# Proving the continuation of the Cayley-Hamilton theorem from Schur's triangularization theorem

The Cayley-Hamilton theorem says that every square matrix can satisfy its own characteristic equation, $p(\lambda) = 0$, or $p(\mathbf{A}) = \mathbf{0}$.

The question is to show how the Cayley-Hamilton theorem follows from Schur's triangularization theorem.

If $\sigma(\mathbf{A}) = \{\lambda_1, \lambda_2 \ldots, \lambda_k \}$, with $\lambda_i$ repeated $a_i$ times, then there is a unitary $\mathbf{U}$ such that

$\mathbf{U*AU} = \mathbf{T} = \begin{pmatrix} \mathbf{T}_1 & \star & \cdots & \star \\ & \mathbf{T}_2 & \cdots & \star \\ & & \ddots & \vdots \\ & & & \mathbf{T}_k \end{pmatrix}$, where $\mathbf{T}_i = \begin{pmatrix} \lambda_i & \star & \cdots & \star \\ & \lambda_i & \cdots & \star \\ & & \ddots & \vdots \\ & & & \lambda_i \end{pmatrix}_{a_i \times a_i}$.

Moreover, $( \mathbf{T}_i - \lambda_i\mathbf{I})^{a_i} = \mathbf{0}, so (\mathbf{T} - \lambda_i \mathbf{I})^{a_i}$ has the form

$( \mathbf{T}_i - \lambda_i\mathbf{I})^{a_i} = \begin{pmatrix} \star & \cdots & \star & \cdots & \star \\ & \ddots & \vdots & & \vdots \\ & & \mathbf{0} & \cdots & \star \\ & & & \ddots & \vdots \\ & & & & \star \end{pmatrix} \leftarrow \mathrm{i^{th} \ row \ of \ blocks}$

After this step, I am not sure how to use the characteristic equation to derive the final answer. Anyone know how to solve it?

• Have you tried mathematical induction? Apr 10, 2015 at 22:25

Take a column vector $v_k:=v$ and split it up into $k$ pieces corresponding to the blocks of the matrix. Since $(\mathbf{T} - \lambda_k \mathbf{I})^{a_k}$ has its lower right block equal to zero, $v_{k-1}:=(\mathbf{T} - \lambda_k \mathbf{I})^{a_k} v_k$ has last piece zero. Then, since $(\mathbf{T} - \lambda_{k-1} \mathbf{I})^{a_{k-1}}$ has a block in its $k-1$st row and $k-1$st column of blocks which is zero, $$v_{k-2}:=(\mathbf{T} - \lambda_{k-1} \mathbf{I})^{a_{k-1}} v_{k-1}=(\mathbf{T} - \lambda_{k-1} \mathbf{I})^{a_{k-1}}(\mathbf{T} - \lambda_k \mathbf{I})^{a_k} v_k$$ has its $k-1$st and $k$th pieces equal to zero. You can continue in this way for a total of $k$ steps until you find that $$v_0=\left(\prod_{1\le i\le k}(\mathbf{T} - \lambda_{i} \mathbf{I})^{a_{i}}\right) v=0.$$ Multiplying by $\mathbf{U}$ then gives $$\left(\prod_{1\le i\le k}(\mathbf{A} - \lambda_{i} \mathbf{I})^{a_{i}}\right) \mathbf{U} v=0,$$ which, since $v$ was arbitrary and $\mathbf{U}$ is invertible, proves that $$p(\mathbf{A})=\prod_{1\le i\le k}(\mathbf{A} - \lambda_{i} \mathbf{I})^{a_{i}}=\mathbf{0}.$$