A linear transformation (also called a linear map) is a map between vector spaces that preserves the vector space operations. More precisely, if \(V\) and \(W\) are vector spaces, a map \(T:V\rightarrow W\) is called a linear transformation if

1. \(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) for any \(\vec{v},\vec{w} \in V\text{,}\) and

2. \(T(c\vec{v}) = cT(\vec{v})\) for any \(c \in \IR,\) and \(\vec{v} \in V\text{.}\)

Given a linear transformation \(T:V\to W\text{,}\) \(V\) is called the domain of \(T\) and \(W\) is called the co-domain of \(T\text{.}\)

Let \(T : \IR^3 \rightarrow \IR^2\) be given by

To show that \(T\) is a linear transformation, we must verify that \(T(\vec{v}+\vec{w}) = T(\vec{v})+T(\vec{w})\) by computing

Therefore \(T\) is a linear transformation.

Let \(S : \IR^2 \rightarrow \IR^4\) be given by

To show that \(S\) is not linear, we only need to find one counterexample.

Since the resulting vectors are different, \(S\) is not a linear transformation.

A map between Euclidean spaces \(T:\IR^n\to\IR^m\) is linear exactly when every component of the output is a linear combination of the variables of \(\IR^n\text{.}\)

For example, the following map is definitely linear because \(x-z\) and \(3y\) are linear combinations of \(x,y,z\text{:}\)

But the map below is not linear because \(x^2\text{,}\) \(y+3\text{,}\) and \(y-2^x\) are not linear combinations (even though \(x+y\) is):

Let \(D:\P\to\P\) be the derivative map defined by \(D(f(x))=f'(x)\) for each polynomial \(f \in \P\text{.}\) We recall from calculus that

Which of the following can we conclude from these calculus rules?

1. \(\P\) is not a vector space

2. \(D\) is a linear map

3. \(D\) is not a linear map

Let the polynomial maps \(S: \P_4 \rightarrow \P_3\) and \(T: \P_4 \rightarrow \P_3\) be defined by

Compute \(S(x^4+x)\text{,}\) \(S(x^4)+S(x)\text{,}\) \(T(x^4+x)\text{,}\) and \(T(x^4)+T(x)\text{.}\) Based on these computations, can you conclude that either \(S\) or \(T\) is definitely not a linear transformation?

If \(L:V\to W\) is a linear transformation, then \(L(\vec z)=L(0\vec v)=0L(\vec v)=\vec z\) where \(\vec z\) is the additive identity of the vector spaces \(V,W\text{.}\)

Put another way, an easy way to prove that a map like \(T(f(x)) = f'(x)+x^3\) can not be linear is to check that

Showing \(T:V\to W\) is not a linear transformation can be done by finding an example for any one of the following.

• Show \(T(\vec z)\not=\vec z\) (where \(\vec z\) is the additive identity of \(V\) and \(W\)).

• Find \(\vec v,\vec w\in V\) such that \(T(\vec v+\vec w)\not=T(\vec v)+T(\vec w)\text{.}\)

• Find \(\vec v\in V\) and \(c\in \IR\) such that \(T(c\vec v)\not=cT(\vec v)\text{.}\)

Otherwise, \(T\) can be shown to be linear by proving the following in general.

• For all \(\vec v,\vec w\in V\text{,}\) \(T(\vec v+\vec w)=T(\vec v)+T(\vec w)\text{.}\)

• For all \(\vec v\in V\) and \(c\in \IR\text{,}\) \(T(c\vec v)=cT(\vec v)\text{.}\)

Note the similarities between this process and showing that a subset of a vector space is or is not a subspace.

Continue to consider \(S: \P_4 \rightarrow \P_3\) defined by

Part 1.

Verify that

Continue to consider \(S: \P_4 \rightarrow \P_3\) defined by

Part 2.

Verify that \(S(cf(x))\) is equal to \(cS(f(x))\) for all real numbers \(c\) and polynomials \(f \in \P_4\text{.}\)

Continue to consider \(S: \P_4 \rightarrow \P_3\) defined by

Part 3.

Is \(S\) linear?

Let polynomial maps \(S: \P \rightarrow \P\) and \(T: \P \rightarrow \P\) be defined by

Part 1.

Note that \(S(0)=0\) and \(T(0)=0\text{.}\) So instead, show that \(S(x+1)\not= S(x)+S(1)\) to verify that \(S\) is not linear.

Let polynomial maps \(S: \P \rightarrow \P\) and \(T: \P \rightarrow \P\) be defined by

Part 2.

Prove that \(T\) is linear by verifying that \(T(f(x)+g(x))=T(f(x))+T(g(x))\) and \(T(cf(x))=cT(f(x))\text{.}\)