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Transformation Matrix

Petrus

Well-known member
Feb 21, 2013
739
T is a surjective linear transformation \(\displaystyle T: \mathbb{R^4}-> \mathbb{R^2}\). Decide dim ker T. How many free variables do I get if I solve equation system \(\displaystyle T(x)=y\) for a vector \(\displaystyle y \in \mathbb{R^2}\)? Construct a transformation matrix belonging to a surjective linear transformation \(\displaystyle T:\mathbb{R^4}->\mathbb{R^2}\)

My progres:
Dim ker T=\(\displaystyle 4-2=2\)
Dim ker T=free variables that mean we got 2 free variables
I'm stuck at transformation matrix

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited by a moderator:

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Let's keep it simple and try things we know. Consider the canonical basis for $\mathbb{R}^4$ and $\mathbb{R}^2$. Let's define $T: \mathbb{R}^4 \to \mathbb{R}^2$ with $T(1,0,0,0) = (1,0)$ and $T(0,1,0,0) = (0,1)$. Well, we've got the image cleared. Where else could we define the other vectors $(0,0,1,0), (0,0,0,1)$ to go? ;)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Well the matrix for $T$ will be a 2x4 matrix. Most 2x4 matrices you might come up with would do, as long as neither row was 0, and the second row was not a scalar multiple of the first. An easy one to come up with is this:

$[T] = \begin{bmatrix}1&0&0&0\\0&1&0&0 \end{bmatrix}$

which is the matrix relative to the standard bases for $\Bbb R^4$ and $\Bbb R^2$ for the surjective linear transformation:

$T(x_1,x_2,x_3,x_4) = (x_1,x_2)$
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks!
May I ask one thing did we Really need to know it is surjective? I mean what I know is that T got an right inverse. I still Dont understand WHY would I need to know it's surjective?

Regards,
\(\displaystyle |\pi\rangle\)
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
It is surjective because you have the linear transformation going from a bigger space to a smaller. If you had instead $T: \mathbb{R}^2 \to \mathbb{R}^4$ all you could have tops was an injective transformation, therefore you'd have a left inverse. ;)
 

Petrus

Well-known member
Feb 21, 2013
739
It is surjective because you have the linear transformation going from a bigger space to a smaller. If you had instead $T: \mathbb{R}^2 \to \mathbb{R}^4$ all you could have tops was an injective transformation, therefore you'd have a left inverse. ;)
Hmm I start to think that the injective transformation for transformation matrix \(\displaystyle T^{-1}\), but 2x4 Dont have determinant hence there is no inverse but is it correctly understand?

Edit:I keep forgeting it got right inverse but that right inverse is it that transformation matrix for
$T: \mathbb{R}^2 \to \mathbb{R}^4$
Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,910
T is a surjective linear transformation \(\displaystyle T: \mathbb{R^4}-> \mathbb{R^2}\). Decide dim ker T. How many free variables do I get if I solve equation system \(\displaystyle T(x)=y\) for a vector \(\displaystyle y \in \mathbb{R^2}\)? Construct a transformation matrix belonging to a surjective linear transformation \(\displaystyle T:\mathbb{R^4}->\mathbb{R^2}\)

My progres:
Dim ker T=\(\displaystyle 4-2=2\)
Dim ker T=free variables that mean we got 2 free variables
I'm stuck at transformation matrix

Regards,
\(\displaystyle |\pi\rangle\)
Since T is surjective (1,0) and (0,1) each must have at least 1 original.
Let's call those originals $\vec b_1$ respectively $\vec b_2$.
They will be independent of each other.
Now find 2 independent vectors in the kernel and let's call them $\vec b_3$ respectively $\vec b_4$.

Let's call B the matrix:
$$B=\begin{pmatrix}\vec b_1&\vec b_2&\vec b_3&\vec b4\end{pmatrix}$$
The matrix B is invertible because all vectors are independent.

Then:
$$TB= \begin{pmatrix}1&0&0&0 \\ 0&1&0&0\end{pmatrix}$$
And:
$$T=TBB^{-1}= \begin{pmatrix}1&0&0&0 \\ 0&1&0&0\end{pmatrix}B^{-1}$$