# Linear Algebra.Linear Transformation.Help

• March 5th 2009, 09:50 AM
ypatia
Linear Algebra.Linear Transformation.Help
Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.
• March 5th 2009, 11:25 AM
Inverse Transformation
Hello ypatia
Quote:

Originally Posted by ypatia
Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.

Do you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.

In the question you have here, the transformation can be written in matrix form as

$\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}$

So, to find the inverse transformation, you'll need the inverse of the matrix $\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}$, which is $\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}$

To find its effect on $(x, y, z)$, just do the matrix multiplication:

$\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}$

So there's your answer: $T^{-1}:(x,y,z) \rightarrow (z,y-z,x-y)$

• March 5th 2009, 11:43 AM
ypatia
Quote:

Hello ypatiaDo you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.

In the question you have here, the transformation can be written in matrix form as

$\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}$

So, to find the inverse transformation, you'll need the inverse of the matrix $\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}$, which is $\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}$

To find its effect on $(x, y, z)$, just do the matrix multiplication:

$\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}$

So there's your answer: $T^{-1}:(x,y,z) \rightarrow (z,y-z,x-y)$

Is this right?? "T is not injective and, as such, it does not have an inverse. For an example of the lack of injectivity, T(1,0,1) -->(2,1,1) and T(2,-1,1) -->(2,1,1)." Because if this is right contradicts to the above solution of T^-1. Thanks in advance
• March 5th 2009, 12:20 PM
Plato
The matrix is $
\left( {\begin{array}{*{20}c}
1 & 1 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\

\end{array} } \right)
$
which is singular.
• March 5th 2009, 01:14 PM
Transformation matrix
Quote:

Originally Posted by Plato
The matrix is $
\left( {\begin{array}{*{20}c}
1 & 1 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\

\end{array} } \right)
$
which is singular.

Sorry, yes. I mapped $(x,y,z)$ onto $(x+y+z, x+y, x)$, of course.