A square matrix \(N\) over the complex numbers is nilpotent just in case \(N^a=0\) for some positive integer \(a\). Show that every nilpotent \(N\) is similar to a matrix of the form \[\begin{bmatrix}N_1&&&\\&N_2&&\\&&\ddots&\\&&&N_s\end{bmatrix}\] where each \(N_i\) is a square matrix of the form \[\begin{bmatrix}0&&&&\\1&0&&&\\&1&\ddots&&\\&&\ddots&0&\\&&&1&0\end{bmatrix}.\]
Solution
Let \(T\) be the operator represented by \(N\) on \(V=\mathbb C^n\). We prove the assertion by induction on \(n\). For \(n=0\) the empty direct sum is the required form. Suppose \(n>0\) and the assertion holds in smaller dimensions. Let \(r\geq1\) be the least integer with \(T^r=0\).
1. There is an invariant subspace \(U\) with basis\(v,Tv,\ldots,T^{r-1}v\).
Minimality of \(r\) gives \(v\in V\) with \(T^{r-1}v\ne0\). If \(\sum_{j=0}^{r-1}a_jT^jv=0\) is a nonzero linear relation, let \(k\) be the least index with \(a_k\ne0\). Applying \(T^{r-1-k}\) gives \(a_kT^{r-1}v=0\): every term with larger index is killed by \(T^r=0\). This is a contradiction. The chain is therefore independent, and its span \(U\) is invariant because \(T(T^{r-1}v)=0\).
2. The subspace \(U\) has a \(T\)-invariant complement.
Extend the chain in step 1 to a basis of \(V\). Define a linear functional \(\ell:V\to\mathbb C\) to be \(1\) on \(T^{r-1}v\) and \(0\) on every other vector of this basis. Define \[ \pi(w)=\sum_{j=0}^{r-1}\ell\bigl(T^{r-1-j}w\bigr)T^jv. \] For \(0\leq k<r\), the coefficient of \(T^jv\) in \(\pi(T^kv)\) is \(\ell(T^{r-1-j+k}v)\). It is \(1\) when \(j=k\), \(0\) when \(k<j\) by the definition of \(\ell\), and \(0\) when \(k>j\) because then the exponent is at least \(r\). Thus \(\pi|_U\) is the identity.
Moreover, \(T^r=0\) gives \[ \pi(Tw)=\sum_{j=1}^{r-1}\ell(T^{r-j}w)T^jv =T\pi(w). \] These sums are empty and both sides are zero when \(r=1\). Consequently \(W=\ker\pi\) is \(T\)-invariant. Since \(\pi\) maps into \(U\) and restricts to its identity, every \(w\in V\) has the unique decomposition \[ w=\pi(w)+(w-\pi(w))\in U\oplus W. \] Indeed, the second term lies in \(\ker\pi\) and \(U\cap\ker\pi=0\).
3. Combining bases of \(U\) and \(W\) produces the asserted blocks.
On the ordered basis \(v,Tv,\ldots,T^{r-1}v\) of \(U\), the matrix of \(T|_U\) has ones immediately below the diagonal and all other entries zero. The restriction \(T|_W\) is nilpotent and \(\dim W=n-r<n\). By induction it has a basis with the required block diagonal matrix. Concatenating these two bases gives a basis of \(V\) in which \(T\) has the stated form. If \(S\) has these basis vectors as columns in the original coordinates, then \(S\) is invertible and \(S^{-1}NS\) is exactly that block diagonal matrix.