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

1. ## 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 $A$ is row equivalent to an m x n matrix $B$, then the row space of $A$ is equal to the row space of $B$.

The proof is like this:

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

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

But it is also true that the rows of $A$ can be obtained from the rows of $B$ 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 $B$ can be written as linear combination of row vectors of $A$ how it makes the statement "The row vectors of $B$ lie in the row space of $A$, and the subspace spanned by the row vectors of $B$ is contained in the row space of $A$" true?

Is it possible to elaborate on this?

2. ## 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.

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

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.