Why every Jordan block $J_k(\lambda)$ has a one-dimensional eigenspace associated with the eigenvalue $\lambda$

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Hello Linear Algebra experts,

My apologies in advance for asking these fundamental questions of Jordan canonical forms (I am new into this and also preparing for my exam). I am stuck with the following problems, and would appreciate very much for your help.

(a) Why every Jordan block $J_k(\lambda)$ has a one-dimensional eigen-space associated with the eigenvalue $\lambda$?

(b) Why $\lambda$ has geometric multiplicity $1$ and algebraic multiplicity $k$ as an eigenvalue of $J_k(\lambda)$?

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Since$$J_k(\lambda)=\begin{pmatrix}1&\lambda&0&0&\ldots&0\\0&1&\lambda&0&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\ldots&1\end{pmatrix}:$$

a) It is clear that$$J_k(\lambda).\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix}=\begin{pmatrix}1\\1\\\vdots\\1\end{pmatrix};$$

b) $J_k-\operatorname{Id}_n$ has rank $n-1$ and therefore the space of the solutions of the equation $\bigl(J_k(\lambda)-\operatorname{Id}_n\bigr).v=v$ is $1$-dimensional.