Understand Linear Algebra: Im(BF) in Im(B)?

[Payam30]

Hi,
Im actually doing some systems theory and it requires some basic linear algebra stuff that I totally forgotten. Anyway:according to my prof. :
For any matrix F. Im(BF) is contained in Im(B).


here is my question:
so Im(FB) is still in Im(B)? or is it true only when we multiply a matrix to the right of the origin matrix?

if Im(B) is in Im(T) and Im(AB) is in Im(T), is Im(A) in Im(T) as well? and is Im(BA) in Im(T). imagin that the dimension ab matrices are appropriate. and they are constant real matrices

 

[Fredrik]

Assuming that B and F are matrices, and that Im(B) denotes the range of B (I would write it as ran B)...

If x is in the range of BF, then there's a y such that x=(BF)y=B(Fy). This implies that x is in the range of B.

So yes, Im(BF) is a subset of Im(B).

 

[Payam30]

Assuming that B and F are matrices, and that Im(B) denotes the range of B (I would write it as ran B)...

If x is in the range of BF, then there's a y such that x=(BF)y=B(Fy). This implies that x is in the range of B.

So yes, Im(BF) is a subset of Im(B).

Hi
Thanks for your answer. What about the other quations? thanks.

 

[micromass]

so Im(FB) is still in Im(B)?

This is not necessarily true. For example, in $\mathbb{R}^2$, let $B$ a projection on the X-axis and let $B$ be a suitable rotation.

if Im(B) is in Im(T) and Im(AB) is in Im(T), is Im(A) in Im(T) as well?

Not necessarily true. in $\mathbb{R}^2$, let $B$ and $T$ be projections on the X-axis and let $A$ be the identity.

and is Im(BA) in Im(T).

This is true.