If $a$ is an eigenvalue of square matrix $A$ with algebraic multiplicity $k$, how can I prove that $\operatorname{dim}(\operatorname{ker}(A-aI)^k)=k$?
p.s. I thought that if the minimal polynomial has $(t-a)^x$, it should be that $x≤k$. And considering the Jordan form, $\operatorname{dim}(\operatorname{ker}(A-aI)^x)$ equals the number of a in the diagonal, which is k. So I was able to prove that $\operatorname{dim}(\operatorname{ker}(A-aI)^k) ≥ \operatorname{dim}(\operatorname{ker}(A-aI)^x)=k$.
Algebraic multiplicity of an eigenvalue and the dimension of kernel
linear algebra
Best Answer
As you proved that $\dim(\ker (A-aI)^k) \geq k$, it remains to show $\dim(\ker (A-aI)^k) \leq k$.
First, observe that $A$ maps $\ker(A-aI)^k$ into itself, since $A$ and $(A-aI)^k$ commutes. Let $\{v_{1}^{1}, ~\dots~, v_{1}^{m_1}\}$ be a basis of $\ker(A-aI)$. Extend this to $\{v_{1}^{1},~ \dots~, v_{1}^{m_1}, v_2^{1}, ~\cdots~, v_2^{m_2}\}$, a basis for $\ker(A-aI)^2$. Repeat this process until obtain $\mathfrak B_k=\{v_{1}^{1},~ \dots~, v_{1}^{m_1}, v_2^{1}, ~\cdots~, v_2^{m_2}, ~ \cdots ~, v_k^{1}, ~ \cdots ~, v_k^{m_k}\}$, a basis for $\ker(A-aI)^k$. By construction, $\mathfrak B_s=\{v_{1}^{1},~ \dots~, v_{1}^{m_1}, v_2^{1}, ~\cdots~, v_2^{m_2}, ~ \cdots ~, v_s^{1}, ~ \cdots ~, v_s^{m_s}\}$ is a basis for $\ker(A-aI)^s$ $(1\leq s \leq k)$.
Note that $A-aI$ maps $v_s^{i} \in \mathfrak B_s$ into $\text{span}(\mathfrak B_{s-1})$, because it sends $\ker(A-aI)^{s}$ into $\ker(A-aI)^{s-1}$. In other words, $A|_{\ker(A-aI)^k}$ is a upper-triangular matrix with all diagonal entries $a$, with respect to the basis $\mathfrak B_k$.
Extend $\mathfrak B_k$ to $\mathfrak B$, a basis for the vector space which is a domain of $A$. Then $A$ has a matrix representation with respect to $\mathfrak B$ as follows:
$$A = \begin{bmatrix} [A|_{\ker(A-aI)^k}]_{\mathfrak B_k}^{\mathfrak B_k} & B \\ 0 & C\end{bmatrix} $$
Thus $(t-a)^m$ divides the characteristic polynomial of $A$, where $m = |\mathfrak B_k|= \dim(\ker (A-aI)^k)$. Hence $\dim(\ker (A-aI)^k) \leq k$.