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Matrices as Linear Tranformations (What is Linear Transformation…
Matrices as Linear Tranformations
What is Linear Transformation
Transformation is a fancy word for function
f(x), it takes in an input and spits out output
In context of LA, we like to think of transformation that takes in vector, and spits out another vector
Why then do we use the word "Transformation", instead of function if they mean the same thing?
One reason why we use the term "Transformation", is to suggest movement
E.g. If a transformation take some input vector to some output vector, we imagine the input vector moving to the output vector
Then to understand the transformation as a whole, we might imagine watching every input vector over to its corresponding output vector
So instead of watching this transformation as arrow (output), we rather see the output vector as points on plane
And we watch every point moving to some other point...in space
Transformations in 2D
Nice to keep a copy of the old grid in background
Limitation: LA transformation limits itself to a special type of transformation:
All Lines must remain lines, without getting curved
The origin must remain fixed in place
Gridlines MUST be parallel and evenly spaced
THE MOST IMPORTANT TAKEAWAY IN LA:
The idea of a Linear Transformation and its relation to Matrices
We going to Matrix multiplication that does not rely on memorization
Not a LA Transformation:
As the lines are all curvy
Also not a LA:
As it moves the origin
How can we describe these animations numerically
What formula can you give a computer
So that you give it the coordinates of a vector, it can give you the coordinates of where the vector lands?
The answer is that, you only need to record where the two basis vectors i(hat) and j(hat) lands
Consider the following:
\( \begin{pmatrix} -1 \\ 2 \\ \end{pmatrix} \) \
Meaning: v = -1i + 2j
If we place some transformation now, and place where all three these vectors go
The property that all he gridlines remains parallel and evenly spaced has a really important consequence
So i(hat) will still be -1 (move to -1 on the new grid), and j(hat) will be 2 times
So we can deduce where v must, base solely on where i(hat) and j(hat) landed
This is also the reason why we keep an old copy of the original grid in backend
And new coordinates for i(hat):
\( \begin{pmatrix} 1 \\ -2 \\ \end{pmatrix} \)
And new condinates for j(hat):
\( \begin{pmatrix} 3 \\ 0 \\ \end{pmatrix} \)
= -1 \( \begin{pmatrix} 1 \\ -2 \\ \end{pmatrix} \) + 2 \( \begin{pmatrix} 3 \\ 0 \\ \end{pmatrix} \)
= \( \begin{pmatrix} 5 \\ 2 \\ \end{pmatrix} \)
This is amazing techique we can use to deduce where any vectors land, as long as we know record (old plane copy) of where i(hat) and j(hat) land
Another way of showing it:
If any vector is given, using this formula, I can tell where any vector lands as a result
What this means that any 2D Linear Transformation can be described by just 4 numbers
The two coordinates where i(hat) lands
And the two coordinates where j(hat) lands
And we package these coordinates into a 2 x 2 matrix
Where the columns are the two special vectors where i(hat) and j(hat) lands
So now we given
\( \begin{pmatrix} 3 \ 2 \\ -1 \ 1 \\ \end{pmatrix} \)
And \( \begin{pmatrix} 5 \\ 7 \\ \end{pmatrix} \)
And we want to know where Linear Transformation takes that vector,
We can take the coordinates of the vector (5 and 7), and multiply them with the corresponding columns of matrix , and add together what you get
= 5 \( \begin{pmatrix} 3 \\ -2 \\ \end{pmatrix} \) + 7 \( \begin{pmatrix} 2 \\ 1 \\ \end{pmatrix} \)
Always interpret the first column (a, c) as the place where the first basis vector lands and the second column (b,d) as where the second basis vector lands
So when we apply this vector to some vector (x,y) what do you get?
It will be the following:
Will will result in further:
This is how students are thought to memories it:
By writing it as a function
And show them the crucial part that makes it feel intuitive
Isn't it better to think of these vectors rather as the "Transformed" versions of your basis vectors
And to think of the result as the appropriate linear combination of those vectors
