Let A be an m×n matrix, and C=AB. Show that: a) Null(B) is a subspace of Null(C). b) Null(C)⊥ is a subspace of Null(B)⊥ and, consequently, Col(CT) is a subspace of Col(BT).

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escumantsu · Accepted Answer

Step 1 A be is an m×n matrix, B is an n×r matrix, and C=AB. So A be an m×r matrix. a). To show that Null(B)⊥ is a subspace of Null(C)⊥. Let x∈Null(B)⊥ then Bx=0 ⇒ABx=A.0 ⇒Cx=0. Therefore, x∈Null(C)⊥, hence Null(B)⊥ is a subpace os Null(C)⊥. Step 2 b) Now to show Null(C)⊥ is a subspace of Null(B)⊥. Using part (a), it is enough to show that if S,V are subspaces of vector space W and if S is a subspace of the vector space V, then V⊥ is a subspace by S⊥. Sperp={x∈X:⟨x, s⟩=0 for all s∈S} Vperp={x∈X:⟨x, v⟩=0 for all v∈V} Suppose x∈V⊥. Then x is orthogonal to each vectors of V, hence will be orthogonal to each vectors of S, since S⊂V⇒x∈S⊥. Therefore, Vperp is a subspace by S⊥ and part (a) Null(C)⊥ is a subspace of Null(B)⊥. Step 3 c) To Show Col(CT) is a subspace of Col(BT). Here we will show that for any matrix Am×n we have Col(AT)=Null(A)⊥. Now we know that Col(AT) and Null(A)⊥ are subspaces of Rn. Now let y∈Col(AT), then there exists x∈Rm such that y=ATx. Now Let z∈Nl―(A), then ⟨z, y⟩=⟨z, ATx⟩=(ATx)Tz=xTAz=0 Therefore, Col(AT) is a subspace of Null(A)⊥ Once again by rank Nullit y theorem we have

Let A be an m\times n matrix, and C=AB.Show that:a) Nul

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