A linear operator on 2-space is a transformation of the vectors in 2-space which preserves sums and scalar multiples. Let's look at what "preserves" means in this context. Call the linear operator T, and write transform of any vector w in 2-space as T(w).

The same ideas work in 3-space as well, so to summarize: a linear operator is a transformation T of vectors in 2-space or 3-space which preserves sums and scalar multiples, i.e. which has the properties that

• for any two vectors u and v,
                T(u + v) = T(u) + T(v)

• for any vector u and any scalar c,
                T(cu) = cT(u).

An example of a linear operator in 2-space

Suppose that the transform of any vector xi + yj in 2-space is defined by the rule

T(xi + yj) = (x + y)i + 2xj.

So, for example, T(3i + 2j) = 5i + 4j., i.e. T transforms 3i + 2j into 5i + 4j. To show that T is indeed a linear operator, you have to check that it preserves all sums and all scalar multiples.

To show that T preserves all sums, take two generic 2-space vectors u = u1i + u2j and
v = v1i + v2j and check that the equation  T(u + v) = T(u) + T(v) holds. On the left-hand side (sum first and then transform), you get

T(u + v) = T( [u1i + u2j] + [v1i + v2j] )
             = T( [u1 + v1]i +[u2 + v2]j )
             = ( [u1 + v1] + [u2 + v2] )i + 2[u1 + v1]j.

On the right hand side (transform first, then sum), you get

T(u) + T(v) = T(u1i + u2j) + T(v1i + v2j)
                 = {[u1 + v1]i + 2u1j } + {[u2 + v2]i + 2v1j}
                 = ([u1 + v1] + [u2 + v2])i + [2u1 + 2v1]j.

You get the same answer for both sides, so the condition  T(u + v) = T(u) + T(v) holds.

To show that T preserves all scalar multiples, take a generic 2-space vector u = u1i + u2j and an arbitrary scalar c, and check that the equation T(cu) = cT(u) holds. On the left-hand side (first take the scalar multiple and then transform), you get

T(cu) = T(c[u1i + u2j])
         = T(cu1i + cu2j)
         = (cu1 + cu2)i + 2cu1j.

On the right-hand side (transform first, then take the scalar multiple), you get

cT(u) = cT(u1i + u2j)
        = c{(u1 + u2)i + 2u1j}
        = c(u1 + u2)i + 2cu1j

You got the same answer for both sides, so the condition T(cu) = cT(u) holds.

Since both conditions hold for any vectors and any scalars, T is a linear operator on 2-space.

An example of a transformation which is not a linear operator in 2-space

Suppose that the transform of any vector xi + yj in 2-space is defined by the rule

T(xi + yj) = y2i + xj.

To show that T is not a linear operator, you need to find a counterexample, i.e. you need to find an example for which at least one of the two conditions for linear operators fails for at least some vectors.

You can find a counterexample by "educated guessing". For example, the second component of T(xi + yj) is the same as in the previous example but the first has a square in it. It looks like if you try to transform a scalar multiple of a vector u = xi + yj, the scalar will come out squared, so the transformation won't be linear. Now you need find a specific counterexample that shows this problem. Try for a simple vector with a non-zero y-component (to get the y2 part), say u = j.

The left-hand side of the relation T(cu) = cT(u) is T(cj) = c2i, while the right-hand side is cT(j) = ci. For any c other than 0 or 1, c2 ≠ c and the two sides are not equal, so T is not a linear operator.

In this particular case, you could also have shown that T doesn't preserve sums for some pair of vectors.

In general, if you can't see a good counterexample easily, a useful strategy to find one is to try to show that T is a linear operator and look at why you can't show it for a hint.

Three other useful facts about linear operators.

Since they preserve sums and scalar multiples, linear operators also preserve quantities defined in terms of sums and scalar multiples.

Suppose T is a linear operator on 2-space or on 3-space. You can prove that T preserves the zero vector, negatives of vectors and differences of vectors by expressing each in terms of sums and scalar multiples.

1. T must preserve the zero vector: T(0) = 0.

Proof: Since 0 = 0u for any vector u, then

T(0) = T(0u) = 0T(u) = 0.

2. T must preserve negatives: T(–u) = – T(u).

Proof: Since -u = [-1]u for any vector u, then

T(–u) = T([–1]u) = [–1]T(u) = –T(u).

3. T must preserve differences: T(u – v) = T(u) – T(v).

Proof: Since u - v = u +[-v] for any vectors u and v, then

T(u – v) = T(u + [–v]) = T(u) + T(–v) = T(u) – T(v).

These facts are useful for recognizing when a transformations is not a linear operator. For example, the transformation T(xi + yj + zk) = (x + 1)i + yj + zk on 3-space is not a linear operator, since it takes the zero vector into i, not 0.

Since a linear operator preserves sums and scalar multiples, it preserves generalized sum-scalar multiple expressions, i.e. linear combinations. For example, if u, v and w are vectors in 3-space with u = 3v + 4w and T is a linear operator on 3-space, then

T(u) = T(3v + 4w)
       =T(3v) + T(4w)
       = 3T(v) + 4T(w).

In general, a linear operator transforms a linear combination of vectors into the same linear combination of the transforms of those vectors.

Another very important property of linear operators: a linear operator is completely determined by its effect on any basis. This means that if you know how a linear operator T transforms the vectors in some basis, you can find out how it transforms any vector, you just need to be able to express that vector in terms of the basis.

For example, if {b1, b2, b3} is a basis of 3-space and you know what T(b1), T(b2) and T(b3) are, then for any vector v = c1b1 + c2b2 + c3b3, you know what T(v) is:

T(v) = c1T(b1) + c2T(b2) + c3T(b3).