Row vectors of B and the subspace spanned by row vectors of B lie in row space A:Why?

I've some questions about the following theorem. The theorem is:

If an m x n matrix is row equivalent to an m x n matrix , then the row space of is equal to the row space of .

The proof is like this:

Because the rows of can be obtained from the rows of by elementary row operations (scalar multiplication and addition), it follows that the row vectors of can be written as linear combinations of the row vectors of .

The row vectors of lie in the row space of , and the subspace spanned by the row vectors of is contained in the row space of . (<-----My question why's that the row vectors of lie in the row space of , and why's the subspace spanned by the row vectors of is contained in the row space of ?)

But it is also true that the rows of can be obtained from the rows of by elementary row operations. So, you can conclude that the two row spaces are subspaces of each other, making them equal.

-----------End of proof---------

Can anyone kindly answer the marked question in above proof?

If the row vecors of can be written as linear combination of row vectors of how it makes the statement "The row vectors of lie in the row space of , and the subspace spanned by the row vectors of is contained in the row space of " true?

Is it possible to elaborate on this?

Re: Row vectors of B and the subspace spanned by row vectors of B lie in row space A:

the row space of A IS linear combinations of the rows of A. we got every row-vector of B by adding together scalar multiples of rows of A.

that is precisely how you form linear combinations of vectors.

Re: Row vectors of B and the subspace spanned by row vectors of B lie in row space A:

Quote:

Originally Posted by

**Deveno** the row space of A IS linear combinations of the rows of A. we got every row-vector of B by adding together scalar multiples of rows of A.

that is precisely how you form linear combinations of vectors.

Thanks a thousand time Deveno. You showed me the way again. Thank you for that.