# Examining a transformation

• February 13th 2009, 06:11 PM
arbolis
Examining a transformation
I couldn't think a lot on this problem and I'd like to know whether my way of thinking is not wrong.
Determine whether the following transformation is invertible and in case of being invertible, give its inverse.
Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(v)=2v-(1,1,1)$.
My attempt : Well I think that it has an inverse and that it's $\frac{v}{2}+\left( \frac{1}{2},\frac{1}{2},\frac{1}{2} \right)$.
But I'd love to know how I could find a general way to determine if a transformation has an inverse. I guess by writing its matrix with respect to the canonical basis (or any other basis) and check out if the matrix is invertible... But I'm not sure. Nor I'm sure how to write such a matrix in this example.
• February 13th 2009, 09:03 PM
ThePerfectHacker
Quote:

Originally Posted by arbolis
I couldn't think a lot on this problem and I'd like to know whether my way of thinking is not wrong.
Determine whether the following transformation is invertible and in case of being invertible, give its inverse.
Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(v)=2v-(1,1,1)$.
My attempt : Well I think that it has an inverse and that it's $\frac{v}{2}+\left( \frac{1}{2},\frac{1}{2},\frac{1}{2} \right)$.
But I'd love to know how I could find a general way to determine if a transformation has an inverse. I guess by writing its matrix with respect to the canonical basis (or any other basis) and check out if the matrix is invertible... But I'm not sure. Nor I'm sure how to write such a matrix in this example.

Let $B = \{ \b{i},\b{j},\b{k}\}$ and now compute $T\b{i},T\b{j},T\b{k}$.
Can you find the matrix now?
• February 13th 2009, 10:37 PM
math2009
$T(\vec{v})=2\vec{v}-\begin{bmatrix} 1\\1\\1 \end{bmatrix} . \ \vec{v},\vec{v}_1,\vec{v}_1 \in R^3, k\in R$
At first, we need verify if T is linear transformation.
$T(\vec{v}_1+\vec{v}_2)=2(\vec{v}_1+\vec{v}_2)-\begin{bmatrix} 1\\1\\1 \end{bmatrix}$, but $T(\vec{v}_1)+T(\vec{v}_2)=2(\vec{v}_1+\vec{v}_2)-2\begin{bmatrix} 1\\1\\1 \end{bmatrix}$, $T(\vec{v}_1+\vec{v}_2)\neq T(\vec{v}_1)+T(\vec{v}_2)$
$T(k\vec{v})=2k\vec{v}-\begin{bmatrix} 1\\1\\1 \end{bmatrix},\ kT(\vec{v})=2k\vec{v}-k\begin{bmatrix} 1\\1\\1 \end{bmatrix}$ , $T(k\vec{v})\neq kT(\vec{v})$

T is NOT a linear transformation. Your defination is mistake. So you couldn't get inverse matrix. Your solution is inverse function.
Generally, $A\in R^{n\times n},T(\vec{x})=A\vec{x}$ , get rank(A) by rref(A). If rank(A)=n and A is square matrix, then A is invertible.
• February 14th 2009, 04:12 AM
HallsofIvy
math2009, arbolis never said linear transformation, just "transformation". Of course, in this case, it cannot be written as a matrix so ThePerfectHackers's idea doesn't help.

In general a function is invertible if it is one-to-one and onto.

"One-to-one" means that two different values in the domain cannot be mapped into the same value in the range.

"Onto" means that every member of the range has some value of the domain mapped to it.

For this transformation, in component notation, T(x,y,z)= (2x-1, 2y-1, 2z-1). Is it one-to-one? Suppose T(x',y',z')= T(x,y,z). Then 2x'- 1= 2x- 1, 2y- 1= 2y-1, and 2z'-= 2z-1. Add 1 to both sides of each equation, divide both sides of each equation by 2 and we get x'= x, y'= y, and z'= z. Yes, this transformation is one-to-one.

Is it onto? Suppose (u, v, w) is any member of $R^3$. Is there an (x, y, z) so that T(x, y, z)= (u, v, w), we must have 2x- 1= u, 2y- 1= v, 2z- 1= w. Solving for x, y, z, we have x= (u+1)/2, y= (v+1)/2, z= (w+1)/2. Since those are perfectly good numbers, ((u+1)/2, (v+1)/2, (w+1)/2) is a member of $R^3$ that is mapped into (u, v, w). Since (u, v, w) could be any member of $R^3$ so T is onto $R^3$.

Being both one-to-one and onto, T has an inverse. And, we have already found its inverse by going backward from (u, v, w) to ((u+1)/2, (v+1)/2, (w+1)/2). $T^{-1}\left[\begin{array}{c} u \\ v \\ w\end{array}\right]=\left[\begin{array}{c}\frac{u+1}{2} \\ \frac{v+1}{2} \\ \frac{w+1}{2}\end{array}\right]$ or, for V a member of $R^3$, $T^{-1}v= \frac{1}{2}\left(V+ \left[\begin{array}{c} 1 \\ 1 \\ 1\end{array}\right]\right)$, exactly what arbolis said.

(I said that since this transformation is not linear it cannot be written as a matrix. It can be written as a matrix and a difference: $T\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right]\right)= \left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]- \left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right]$: multiply by the matrix and then subtract. Since finding an inverse basically involved "doing everything backwards", its inverse can be expressed as "first add, then multiply by the inverse matrix". Since the matrix is diagonal, its inverse is trivial and we have $T^{-1}\left[\begin{array}{c} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} \frac{1}{2} & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right]\left(\left[\begin{array}{c} x \\ y \\ z \end{array}\right]+ \left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right]\right)$)
• February 14th 2009, 04:29 AM
math2009
HallsofIvy,function is not usually discussed with tranformation
He may have implicit linear tranformation especially in this branch of forum.

arbolis, how do you think ?