Composition of linear maps

Linear mappingsLinear mappings refers to the fundamental concept in linear algebra that involves functions or transformations that preserve the operations of addition and scalar multiplication., a fundamental concept in linear algebra, lay the groundwork for understanding transformations in mathematics and beyond. Among the powerful tools within linear algebra is the concept of composition, which allows us to combine multiple mappings to create new transformations.

Understanding linear mappings

In mathematics, a linear mapping between vector spaces VV and WW is a function (T:VW)( T: V \rightarrow W) that preserves vector addition and scalar multiplication. That is, for any vectors (v1,v2)( \mathbf{v}_1, \mathbf{v}_2) in (V)( V ) and any scalar (c)( c ), the following properties hold:

  1. (T(v1+v2)=T(v1)+T(v2))( T(\mathbf{v}_1 + \mathbf{v}_2) = T(\mathbf{v}_1) + T(\mathbf{v}_2) )

  2. (T(cv1)=cT(v1))( T(c \mathbf{v}_1) = c T(\mathbf{v}_1) )

When we compose linear mappings, we’re essentially applying one linear mapping followed by another.

Note: The expression (T:VW)( T: V \rightarrow W) represents a linear mapping TT from the vector space VV to the vector space WW. In other words, it signifies that TT is a function that takes vectors from the space VVas input and produces vectors in the space WWas output.

What is the composition of linear mappings?

Linear mappings, often represented by matrices, describe transformations of vector spaces. Composition of linear mappings involves applying one mapping followed by another, resulting in a new combined mapping. This process is akin to performing multiple transformations in succession.

Consider the process of assembling furniture, such as a bookshelf, from a flat-pack kit. Each step in the assembly instructions represents a linear mapping, guiding the transformation of individual components into the final product.

Understanding the process

Suppose we have two linear mappings, (T:VW)( T: V \rightarrow W ) and (S:WU)( S: W \rightarrow U ). The composition of these mappings, denoted (ST)( S \circ T ), is a new mapping from (V)( V ) to (U)( U ) defined as:

              (ST)(v)=S(T(v))(S∘T)(\mathbf{v})=S(T(\mathbf{v})), for all vectors vv in (V)( V )

The composition of two linear mappings.
The composition of two linear mappings.

In other words, we first apply the linear mapping (T)( T ) to the input vector (v)( \mathbf{v} ), and then apply the linear mapping (S)( S )to the result of (T(v))( T(\mathbf{v}) ).

Note: The composition of two linear mappings is itself a linear mapping. This is because applying a linear mapping preserves the properties of linearity, so composing linear mappings results in another linear mapping.

Let's look into examples to further understand the concept better!

Example 1

Let (T:R2R2)( T: \mathbb{R}^2 \rightarrow \mathbb{R}^2 ) be defined by (T(x,y)=(2x,y))( T(x, y) = (2x, y) ), and let (S:R2R2)( S: \mathbb{R}^2 \rightarrow \mathbb{R}^2 ) be defined by (S(x,y)=(x,y+3))( S(x, y) = (x, y + 3) ).

To find the composition (ST)( S \circ T ), we first apply (T)( T ) and then (S)( S ) to the result:

  1. (ST)(x,y)=S(T(x,y))(S \circ T)(x,y) =S(T(x,y))

  2. Since T(x,y)=2x,yT(x,y)= 2x,y, our new equation will be: (ST)(x,y)=S(T(x,y))=S(2x,y)(S \circ T)(x,y) =S(T(x,y)) = S(2x,y)

  3. Lastly, since S(x,y)=(x,y+3)S(x, y) = (x, y + 3) then S(2x,y)=2x,y+3S(2x,y) = 2x, y+3

So, the composition (ST)( S \circ T ) is the linear mapping that doubles the xx-coordinate and adds 3 to the yy-coordinate.

Example 2

Let (T:R3R2)( T: \mathbb{R}^3 \rightarrow \mathbb{R}^2 ) be defined by (T(x,y,z)=(x+y,2yz))( T(x, y, z) = (x + y, 2y - z)), and let (S:R2R3)( S: \mathbb{R}^2 \rightarrow \mathbb{R}^3) be defined by (S(u,v)=(uv,3u+v,2v))( S(u, v) = (u - v, 3u + v, 2v)).

To find the composition (ST)( S \circ T), we first apply (T)( T ) and then (S)( S ) to the result:

  1. (ST)(x,y,z)=S(T(x,y,z)(S \circ T)(x,y,z) =S(T(x,y,z)

  2. Since T(x,y,z)=(x+y,2yz)T(x,y,z) = (x+y, 2y-z), S(T(x,y,z))S(T(x,y,z)) can be rewritten as S(x+y,2yz)S(x+y,2y-z)

  3. Since S(u,v)=(uv,3u+v,2v)S(u, v) = (u - v, 3u + v, 2v), then: S(x+y,yz)=(((x+y)(2yz)),(3(x+y)(yz)),(2(yz)))S(x+y,y-z)= ( ((x+y)-(2y-z)), (3(x+y) - (y-z)), (2(y-z)) )
    We can solve the right hand side to get:
    S(x+y,yz)=(xy+z,3x+5yz,4y2z)S (x+y, y-z) = (x - y + z, 3x + 5y - z, 4y - 2z)

  4. As (ST)(x,y,z)=S(T(x,y,z))(S \circ T)(x,y,z) =S(T(x,y,z)), our final answer is:
    (ST)(x,y,z)=(xy+z,3x+5yz,4y2z)(S \circ T)(x,y,z) = (x - y + z, 3x + 5y - z, 4y - 2z)

So, the composition (ST)( S \circ T ) is a linear mapping from (R3)( \mathbb{R}^3 ) to (R3)( \mathbb{R}^3 ) that involves some linear combinations of the input coordinates.

Quiz

Test your knowledge by doing the quiz below!

Choose the correct answer.

1

What is the result of composing the linear mappings T:R3R2T:R^3→R^2 and S:R2R3S:R^2→R^3 given by the definitions: T(x,y,z)=(x+y,2yz)T(x,y,z)=(x+y,2y−z) and S(u,v)=(uv,3u+v,2v)S(u,v)=(u−v,3u+v,2v)?

A)

(x+3y−z, 5u+2v, 2v)

B)

(x−y, 3u+2v, 2v)

C)

(x+2y−z, 3u+3v, 2v)

D)

(x+y−z, 3u+3v, 2v)

Question 1 of 20 attempted

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