Linear Algebra Topic Orthogonal vector and complements Given

Linear Algebra: Topic: Orthogonal vector and complements

Given the transformation A R^5 rightarrow R^4 with matrix [1 0 0 0 0 0 1 -3 0 0 0 0 0 0 1 0 0 0 0 0] verify that: (Row A) =Nul A (Col A) = Nul A^T

Solution

We have Row(A) = Span{ (1,0,0,0,0), ( 0,1,-3,0,0), ( 0,0,0,0,1)} and since the 3^rd column of A is -3 times its 2^nd column , hence Col(A) = Span{( (1,0,0,0)^T,(0,1,0,0)^T,(0,0,1,0)^T}.

Null(A) is the set of solutions of the equation AX = 0. If X = (x₁,x₂,x₃,x₄,x₅)^T , then this equation is equivalent to the linear system x₁=0,x₂-3x₃= 0,and x₅=0.Let x₃=r and x₄= t.Then X = (0,3r,r,t,0)^T = r(0,3,1,0,0)^T + t (0,0,0,1,0)^T so that Null (A) = Span{ (0,3,1,0,0)^T, (0,0,0,1,0)^T}. Let an arbitrary vector in (Row(A) )   be ( a,b,c,d,e)^T . Now,

Since ( a,b,c,d,e)^T is orthogonal to all the vectors in Row(A), hence a = 0, e = 0 and b = 3c and d is arbitrary. Let c = r and t = t. Then ( a,b,c,d,e)^T = ( 0, 3r,r, t ,0)^T = r(0,3,1,0,0)^T+ t (0,0,0,1,0)^T . Hence, (Row(A) )   = Span{ (0,3,1,0,0)^T, (0,0,0,1,0)^T} = Null(A).

Null(A^T) is the set of solutions of the equation A^T Y = 0. If Y = (y₁,y₂,y₃,y₄)^T, then, this equation is equivalent to the linear system y₁ = 0, y₂ = 0, -3y₃ = 0, and y₃ = 0, y₄ is arbitrary. Therefore, Null(A^T) = Span{ (0,0,0,1)^T}.

       Let an arbitrary vector in (Col(A)) be (p,q,r,s)^T. Then :

     Since (p,q,r,s)^T is orthogonal to all the vectors in Col(A), hence p=0,q= 0 and r = 0,s is arbitrary. Therefore, (Col(A)) = Span{(0,0,0,1)^T) = Null(A^T).