Uniqueness of the Jordan Form of a Matrix: Hoffman and Kunze, page 246

Question

I am having trouble understanding one line in Hoffman and Kunze's proof of the uniqueness of the Jordan form $A$ of a matrix, on page 246. Rather than describe all the details from the ground up, I will insert a picture of the portion of the proof I am having trouble with:

Why is $W_i$ the kernel of $(T - c_iI)^n$, where $n = \dim V$? I do not understand how the rationale given as proof in the next sentence yields the result. I do understand that $A_i - c_iI$ is nilpotent, and that $A_j - c_iI$ is nonsingular for $j \neq i$, I just do not see how these facts tell us what $W_i$ is.

Answer 1

Edit: Rewritten to align more closely with the text in the book.

Every element $v \in V$ has a unique decomposition of the form $$ v = w_1 + \dots + w_k,$$ where $w_j \in W_j$ for each $j$.

The image of $v$ under $(T - c_iI)^n$ is given by $$ (T - c_iI)^n v= (A_1 - c_i I)^n w_1 + \dots (A_k - c_i I)^n w_n.$$ Since $(A_j - c_i I)^n w_j \in W_j$ for each $j$, this equation is in fact the unique decomposition of $(T - c_iI)^n v$ as a sum of elements in $W_1, \dots, W_k$.

Thus $$ (T - c_iI)^nv = 0 \iff (A_j - c_i I)^n w_j = 0 \text{ for all } j.$$

Our task is to show that $\text{Ker}(T - c_iI)^n = W_i$. This is equivalent to showing that:

• $\text{Ker}(A_i - c_i I)^n = W_i$.
• $\text{Ker}(A_j - c_i I)^n = \{ 0 \} $ for $j \neq i$.

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Proving that $\text{Ker}(A_i - c_i I)^n = W_i$

The book says that $A_i - c_i I$ is nilpotent. This is true, but it is not enough. What we really want to show is that $(A_i - c_iI)^n$ is the zero matrix.

To prove this, note that $A_i$ is a upper triangular matrix, where every entry along the diagonal is equal to $c_i$. So $A_i - c_iI$ is a strictly upper triangular matrix (i.e. it is an upper triangular matrix where every entry along the diagonal is $0$). Therefore, if $d_i$ is the dimension of $W_i$, then $(A_i - c_iI)^{d_i}$ is the zero matrix. Since $d_i \leq n$, $(A_i - c_iI)^{n}$ is also the zero matrix.

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Proving that $\text{Ker}(A_j - c_i I)^n = \{ 0 \}$ for $j \neq i$

The book says that $A_j - c_iI$ is non-singular, but let's be clear about why this is and how this is relevant. $A_j$ is an upper diagonal matrix, where every entry along the diagonal is equal to $c_j$. So $A_j - c_i I$ is an upper diagonal matrix, where every entry along the diagonal is equal to $(c_j - c_i)$. Hence the determinant of $A_j - c_jI$ is $(c_j - c_i)^{d_j}$, where $d_j$ is the dimension of $W_j$. The determinant of $(A_j - c_jI)^n$ is therefore $(c_j - c_i)^{d_j n}$. Since $c_j \neq c_i$, this determinant is non-zero. So $(A_j - c_i I)^n$ is non-singular, and hence $\text{Ker}(A_j - c_i I)^n = \{ 0 \}$.