Topic 4.12 Notes – Linear Transformations and Matrices

How a linear map acts on the plane

A linear transformation $L:\mathbb R^2\to\mathbb R^2$ takes a vector like $(x,y)$ and sends it to another vector by a linear rule. In this topic, that rule has the form

$$L(x,y)=(ax+by,\;cx+dy)$$

That means each output piece is built from constant multiples of $x$ and $y$, then added together.

A vector has to be written as a column for matrix multiplication:

$$\begin{bmatrix}x\\y\end{bmatrix}$$

What counts as linear:

• $L(x,y)=(3x-2y,\;x+4y)$
• $L(x,y)=(-y,\;2x)$

What does not count:

• added constants, like $(3x-2y+5,\;x+4y)$
• products, like $(xy,\;x+y)$
• powers, like $(x^2,\;y)$

One fast check is the zero-vector test. Every linear transformation sends

$$\mathbf 0=\begin{bmatrix}0\\0\end{bmatrix} \quad\text{to}\quad \mathbf 0$$

So if $L(\mathbf 0)\ne \mathbf 0$, the rule is definitely not linear.
Be careful, though. If $L(\mathbf 0)=\mathbf 0$, that does not prove linearity by itself.

A common trap is a translation:

$$T(x,y)=(x+h,\;y+k)$$

This is not linear unless $h=k=0$, because the origin moves.

The Matrix Form of a Linear Transformation

Every linear transformation here matches exactly one $2\times2$ matrix $A$ so that

$$L(\mathbf v)=A\mathbf v$$

If

$$L(x,y)=(ax+by,\;cx+dy)$$

then the matrix is

$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

So you can read the matrix directly from the coefficients of $x$ and $y$ in each output component.

• first output $ax+by$ gives the first row $[a\;\;b]$
• second output $cx+dy$ gives the second row $[c\;\;d]$

Example:

$$L(x,y)=(2x-3y,\;5x+y) \quad\Rightarrow\quad A=\begin{bmatrix}2&-3\\5&1\end{bmatrix}$$

This matrix form explains why linear transformations preserve structure:

• $L(\mathbf u+\mathbf v)=L(\mathbf u)+L(\mathbf v)$
• $L(k\mathbf u)=kL(\mathbf u)$
• $L(r\mathbf u+s\mathbf v)=rL(\mathbf u)+sL(\mathbf v)$

“Unique matrix” means one whole transformation rule gives one matrix. It does not mean one input-output pair is enough to find the matrix.

Transforming One Vector

Here’s the full move.

1. Write the vector as a $2\times1$ column.
2. Put the matrix on the left.
3. Multiply using row-by-column dot products.

Example:

$$A=\begin{bmatrix}3&-2\\1&4\end{bmatrix}, \qquad \mathbf v=\begin{bmatrix}2\\-1\end{bmatrix}$$

Then

$$A\mathbf v= \begin{bmatrix}3&-2\\1&4\end{bmatrix} \begin{bmatrix}2\\-1\end{bmatrix} = \begin{bmatrix}3(2)+(-2)(-1)\\1(2)+4(-1)\end{bmatrix} = \begin{bmatrix}8\\-2\end{bmatrix}$$

So the output vector is $\begin{bmatrix}8\\-2\end{bmatrix}$. On a graph, the transformation sends $\mathbf v$ from $(2,-1)$ to $(8,-2)$.

Quick checks:

• dimensions work because $(2\times2)(2\times1)=2\times1$
• the order is $A\mathbf v$, never $\mathbf vA$

Common mistakes:

• writing the vector as a row
• reversing the order
• multiplying entries straight across instead of using row-by-column
• giving a scalar instead of a vector

Transforming Several Vectors at Once

If you have $n$ vectors, put them into a $2\times n$ matrix by columns:

$$V=\begin{bmatrix}\mathbf v_1&\mathbf v_2&\cdots&\mathbf v_n\end{bmatrix}$$

Then multiplying $AV$ transforms all of them at once:

$$AV=\begin{bmatrix}A\mathbf v_1&A\mathbf v_2&\cdots&A\mathbf v_n\end{bmatrix}$$

So each output column matches the same input column.

Example setup:

$$V=\begin{bmatrix}2&0&-1\\-1&3&2\end{bmatrix}$$

This means the input vectors are $\begin{bmatrix}2\\-1\end{bmatrix}$, $\begin{bmatrix}0\\3\end{bmatrix}$, and $\begin{bmatrix}-1\\2\end{bmatrix}$.

Dimension check:

$$(2\times2)(2\times n)=2\times n$$

This is useful for transforming the vertices of a figure all at once.

Common mistakes:

• putting vectors in rows
• mixing entries from different columns
• matching an output with the wrong input column

What to Watch for on Problems

This topic is about finding output vectors from a given $2\times2$ matrix and either one vector or a matrix of vectors.

It does not yet include finding a matrix from partial information, composing transformations, inverses, or classifying geometry in detail.