How many Jordan blocks correspond to $\lambda$? What is the intuition?

Question

Let $V$ be a finite-dimensional vector space over $\mathbb{F}$ and $T: V → V$ be linear. Assume that $m_T(t)$ the minimal polynomial splits over $\mathbb{F}$. Show that for any eigenvalue $λ$ of $T$, the number of Jordan blocks corresponding to $λ$ in the Jordan canonical form of $T$ is $\dim\ker(T-\lambda I)$

What is the intuition behind this? I fail to see why this must hold, it seems totally non-intuitive and out of the blue. Could someone help me come up with a proof, and possibly drive some much-needed intuition along the way?

I know that since the minimal polynomial splits over $\mathbb{F}$, $V$ can be written as a direct sum of the generalized eigenspaces $V(\lambda_i)$. Also, $(T-\lambda_iI)$ is nilpotent on $V(\lambda_i)$, so the restriction $(T-\lambda_iI)\vert_{V(\lambda_i)}:V(\lambda_i)\to V(\lambda_i)$ has a Jordan basis (its matrix can be brought to the Jordan normal form). The number of elements in this basis are equal to $\dim V(\lambda_i) = \dim\ker (T-\lambda_iI)^{\dim V}$ which is not necessarily equal to $\dim\ker (T-\lambda_iI)$.

What now? I'm stuck.

Answer 1

Let $J$ be the Jordan form of $T$. In other words, $J$ is in Jordan form and there is an invertible map $S: \Bbb R^n \to V$ for which $$ T = SJS^{-1}. $$ With that, it follows that $$ \begin{align} \dim \ker(T - \lambda I_V) &= \dim \ker(SJS^{-1} - \lambda SI_n S^{-1}) \\ &= \dim \ker(S[J - \lambda I_n]S^{-1}) = \dim \ker(J - \lambda I_n). \end{align} $$ Now, the question is how $\dim \ker(J - \lambda I_n)$ relates to the structure of the blocks of $J$. Without loss of generality, suppose that the blocks of $J$ are ordered so that $$ J = \pmatrix{J_1 &0\\0 & J_2}, $$ where $J_1$ consists of all blocks associated with $\lambda$ (and the remaining blocks form $J_2$). We see that $J_2 - \lambda I$ is invertible, which means that $\dim \ker(J - \lambda I) = \dim \ker (J_1 - \lambda I)$.

However, note that $J_1 - \lambda I$ is already in reduced row-echelon form: at the start of each Jordan block, there is a zero-column, and all other columns are pivot-columns. Thus, $\dim \ker(J_1 - \lambda I)$ is simply the total number of Jordan blocks in $J_1$.

Answer 2

Consider a $3\times 3$ Jordan block $$\begin{pmatrix}\lambda&0&0\\ 1&\lambda&0\\ 0&1&\lambda \end{pmatrix}.$$ Now subtract the matrix $\lambda I$ to get the matrix $$\begin{pmatrix}0&0&0\\ 1&0&0\\ 0&1&0\end{pmatrix}.$$ Can you see why the kernel of this matrix has dimension $1$? And if we subtract $\mu I$, where $\mu\neq\lambda$, we get the matrix $$\begin{pmatrix}\lambda-\mu&0&0\\ 1&\lambda-\mu&0\\ 0&1&\lambda-\mu\end{pmatrix}.$$ Can you see why the kernel of this matrix has dimension $0$?

This applies to all Jordan blocks. The kernel of $J_\lambda-\lambda I$ for any Jordan block $J_\lambda$ has dimension $1$, always. And the kernel of $J_\lambda-\mu I$ where $\mu\neq0$ has dimension $0$, always. So if we have a block diagonal matrix consisting of only Jordan blocks, each block with eigenvalue $\lambda$ contributes exactly one dimension to the kernel of $T-\lambda I$, and there are no other contributions from the rest of the blocks. So $\dim\ker(T-\lambda I)$ is the number of blocks corresponding to the eigenvalue $\lambda$.