Past exam of the linear algebra course of the University of Oxford / 2021 / 2

In the monomial basis $\mathcal{B}=(1,x,x^2,x^3)$ of the complex vector space $V=\mathbb{C}[x{]}_{\le 3}$, the binomial theorem givesThe characteristic polynomial of this upper triangular matrix is the product of $t$ minus its diagonal entries, so ${\chi }_T(t)=(t-1{)}^4$.

For the minimal polynomial, set $\Delta =T-I$, the forward difference with step one. For a nonconstant polynomial of polynomial degree $d$, the binomial theorem shows that $f(x+1)-f(x)$ has polynomial degree $d-1$. Constant polynomials have zero forward difference. Consequently ${\Delta }^4=0$ on $V$. To prove that the nilpotency index is exactly four, computeIt follows thatIf $T$ were diagonalizable, its only eigenvalue would be $1$, and its diagonal matrix in an eigenbasis would be $I$. This would make $T=I$, whereas $T(x)=x+1\ne x$. Thus $T$ is not diagonalizable. Equivalently, its minimal polynomial has a repeated root.

By Jordan normal form, each complex nilpotent matrix of size six can be put by matrix similarity into a block diagonal matrix whose Nilpotent Jordan blocks have sizesA Nilpotent Jordan block of size $s$ has matrix rank $s-1$ and minimal polynomial $t^s$. Therefore, for the whole matrix,Thus the given matrix rank determines the number $k$ of Jordan blocks, and the given minimal polynomial determines the largest size $s_1$. For size six these two data determine the entire integer partition. Here is the complete list, arranged by the largest size:To see that the list is exhaustive, choose the largest part $s_1$ between one and six and partition the remainder $6-s_1$ into parts at most $s_1$. The resulting rows all have distinct pairs $(s_1,k)$. Hence $A$ and $B$ have the same Jordan normal form $J$.

Explicitly, choose invertible matrices $X,Y$ such that $A=XJ{X}^{-1}$ and $B=YJ{Y}^{-1}$. ThenThis proves matrix similarity and illustrates the classification of six-dimensional nilpotent matrices by rank and nilpotency index.

We construct a cube root of a matrix on each Jordan block. Since $A$ is an invertible matrix over the complex numbers, its Jordan normal form isConsider one block $J_m(\lambda )=\lambda I+N$, where $N=J_m(0)$ is a Nilpotent Jordan block and $N^m=0$. Choose a complex cube root $\beta$ of $\lambda$, so ${\beta }^3=\lambda$.

We need a polynomial $q(z)$ with constant term one satisfyingThis can be constructed using only polynomial multiplication. Write $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$. Once $c_1,\dots ,c_{j-1}$ have been chosen, the coefficient of $z^j$ in $q(z{)}^3$ is $3c_j$ plus an expression depending only on those earlier coefficients. Choose $c_j$ to make this coefficient one at $j=1$ and zero at $2\le j<m$. Division by three is possible over the complex numbers, so the recursion always succeeds. Its first coefficients areThis is the truncated binomial cube root of the identity plus a nilpotent matrix; equivalently $c_j=\left(\genfrac{}{}{0ex}{}{1/3}{j}\right)$, where the generalized binomial coefficient is $\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$ and the empty product is one.

Substitute $z=N/\lambda$. The polynomial remainder is a multiple of $N^m$, hence zero. It follows thatFor $m=1$, take $q=1$, so the same construction covers single-entry blocks. Assemble the block diagonal matrix $C={⨁}_r{C}_{m_r}({\lambda }_r)$. Taking matrix powers respects these blocks, and conjugating back givesThus every invertible matrix over the complex numbers has at least one cube root of a matrix. The construction uses only finitely many matrix powers.

Take the Nilpotent Jordan blockThenSuppose that a complex matrix $B$ of size three satisfied $B^3=A$. We would have $B^9=A^3=0$, so $B$ would be a nilpotent matrix. Every eigenvalue of $B$ must then be zero: if $Bv=\theta v$ with $v\ne 0$, applying $B^9$ gives ${\theta }^{9}v=0$. Its characteristic polynomial is therefore $t^3$, and the Cayley-Hamilton theorem gives $B^3=0$. This contradicts $B^3=A\ne 0$. HenceThe obstruction is that a nilpotent matrix of size three cannot have nilpotency index greater than three.