Problem of rank

**Xingyuan** · December 12th 2009, 11:36 PM

Assume $A,B \in M_n(\mathbb{K})$

show that:

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

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

**NonCommAlg** · December 13th 2009, 11:27 AM

Originally Posted by Xingyuan

Assume $A,B \in M_n(\mathbb{K})$

show that:

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

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

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

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

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

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

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

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

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

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

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

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