Suppose b is an n-dimensional column vector that can be decomposed as b=b_1+b_2

abkapseln87

abkapseln87

Answered question

2022-09-29

Suppose b is an n-dimensional column vector that can be decomposed as b = b 1 + b 2 . In this decomposition, b 1 and b 2 hold the non-zero components of b respectively (e.g. if b= [ 1 2 3 ] , then this decomposition can be b 1 = [ 1 0 3 ] and b 2 = [ 0 2 0 ] ).
Suppose A is a real matrix of order n and let C=[ b, A b , ... , A n 1 b], C 1 =[ b 1 , A b 1 , ... , A n 1 b 1 ] and C 2 =[ b 2 , A b 2 , ... , A n 1 b 2 ]. If we have rank([ C 1 , C 2 ])=n, can we prove rank(C)=n?

Answer & Explanation

Xavier Jennings

Xavier Jennings

Beginner2022-09-30Added 9 answers

The answer is no. As an example, consider
A = ( 1 1 1 1 ) , b = ( 1 1 ) , b 1 = ( 1 0 ) , b 2 = ( 0 1 ) .
We find that C 1 = [ b 1     A b 1 ] and C 2 = [ b 2     A b 2 ] each have full row rank n=2, as does the block matrix [ C 1     C 2 ]. However, C = [ b     A b ] has a zero column and therefore does not have full rank.

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