MA2.101 - Linear Algebra | Linear Algebra Lecture 7

Linear Transformation

A linear transformation is a function $f: U \rightarrow V$ for two vector spaces $U$ and $V$ defined over the field $F$, such that:

• $f(\alpha + \beta) = f(\alpha) + f(\beta)$
• $f(c \alpha) = c f(\alpha)$

where $\alpha, \beta \in U$, $c \in F$.

Theorem: If $U$ and $V$ are linear spaces on the same field then $f: U \rightarrow V$ is a linear mapping if and only if
\(f(a \cdot \bar\alpha + b \cdot \bar\beta) = a f(\bar\alpha) + b f(\bar\beta)\)

Proof: This is an if and only statement, therefore the proof has two parts, for vector spaces $U$ and $V$ and $\bar\alpha, \bar\beta \in U$ and $a, b \in F$:

1. $f:U \rightarrow V$ is a linear mapping $\Rightarrow$ $f(a\cdot \bar\alpha + b \bar\beta) = a \cdot f(\bar\alpha) + b \cdot f(\bar\beta)$
2. $f(a\cdot \bar\alpha + b \bar\beta) = a \cdot f(\bar\alpha) + b \cdot f(\bar\beta)$ $\Rightarrow$ $f: U \rightarrow B$ is a linear mapping.

The first part:
Given that $f: U \rightarrow V$ is a linear mapping, we get two stipulations:

1. $f(\bar\alpha + \bar\beta) = f(\bar\alpha) + f(\bar\beta)$ for arbitrary $\bar\alpha, \bar\beta \in U$
2. $f(c \cdot \bar\alpha) = c \cdot f(\bar\alpha)$ for $c \in F$ and $\bar\alpha \in U$

Now, let us consider $f(a \cdot \bar\alpha + b \cdot \bar\beta)$, by closure of scalar multiplication we know that $a \cdot \bar\alpha, b \cdot\bar\beta \in U$ thus by the first stipulation we can say \(f(a \cdot \bar\alpha + b \cdot \bar\beta) = f(a \cdot \bar\alpha) + f(b \cdot \bar\beta) \\\)

and using the second stipulation we get (for each term on the right):

Hence \(f(a \cdot \bar\alpha + b \cdot \bar\beta) = a \cdot f(\bar\alpha) + b \cdot f(\bar\beta)\)

The second part:
Given that

for arbitrary vectors in $U$ and arbitrary scalars in $F$
since $1 \in F$ we let $a = b = 1$, after making that substitution we get \(f(\bar\alpha + \bar\beta) = f(\bar\alpha) + f(\bar \beta)\)

for arbitrary vectors in $U$, this is the first stipulation required for a linear transformation.

since $0 \in F$, we let $b = 0$, after making this substitution in the given statement, we have

for abitrary vectors in $U$ and arbitrary scalars in $F$. We have obtained the second stipulation of a linear transformation

Having obtained the two required stipulations for linear transformation, we claim that $f$ is a linear transformation. Hence proved.

Theorem: If there is a set of linearly dependent vectors in $U$, then the set of the images of the elements of that set in $V$ will also be linearly dependent.

Proof:
Let the set in the domain be \(\{ \alpha_1, \alpha_2, \alpha_3, ... \alpha_n \}\), and we have corresponding scalars in $F$ \(\{ a_1, a_2, a_3, ..., a_n \}\) such that
\(a_1 \alpha_1 + a_2 \alpha_2 + a_3 \alpha_3 + ... + a_n \alpha_n = \phi\)
where not all $a_i$ are zero.

Taking both sides of the equation as the input for the function $f$, we have

Since at least one of the $a_i$’s must be non-zero, and all $f(\alpha_i) \in V$,
we can say that the set \(\{ f(\alpha_1), f(\alpha_2), f(\alpha_3), ... f(\alpha_n) \}\), is a linearly dependent set of vectors.