# Simple lineal algebra question, why the dim ImT=3?

The problem is to found the kernel and Image of T, and its nullity and range. Be T(x,y,z)=(x-y+2z,2x+y,-x-2y+2z|x,y,z $\in \Bbb R$) V=W=$\Bbb R^{3}$ and K=$\Bbb R$. For now I saw that the Kernel of T={0}, but for the ImT I got the following:

\begin{align}\operatorname{Im}T&=\{x-y+2z,2x+y,-x-2y+2z \mid x,y,z \in \Bbb R \} \\ &=\{(x,2x,-x)+(-y,y,-2y)+(2z,0,2z) \mid x,y,z \in \Bbb R \} \\ &=\{x(1,2,-1)+y(-1,1,-2)+z(2,0,2) \mid x,y,z \in \Bbb R\} \\ &=\langle\{(1,2,-1),(-1,1,-2),(2,0,2)\}\rangle\end{align}. But (1,2,-1),(-1,1,-2),(2,0,2) are lineal dependent, so I could take the generator of V as $\langle${(-1,1,-2),(2,0,2)}$\rangle$, and both elements are linean independent so they form a base for ImT. So dim ImT=2

But the dimension theorem states that for V,W vectorial spaces so T:V$\rightarrow$W lineal If V is of finite dimension then ImT and KerT are of finite dimension, and also: dim V=dim KerT+dim ImT. But for my result i get that dim V=dim KerT+dim ImT is equal to 3=0+2 ! Why? Where did I make a mistake

• Learn to MathJax, friend. See how much nicer the part I edited looks? – user137731 Oct 10 '16 at 4:31
• Thanks, for the link I will fix those things for later publications – Mounice Oct 10 '16 at 4:36

Writing $T$ as a matrix:

$$T = \left[ \begin{array}{ccc} 1 & -1 & 2 \\ 2 & 1 & 0 \\ -1 & -2 & 2 \\ \end{array} \right]$$

Which has row reduced form

$$rref(T) = \left[ \begin{array}{ccc} 1 & 0 & \frac{2}{3} \\ 0 & 1 & -\frac{4}{3} \\ 0 & 0 & 0 \\ \end{array} \right].$$

From this, we see that the rank of T is 2, so $\dim Ker(T)= 3-2=1$ and $\dim Im(T) = 2$. A basis for $Im(T)$ is given by

$$\left\{ \begin{bmatrix} 1 \\ 2\\-1 \end{bmatrix} , ~ \begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix} \right\}$$

and a basis for $Ker(T)$ is given by

$$\left\{ \begin{bmatrix} -2/3 \\ 4/3 \\ 1 \end{bmatrix} \right\}$$.