have no components in the last $n-k$ coordinates. ThusHere $A$ is the matrix of $T{|}_W$, $D$ is the matrix of the induced map on the quotient vector space $V/W$, and $B$ is unrestricted. The chosen complementary span need not itself be invariant, so $D$ should not in general be called a restriction to that complement. The block-triangular determinant gives

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.