Orthogonal vector spaces and matrices

[julie94]

Hi everyone,

I would need to get some help on the following question

Let A (m*n)
Let B (m*p)

Let L(A) be the span of the columns of A.
L(A) is orthogonal to L(B) <=> A'B=0

I suppose that the => direction is pretty obvious, since A is in L(A)
and B in is L(B).

Now I am not sure how to attack the <= statement. I guess that every
vector in L(A) is generated by the rows of A, and every vector in L(B)
is generated by the columns of B.
Therefore A'B=0, means Sum(i, j, constant_i*row_i of
A*constant_j*row_j of B).

How can I improve my argument?

 

[jbunniii]

Hi everyone,

I would need to get some help on the following question

Let A (m*n)
Let B (m*p)

Let L(A) be the span of the columns of A.
L(A) is orthogonal to L(B) <=> A'B=0

I suppose that the => direction is pretty obvious, since A is in L(A)
and B in is L(B).

Not exactly. Each column of A is in L(A), and each column of B is in L(B).

I assume by A' you mean $A^T$, the transpose of A, or if you're working over the complex field, it would have to be $A^H$, the conjugate transpose of A. If so, then each element of A'B is formed by computing the inner product between one column of A and one column of B. By the orthogonality assumption, each inner product is 0, and therefore so is A'B.

Now I am not sure how to attack the <= statement. I guess that every
vector in L(A) is generated by the rows of A, and every vector in L(B)
is generated by the columns of B.
Therefore A'B=0, means Sum(i, j, constant_i*row_i of
A*constant_j*row_j of B).

How can I improve my argument?

That's the right idea, though you mean "every vector in L(A) is generated by the columns of A", or "every vector in L(A) is generated by the rows of (A')".

I suggest writing down an expression for the (i,j)'th element of A'B and setting it equal to zero. This will show that any column of A is orthogonal to any column of B. You then need to show that this implies that any element of L(A) is orthogonal to any element of L(B). (Hint: by definition, the columns of A are a basis for L(A), and similarly for B.)