Problem of rank

• Dec 12th 2009, 11:36 PM
Xingyuan
Problem of rank
Assume $\displaystyle A,B \in M_n(\mathbb{K})$

show that:

If $\displaystyle AB=BA=0$ and $\displaystyle rank(A^2)=rank(A)$

then $\displaystyle rank(A+B)=rank(A)+rank(B)$
• Dec 13th 2009, 11:27 AM
NonCommAlg
Quote:

Originally Posted by Xingyuan
Assume $\displaystyle A,B \in M_n(\mathbb{K})$

show that:

If $\displaystyle AB=BA=0$ and $\displaystyle rank(A^2)=rank(A)$

then $\displaystyle rank(A+B)=rank(A)+rank(B)$

let $\displaystyle V=\mathbb{K}^n$ and $\displaystyle T,S : V \longrightarrow V$ be the corresponding linear transformations defined by $\displaystyle Tv=Av, \ Sv = Bv.$ let's put $\displaystyle K_1=\ker T, \ K_2=\ker S,$ and $\displaystyle W_1=\text{im}(T).$

since $\displaystyle ST=0,$ we have $\displaystyle W_1 \subseteq K_2.$ thus, using the rank nullity theore, we have $\displaystyle K_1 + K_2 \supseteq K_1 \oplus W_1=V$ and hence $\displaystyle \dim (K_1+K_2)=n. \ \ \ \ \ \ (1)$

on the other hand, since $\displaystyle \text{rank}(T^2)=\text{rank}(T)$ and $\displaystyle \ker T \subseteq \ker T^2,$ we have, again by the rank-nullity theorem, $\displaystyle \ker T^2=\ker T = K_1.$ now suppose $\displaystyle v \in \ker (T+S).$

then $\displaystyle Tv=-Sv$ and thus $\displaystyle T^2v=-TSv=0,$ i.e. $\displaystyle v \in \ker T^2=K_1.$ hence $\displaystyle Tv=0$ and therefore $\displaystyle Sv=-Tv=0.$ so $\displaystyle \ker(T+S)=K_1 \cap K_2. \ \ \ \ \ (2)$

finally, using (1) and (2) and the rank-nullity theorem, completing the proof is easy:

$\displaystyle n=\dim(K_1+K_2)=\dim K_1 + \dim K_2 - \dim(K_1 \cap K_2)$

$\displaystyle =2n-\text{rank}(T)-\text{rank}(S) - \dim(\ker(T+S))$

$\displaystyle =2n-\text{rank}(T)-\text{rank}(S) - (n - \text{rank}(T+S))$

$\displaystyle =n -\text{rank}(T)-\text{rank}(S) + \text{rank}(T+S). \ \ \ \Box$