# Do the vectors (1,3,-2),(2,1,1) belong to R(T)?

• Dec 12th 2009, 08:38 PM
Undefdisfigure
Do the vectors (1,3,-2),(2,1,1) belong to R(T)?
Let T:IR^3-->IR^3 be defined by T(a,b,c) = (a+b, b-2c, a + 2c). Do the vectors (1,3,-2), (2,1,1) belong to R(T)?

Looking through my notes I found some clues to this question. I found that "RT is always span (Tv1,..,Tvn) where v1,..,vn is basis of domain V.

My first approach was to make a column vector (a) = (a+b)
(b) (b-2c)
(c) (a+2c)

and then remove the coefficients like this:

a (1) b(1) c(0) to get a basis. I don't know if I was on the right track.
(0) (1) (-2)
(1) (0) (2)

The second thing I tried was to compute two vectors for R(T) like this:

T(1,3,-2) = (4,7,-3)

T(2,1,1) = (3,-1,4)

And then get that R(T) = span {(4,7,-3), (3,-1,4)}.

Anyways, I didn't finish my answer since I am confused. Also sorry about my notation because I don't know what syntax I need to use to make a matrix on this forum.
• Dec 13th 2009, 05:53 AM
tonio
Quote:

Originally Posted by Undefdisfigure
Let T:IR^3-->IR^3 be defined by T(a,b,c) = (a+b, b-2c, a + 2c). Do the vectors (1,3,-2), (2,1,1) belong to R(T)?

Looking through my notes I found some clues to this question. I found that "RT is always span (Tv1,..,Tvn) where v1,..,vn is basis of domain V.

My first approach was to make a column vector (a) = (a+b)
(b) (b-2c)
(c) (a+2c)

and then remove the coefficients like this:

a (1) b(1) c(0) to get a basis. I don't know if I was on the right track.
(0) (1) (-2)
(1) (0) (2)

The second thing I tried was to compute two vectors for R(T) like this:

T(1,3,-2) = (4,7,-3)

T(2,1,1) = (3,-1,4)

And then get that R(T) = span {(4,7,-3), (3,-1,4)}.

Anyways, I didn't finish my answer since I am confused. Also sorry about my notation because I don't know what syntax I need to use to make a matrix on this forum.

Since $T\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatri x}a+b\\b-2c\\a+2c\end{pmatrix}$, you only have to ask whether there exist $a,b,c,\in\mathbb{R}\,\,\,s.t.\,\,\,\begin{pmatrix} a+b\\b-2c\\a+2c\end{pmatrix}=\begin{pmatrix}\;\;1\\\;\;3\ \-2\end{pmatrix}\,\,or\,\,\begin{pmatrix}2\\1\\1\end {pmatrix}$.

For example, for the first one, we'd get:

** $a+b=1$

** $b-2c=3$

** $a+2c=-2$

Substracting the third eq. from the first one we get $b-2c=3=$ second eq., so the system seems to be congruent (i.e., has at least one solution), for example: $a=2,b=-1,c=-2$, so this first vector does belong to the range of $T$. Now you continue with the other one.

Tonio
• Dec 13th 2009, 01:39 PM
Undefdisfigure
how do I thank you for this reply??
• Dec 13th 2009, 02:00 PM
tonio
Quote:

Originally Posted by Undefdisfigure
how do I thank you for this reply??

You just have...any time!(Wink)

Tonio