Null-space proof for square matrix

[VitaminC]

Homework Statement

Suppose F is a field and A ε Mnn(F) is a square matrix whose kjth entry is denoted
by akj ε F for all 1 ≤ k,j ≤ n. Suppose further that (b1; . . . ; bn) and (c1; . . . ; cn)
belong to the null space of A.

i) Prove that the sum (b1 + c1; b2 + c2; . . . ; bn + cn) also belongs to the null space of A

ii) Suppose d ε F. Prove that the scalar multiple (db1; . . . ; dbn) also belongs to the
null space of A.

Homework Equations

The Attempt at a Solution

For both proofs,

Assume A is a non-singular; the linear system (A,0) has a unique solution
----> is there a proof for this assumption using akj ε F for all 1 ≤ k,j ≤ n ?

Thus, A is non-singular <=> LS(A,0)
<=> leading ones (# of nonzero rows) is equal to the number of rows
<=> A = I (identity matrix)

For i)
Each leading one of the matrix will equal to some corresponding (b1,...,bn) or (c1,...,cn).
Since (b1,...,bn) and (c1,...,cn) fall in the null space of A, the corresponding (b1,...,bn) and (c1,...,cn) of A = I ,should equal 0.

Thus, b1 = 0 and c1 = 0 --> b1 = c1 --> b1 + b1 = 0 --> b1 + c1 = 0 for all (b1,...,bn) and (c1,...,cn). Thus, (b1 + c1, b2 + c2, . . . , bn + cn) is also in the null space of A. For ii) from i), I established that some corresponding (b1,...,bn) will equal 0, when A = I. Thus, multiplying by the scalar 'd' does not change the output.

So, (db1,...,dbn) = (b1,...,bn) --> (db1,...,dbn) is also in the null space of A.

 

[Mark44]

Homework Statement

Suppose F is a field and A ε Mnn(F) is a square matrix whose kjth entry is denoted
by akj ε F for all 1 ≤ k,j ≤ n. Suppose further that (b1; . . . ; bn) and (c1; . . . ; cn)
belong to the null space of A.

i) Prove that the sum (b1 + c1; b2 + c2; . . . ; bn + cn) also belongs to the null space of A

ii) Suppose d ε F. Prove that the scalar multiple (db1; . . . ; dbn) also belongs to the
null space of A.

Homework Equations

The Attempt at a Solution

For both proofs,

Assume A is a non-singular; the linear system (A,0) has a unique solution
----> is there a proof for this assumption using akj ε F for all 1 ≤ k,j ≤ n ?

Thus, A is non-singular <=> LS(A,0)
<=> leading ones (# of nonzero rows) is equal to the number of rows
<=> A = I (identity matrix)

For i)
Each leading one of the matrix will equal to some corresponding (b1,...,bn) or (c1,...,cn).
Since (b1,...,bn) and (c1,...,cn) fall in the null space of A, the corresponding (b1,...,bn) and (c1,...,cn) of A = I ,should equal 0.

Thus, b1 = 0 and c1 = 0 --> b1 = c1 --> b1 + b1 = 0 --> b1 + c1 = 0 for all (b1,...,bn) and (c1,...,cn). Thus, (b1 + c1, b2 + c2, . . . , bn + cn) is also in the null space of A.


For ii) from i), I established that some corresponding (b1,...,bn) will equal 0, when A = I. Thus, multiplying by the scalar 'd' does not change the output.

So, (db1,...,dbn) = (b1,...,bn) --> (db1,...,dbn) is also in the null space of A.

Someone else asked what seems to be the same question - see https://www.physicsforums.com/showthread.php?t=539211.