[proofplan]
Since similar matrices have the same characteristic polynomial, we compute $\chi_T(\lambda) = \det(\lambda I - J)$. The block-diagonal structure of $J$ lets us factor the determinant as a product over blocks. Each Jordan block $J_{m_j}(\lambda_j)$ contributes $(\lambda - \lambda_j)^{m_j}$ because $\lambda I - J_{m_j}(\lambda_j)$ is upper triangular with $(\lambda - \lambda_j)$ on the diagonal. Collecting terms by eigenvalue gives the algebraic multiplicity.
[/proofplan]
[step:Reduce to computing $\det(\lambda I - J)$ via similarity invariance]
Let $n = \dim V$ and fix a basis of $V$ with respect to which the matrix representation of $T$ is $A \in \mathrm{Mat}_n(\mathbb{C})$. The Jordan Normal Form $J$ satisfies $J = P^{-1}AP$ for some invertible matrix $P \in GL_n(\mathbb{C})$. The characteristic polynomial of $T$ is
\begin{align*}
\chi_T(\lambda) &= \det(\lambda I - A).
\end{align*}
Since $\lambda I - A = P(\lambda I - J)P^{-1}$, the multiplicativity of the determinant gives
\begin{align*}
\det(\lambda I - A) &= \det(P) \cdot \det(\lambda I - J) \cdot \det(P^{-1}) = \det(\lambda I - J).
\end{align*}
Therefore $\chi_T(\lambda) = \det(\lambda I - J)$.
[/step]
[step:Factor $\det(\lambda I - J)$ as a product over Jordan blocks]
The Jordan form $J = \bigoplus_{j=1}^k J_{m_j}(\lambda_j)$ is block diagonal:
\begin{align*}
J &= \begin{pmatrix} J_{m_1}(\lambda_1) & & \\ & \ddots & \\ & & J_{m_k}(\lambda_k) \end{pmatrix}.
\end{align*}
Therefore $\lambda I - J$ is also block diagonal with blocks $\lambda I_{m_j} - J_{m_j}(\lambda_j)$. The determinant of a block-diagonal matrix equals the product of the determinants of the diagonal blocks:
\begin{align*}
\det(\lambda I - J) &= \prod_{j=1}^k \det(\lambda I_{m_j} - J_{m_j}(\lambda_j)).
\end{align*}
[/step]
[step:Compute the determinant of each block $\lambda I_{m_j} - J_{m_j}(\lambda_j)$]
A single Jordan block $J_{m_j}(\lambda_j)$ is the $m_j \times m_j$ matrix with $\lambda_j$ on the main diagonal, $1$'s on the superdiagonal, and $0$'s elsewhere. Therefore
\begin{align*}
\lambda I_{m_j} - J_{m_j}(\lambda_j) &= \begin{pmatrix} \lambda - \lambda_j & -1 & 0 & \cdots & 0 \\ 0 & \lambda - \lambda_j & -1 & \cdots & 0 \\ \vdots & & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda - \lambda_j & -1 \\ 0 & 0 & \cdots & 0 & \lambda - \lambda_j \end{pmatrix}.
\end{align*}
This is an upper triangular matrix with every diagonal entry equal to $\lambda - \lambda_j$. The determinant of an upper triangular matrix equals the product of its diagonal entries, so
\begin{align*}
\det(\lambda I_{m_j} - J_{m_j}(\lambda_j)) &= (\lambda - \lambda_j)^{m_j}.
\end{align*}
[guided]
We need to identify the structure of the matrix $\lambda I_{m_j} - J_{m_j}(\lambda_j)$ and compute its determinant. Recall the entry-by-entry definition of a Jordan block: $(J_{m_j}(\lambda_j))_{ab} = \lambda_j$ when $a = b$ (diagonal), $(J_{m_j}(\lambda_j))_{ab} = 1$ when $b = a + 1$ (superdiagonal), and $0$ otherwise. Subtracting from $\lambda I_{m_j}$:
\begin{align*}
(\lambda I_{m_j} - J_{m_j}(\lambda_j))_{ab} &= \begin{cases} \lambda - \lambda_j & \text{if } a = b, \\ -1 & \text{if } b = a + 1, \\ 0 & \text{otherwise.} \end{cases}
\end{align*}
All entries with $a > b$ (below the diagonal) are $0$, so $\lambda I_{m_j} - J_{m_j}(\lambda_j)$ is upper triangular. Explicitly:
\begin{align*}
\lambda I_{m_j} - J_{m_j}(\lambda_j) &= \begin{pmatrix} \lambda - \lambda_j & -1 & 0 & \cdots & 0 \\ 0 & \lambda - \lambda_j & -1 & \cdots & 0 \\ \vdots & & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda - \lambda_j & -1 \\ 0 & 0 & \cdots & 0 & \lambda - \lambda_j \end{pmatrix}.
\end{align*}
For any upper triangular matrix, the determinant equals the product of the diagonal entries. This follows from cofactor expansion along the first column: only the diagonal entry $(\lambda - \lambda_j)$ in position $(1,1)$ is non-zero (all entries below it are $0$), so the determinant reduces to $(\lambda - \lambda_j)$ times the determinant of the $(m_j - 1) \times (m_j - 1)$ upper triangular submatrix obtained by deleting the first row and column. Iterating this argument $m_j$ times:
\begin{align*}
\det(\lambda I_{m_j} - J_{m_j}(\lambda_j)) &= (\lambda - \lambda_j)^{m_j}.
\end{align*}
[/guided]
[/step]
[step:Combine the block contributions and identify the algebraic multiplicity]
Substituting into the product:
\begin{align*}
\chi_T(\lambda) &= \prod_{j=1}^k (\lambda - \lambda_j)^{m_j}.
\end{align*}
To read off the algebraic multiplicity of a given eigenvalue $\mu$, collect all blocks with $\lambda_j = \mu$. The factor $(\lambda - \mu)$ appears with total exponent
\begin{align*}
\sum_{\substack{j=1 \\ \lambda_j = \mu}}^k m_j,
\end{align*}
which is the total size of all Jordan blocks with eigenvalue $\mu$. By definition, the algebraic multiplicity of $\mu$ is the multiplicity of $\mu$ as a root of $\chi_T$, which equals this total size.
[/step]
