Intuition for Geometric Transformations

This might help,

Lets first look at what the matrix does to vectors. A useful way of looking at a matrix is to think of each column as being the result of applying the matrix to one of the basis vectors.

$$ \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{cc} 1 \\ 0 \end{array} \right] =\left[ \begin{array}{cc} a \\ c \end{array} \right] \qquad \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{cc} 0 \\ 1 \end{array} \right] =\left[ \begin{array}{cc} b \\ d \end{array} \right] $$

When you have more than one nonzero component for a column vector the matrix is applied to each piece and the results are added. In other words multiplying a matrix by a column vector adds columns of the matrix weighted by the components of the vector.

$$ \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{cc} 3 \\ 4 \end{array} \right] =\left[ \begin{array}{cc} 3a + 4b\\ 3c + 4d \end{array} \right] $$ Now lets use this to see what your matrix does to the (x,y) ordered pairs,

$$ \left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right] \left[ \begin{array}{cc} x \\ y \end{array} \right] =\left[ \begin{array}{cc} x \\ x+y \end{array} \right] \qquad \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right] \left[ \begin{array}{cc} x \\ y \end{array} \right] =\left[ \begin{array}{cc} x+y \\ y \end{array} \right]. $$

So you can see that when the first transformation is applied to a point it leaves the $x$ coordinate alone and then makes a new $y$ coordinate by adding the old $x$ and $y$ values. You can see this in your figure the shape is deformed vertically but not horizontally.

The second transformation does the same thing by with the roles of $x$ and $y$ reversed. Looking at this you can see that the $y$ coordinates do not change as a result of the transformation.

I hope this helps your intuition a bit. I'm not sure of you background but here are some general guidelines:

• Your matrix will always treat $(0,0)$ as a special point. So when you are looking at what the transformation does you should keep in mind that it won't treat all squares equally.

• When visualizing geometric transformations like this it is helpful to apply it to every point in the xy-grid. This gives you a new deformed grid which gives you an idea of what it does where. Think of this as stretching/compressing the xy plane.

• If your transformation is diagonalizable you should find its eigenvalues and eigenvectors. These provide invaluable information about the transformation.

Some helpful things to look for in a matrix:

The determinant of a matrix is equal to the (signed) area of the parallelogram you get by applying the matrix to a unit length square with its bottom left vertex at the origin. This parallelogram has its vertices at $\mathbf{0}$, $\mathbf{a}$, $\mathbf{b}$, $\mathbf{a}+\mathbf{b}$ where $\mathbf{a}$ and $\mathbf{b}$ are the vectors correspond to the first and second column of the matrix respectively.

The eigenvectors of a matrix correspond to lines through the origin on which vectors will remain after applying the matrix to them. The eigenvalues corresponding to each eigenvector tell you by what factor the distance between the origin and a vector on the line will be multiplied, as it is 'pushed' along the line (So for instance if $(1,1)$ is an eigenvector with corresponding eigenvalue equal to $-4$ then any vector on the line $\alpha(1,1)$ is sent to the same vector, four times further away from the origin, and on the other side of the origin. so $(2,2)$ will get sent to $(-8,-8)$ for examples.)

A matrix is fully determined by how it acts on basis vectors, so if you know where $(0,1)$ and $(1,0)$ are sent to by your matrix, then you can see how the matrix will act on any other vector by simply writing it as a linear combination of basis vectors.

Similarly, if a matrix is acting on the left of vectors, then the matrix will send the vector $(1,0)$ to the vector corresponding to the first column of the matrix, and it will send the vector $(0,1)$ to the vector corresponding to the second column of the matrix.