Question

Vector space/Subspace
Let A=the 2x3 matrix with vectors (1,1,0) and (1,-1,0). Let V(3) be the set of vectors x in R(3) such that Ax=0. Verify that V(3) is a subspace of R(3).

Answer 1

\[\left[\begin{matrix}1 & 1 & 0 \\ 1 & -1 & 1\end{matrix}\right]\]

Answer 2

you just did a problem verifying that 2 by 3 matrices are a vector space, so it should be clear that 1 by 3 matrices do as well. the difference here is that you must verify that these are closed under addition and scalar multiplication
that is, if two vectors \(x,y\) satisfy \(Ax=0\) and \(Ay=0\) then so does their sum and also any scalar multiple

Answer 3

I dont know why these problems with matrices are confusing me. So when you are proving closed under addition are you adding the matrix A plus the two vectors. And maybe I am getting confused on how to show this b/c it says compare V(1) and V(3) from this problem... and V(1) was from a previous problem that said V(1) consisted of all scalar multiples of (1, -1, -2), and the book says V(1) and V(3) are the same... but I am not seeing how so.

Answer 4

ohhh wait, yeah if I multiply that vector by A I get Ax=0

Answer 5

i think i see the confusion.
the vectors in the previous example where 2 by 3 matrices, the scalars were real numbers
in this case the vector are, well, vectors, which are 1 by 3 matrices
the scalars are still real numbers
so what you need to show is that if you add two such vectors (1 by 3 matrices) say \(x\) and \(y\) you get another 1 by 3 matrix (that is obvious) say \(z\) that also satisfy the condition that \(Az=0\)

Answer 6

the definition for your vector space is
"Let V(3) be the set of vectors x in R(3) such that Ax=0" so vectors in this case are vectors, that is things that look like \((a_1,a_2,a_3)\) although perhaps written as a column

Answer 7

so more like \[(a _{1}, -a _{1}, -2a _{1})\]

Answer 8

yeah probably
to be perfectly honest i didn't do it

Answer 9

ohhh okay. well regardless you you still helped me. thank you.

Answer 10

yw