In Chapter 8 (part B) from the book "Linear Algebra Done Right", the author mentioned that if a transformation $T$ has the following block diagonal matrix $A$ with respect to some given bases:
$$\begin{bmatrix}4 & 0 & 0 & 0 & 0 \\ 0 & 2 & -3 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$
Then $T$ has 3 eigenvalues 4, 2, and 1 with multiplicities (or algebraic multiplicity) 1, 2, 2 respectively. But I don't understand why the multiplicities have to be 1, 2, 2?
For your reference, this is the definition of multiplicity in the book:
Suppose $T$ is a the linear functional in $L(V)$. The multiplicity of an eigenvalue $\lambda$ of $T$ is defined to be the dimension of the corresponding generalized eigenspace $G(\lambda, T)$.
Where $G(\lambda, T)$ is the generalized eigenspace of $T$ corresponding to the eigenvalue $\lambda$ of $T$.