Theorem: Every $n \times n$ complex matrix $A$ is similar to a Jordan matrix.

Proof outline:

Step 1: Primary decomposition.

By the fundamental theorem of algebra, the characteristic polynomial factors completely: $p_A(\lambda) = (\lambda - \lambda_1)^{a_1} \cdots (\lambda - \lambda_k)^{a_k}$

By the Primary Decomposition Theorem: $\mathbb{C}^n = G_{\lambda_1} \oplus \cdots \oplus G_{\lambda_k}$

where $G_{\lambda_i} = \ker((A - \lambda_i I)^{a_i})$ has dimension $a_i$.

Step 2: Reduce to single eigenvalue case.

Since each $G_{\lambda_i}$ is $A$-invariant, we can choose a basis for $\mathbb{C}^n$ by combining bases for each $G_{\lambda_i}$. In this basis, $A$ is block diagonal: $A = \begin{bmatrix} A_1 & & \\ & \ddots & \\ & & A_k \end{bmatrix}$

where $A_i = A|_{G_{\lambda_i}}$ acts on $G_{\lambda_i}$ and has only eigenvalue $\lambda_i$.

Step 3: Handle single eigenvalue case.

Focus on $A_1$ acting on $G_{\lambda_1}$ with only eigenvalue $\lambda_1$. Write $A_1 = \lambda_1 I + N$ where $N = A_1 - \lambda_1 I$.

Since $A_1$ has only eigenvalue $\lambda_1$, and $(A_1 - \lambda_1 I)^{a_1} = N^{a_1} = 0$ on $G_{\lambda_1}$, the operator $N$ is nilpotent.

Step 4: Structure of nilpotent operators.

For nilpotent $N$ with $N^m = 0$, we can find Jordan chains. Consider the filtration: $\{0\} \subset \ker(N) \subset \ker(N^2) \subset \cdots \subset \ker(N^m) = G_{\lambda_1}$

Choose vectors to form Jordan chains by working backwards from higher kernels:

• Pick $\mathbf{v}_k \in \ker(N^k) \setminus \ker(N^{k-1})$
• Form chain: $\mathbf{v}_k, N\mathbf{v}_k, N^2\mathbf{v}_k, \ldots, N^{k-1}\mathbf{v}_k$

These chains are linearly independent and span $G_{\lambda_1}$.

Step 5: Convert chains to Jordan blocks.

For Jordan chain $\{\mathbf{v}_k, N\mathbf{v}_k, \ldots, N^{k-1}\mathbf{v}_k\}$, we have: $A_1(\mathbf{v}_j) = (\lambda_1 I + N)(\mathbf{v}_j) = \lambda_1\mathbf{v}_j + N\mathbf{v}_j$

The action of $A_1$ on this chain is:

• $A_1(N^{k-1}\mathbf{v}_k) = \lambda_1 N^{k-1}\mathbf{v}_k$ (since $N^k\mathbf{v}_k = 0$)
• $A_1(N^{j}\mathbf{v}_k) = \lambda_1 N^j\mathbf{v}_k + N^{j+1}\mathbf{v}_k$

In the ordered basis $(N^{k-1}\mathbf{v}_k, \ldots, N\mathbf{v}_k, \mathbf{v}_k)$, this is exactly a Jordan block $J_k(\lambda_1)$.

Step 6: Combine all blocks.

Performing this for each generalized eigenspace and each Jordan chain within, we obtain a basis where $A$ is in Jordan form. ∎