# Matrix proofs

• October 30th 2008, 08:20 AM
Showcase_22
Matrix proofs
Quote:

Let $T : \Re \rightarrow \Re ^3$ be given by $T \begin{pmatrix}
{x}\\
{y}\\
{z}
\end{pmatrix}=\begin{pmatrix}
{x}\\
{y}\\
{0}
\end{pmatrix}$
. This is projection from 3-space onto the xy-plane. Show that T is a linear transformation. What is the matrix associated to T?
I'm not really sure how to prove this is a linear transformation. I couldn't find a definition for a linear transformation either which made the problem more confusing.

I also don't know what it means by "matrix associated to T". Is this all it wants:

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

?

Help would be appreciated!
• October 30th 2008, 08:22 AM
ThePerfectHacker
Quote:

Originally Posted by Showcase_22
I'm not really sure how to prove this is a linear transformation. I couldn't find a definition for a linear transformation either which made the problem more confusing.

I also don't know what it means by "matrix associated to T". Is this all it wants:

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

?

Help would be appreciated!

Yes! (Clapping)
Since $T: \mathbb{R}^3 \to \mathbb{R}^3$ is a linear transformation it means $T(\bold{x}) = A\bold{x}$.
Where $A$ is a $3\times 3$ matrix.
That is the associated matrix with $T$.
• October 30th 2008, 08:27 AM
Showcase_22
Awesome!

but how do I show it's a linear transformation?

Can I just explain that (x,y,0) is a line since it has only one dimension (length)?
• October 30th 2008, 08:29 AM
ThePerfectHacker
Quote:

Originally Posted by Showcase_22
but how do I show it's a linear transformation?

To show that $T$ is a linear transformation you need to show that: $T(\bold{x}+\bold{y}) = T(\bold{x}) + T(\bold{y})$ and $T(k\bold{x}) = kT(\bold{x})$.
Can you do those two steps?
• October 30th 2008, 08:42 AM
Showcase_22
yes I can (and have!)

Cheers! That's some homework I won't have to do at the weekend! =D