# Math Insight

### Linear transformations

A linear transformation (or a linear map) is a function $\vc{T}: \R^n \to \R^m$ that satisfies the following properties:

1. $\vc{T}(\vc{x}+\vc{y}) = \vc{T}(\vc{x}) +\vc{T}(\vc{y})$
2. $\vc{T}(a\vc{x}) = a\vc{T}(\vc{x})$

for any vectors $\vc{x}, \vc{y} \in \R^n$ and any scalar $a \in \R$.

It is simple enough to identify whether or not a given function $\vc{f}(\vc{x})$ is a linear transformation. Just look at each term of each component of $\vc{f}(\vc{x})$. If each of these terms is a number times one of the components of $\vc{x}$, then $\vc{f}$ is a linear transformation.

Therefore, the function $$\vc{f}(x,y,z) = (3x-y, 3z, 0, z-2x)$$ is a linear transformation, while neither $$\vc{g}(x,y,z) = (3x-y, 3z+2, 0,z-2x)$$ nor $$\vc{h}(x,y,z) = (3x-y, 3xz, 0,z-2x)$$ are linear transformations. The function $\vc{h}$ has a nonlinear component $3xz$ that disqualifies it. What about the function $\vc{g}$? It's the second component $3z+2$ that's the problem because the term 2 is a constant that doesn't contain any components of our input vector $(x,y,z)$.

It's easy to see that the function $\vc{g}$ violates the second condition above. In particular, if you set $a=0$ in that second condition, you see that each linear transformation must satisfy $$\vc{T}(\vc{0}) = \vc{0}$$ but $\vc{g}(0,0,0) = (0,2,0,0)$. The condition for a linear transformation is stronger than the condition one learns in grade school for a function whose graph is a line. A single variable function $f(x)=ax+b$ is not a linear transformation unless its y-intercept $b$ is zero.

A useful feature of a feature of a linear transformation is that there is a one-to-one correspondence between matrices and linear transformations, based on matrix vector multiplication. So, we can talk without ambiguity of the matrix associated with a linear transformation $\vc{T}(\vc{x})$.