1. ## Linear Transformation question

According to my book,

There exists a linear transformation $\displaystyle T:R^3 \rightarrow R^3$ that has $\displaystyle (1,-1,2),(2,1,-1),(1,2,-3) \in KerT$ and $\displaystyle (1,-1,2) \in ImT$

They explain that (1,-1,2),(2,1,-1),(1,2,-3) are dependent vectors.

My question is: If you are dealing with the standard basis (or any set of 3 independent vectors), does there exist a linear transformation where $\displaystyle T(1,0,0)=T(0,1,0)=T(0,0,1) =(0,0,0,)$ and have $\displaystyle (1,0,0) \in ImT$

2. ...does there exist a linear transformation where...
I would say no. Not if $\displaystyle T:\mathbb{R}^{3}\to\mathbb{R}^{3}.$ The vectors you have exhibited there are a basis. So, take any vector $\displaystyle \mathbf{x}$ in $\displaystyle \mathbb{R}^{3}.$ Then $\displaystyle \mathbf{x}=a\hat{i}+b\hat{j}+c\hat{k},$ where the hatted vectors are the unit vectors you mentioned. Then

$\displaystyle T\mathbf{x}=T(a\hat{i}+b\hat{j}+c\hat{k})=aT\hat{i }+bT\hat{j}+cT\hat{k}=0.$ Therefore, $\displaystyle (1,0,0)\not\in\text{Image}(T).$

However, this is not the situation you're dealing with in the first part of your post. In the first part, the book says that the three vectors there are dependent. A basis is, by definition, independent. Therefore, all bets are off.

3. Originally Posted by Ackbeet
I would say no. Not if $\displaystyle T:\mathbb{R}^{3}\to\mathbb{R}^{3}.$ The vectors you have exhibited there are a basis. So, take any vector $\displaystyle \mathbf{x}$ in $\displaystyle \mathbb{R}^{3}.$ Then $\displaystyle \mathbf{x}=a\hat{i}+b\hat{j}+c\hat{k},$ where the hatted vectors are the unit vectors you mentioned. Then

$\displaystyle T\mathbf{x}=T(a\hat{i}+b\hat{j}+c\hat{k})=aT\hat{i }+bT\hat{j}+cT\hat{k}=0.$ Therefore, $\displaystyle (1,0,0)\not\in\text{Image}(T).$
So that means that ImageT= (0,0,0) = KernelT ?

However, this is not the situation you're dealing with in the first part of your post. In the first part, the book says that the three vectors there are dependent. A basis is, by definition, independent. Therefore, all bets are off.
I know! That's why I was asking. (:

4. So that means that ImageT= (0,0,0) = KernelT ?
No, that's a mistake in categories. Let's say I had a linear transformation $\displaystyle T:V\to W.$ Then $\displaystyle \text{Image}(T)\subseteq W,$ but $\displaystyle \text{Ker}(T)\subseteq V.$

In words, the image is the set of all vectors that T can "reach" by acting on vectors in its domain. The kernel is the set of all vectors in the domain that get killed by T (sent to the zero vector).

I know! That's why I was asking. (:
Been doing some computations. Let's start with our three vectors in the kernel of T. According to my computations, this implies that T looks like this:

$\displaystyle T=\begin{bmatrix}-a/3 &5a/3 &a\\ -b/3 &5b/3 &b\\ -c/3 &5c/3 &c\end{bmatrix}.$

Setting $\displaystyle T\mathbf{x}=(1,-1,2)^{T}$ yields the augmented matrix

$\displaystyle \left[\begin{matrix}-a/3 &5a/3 &a\\ -b/3 &5b/3 &b\\ -c/3 &5c/3 &c\end{matrix}\left|\begin{matrix}1\\-1\\2\end{matrix}\right].$

The three rows of T are obviously linearly dependent. So you can do an elementary row operation that kills the second and third lines. You get

$\displaystyle \left[\begin{matrix}-a/3 &5a/3 &a\\ 0 &0 &0\\ 0 &0 &0\end{matrix}\left|\begin{matrix}1\\-(b/a)-1\\-(c/a)+2\end{matrix}\right].$

Therefore, in order for this system to have a solution at all, you must have $\displaystyle -b/a=1$ and $\displaystyle c/a=2.$ Thus, $\displaystyle b=-a$ and $\displaystyle c=2a.$ Now then. Your remaining equation is

$\displaystyle -(a/3)x+(5a/3)y+az=1.$ This will be a two-parameter solution. Set $\displaystyle y=s$ and $\displaystyle z=t$. Then $\displaystyle -(a/3)x=1-(5a/3)s-at,$ from which it follows that $\displaystyle x=-3/a+5s+3t.$ Assuming $\displaystyle a\not=0,$ you can definitely find values of $\displaystyle s$ and $\displaystyle t$ to make this work. Hence, any matrix of the form

$\displaystyle T=\begin{bmatrix}-a/3 &5a/3 &a\\ a/3 &-5a/3 &-a\\ -2a/3 &10a/3 &2a\end{bmatrix}=a\begin{bmatrix}-1/3 &5/3 &1\\ 1/3 &-5/3 &-1\\ -2/3 &10/3 &2\end{bmatrix}$

will work. You can check that it still kills all the vectors in the kernel. Let's say you set $\displaystyle a=s=t=1.$ Then the vector

$\displaystyle \mathbf{x}=\begin{bmatrix}-3+5+3\\1\\1\end{bmatrix}$

gets sent to $\displaystyle (1,-1,2)^{T}.$

So the book is correct.