Linear Transformations

Definition of a linear transformation.

A linear transformation from a vector space $V$ to a vector space $W$ (over $K$) is a function $T:V→W$ such that for all $u,v\in V$ and all $\lambda \in K$

• $$T(u+v)=T(u)+T(v)$$

• $$T(\lambda u)=\lambda T(u)$$

What’s the kernel of a linear transformation?

The Kernel of a linear transformation is the set of all vectors in V that are mapped to 0 by $T$.

$$Ker(T)=\left\lbrace v\in V\vert T\left(v\right)=0\right\rbrace\subseteq V$$

What’s the image of a linear transformation?

The image of T as a function.

$Im(T)=\left\lbrace w\in W\vert w=T\left(v\right)\right.$ for some $v\in V\rbrace$

Definition of nullity and rank?

dim(Ker(T)) is called the nullity of T, written $nullity(T)$

dim(Im(T)) is called the rank of T, written $rank(T)$.

These must add up to the total dimension of the vector space.

When is a linear transformation injective?

If and only if $Ker(T)={0}$

If $T:U→V$ and $S:V→W$ are linear transformations what can we say about their composition?

It is also a linear transformation

What does it mean for a linear transformation to be an isomorphism?

It is a bijection. We say two vector spaces $V$ and $W$ are isomorphic, denoted by $V\cong W$ is there exists an isomorphism $T:V→W$

Is the inverse of a linear transformation also a linear transformation?

Yes

What’s the matrix of T with respect to B and C?

Let $V$and $W$ be fin-dim v.s over K with bases $B$ and $C$ respectively and let $T:V→W$ be a linear transformation. The $m\times n$ matrix whose ith column is the co-ordinate vector $[T(v_i)]_c$ is denoted by $_B[T]_c$ is called the matrix of T with respect to B and C.

What does the matrix of a linear transformation do to coordinate vectors $[v]_B$

Let $V$ and $W$ be finite-dimensional vector spaces with bases $\mathcal{B}$ and $\mathcal{C}$, and let $T : V \to W$ be a linear transformation. For all $v \in V$:

$$_{\mathcal{B}}[T]_{\mathcal{C}} \, [v]_{\mathcal{B}} = [T(v)]_{\mathcal{C}}$$

For a basis $B=\{v_1,\ldots,v_{n}\}$ what is $[v_i]_B$

$$e_i$$

How does the matrix of a composition of linear transformations ($_{B}[S\circ T]_{D}$ ) relate to matrix multiplication?

Let $T : U \to V$ and $S : V \to W$ be linear transformations. Then:

$$_{\mathcal{B}}[S \circ T]_{\mathcal{D}} = {_{\mathcal{C}}[S]_{\mathcal{D}}} \, {_{\mathcal{B}}[T]_{\mathcal{C}}}$$

Proof: Let $\mathcal{B} = \{v_1, \ldots, v_n\}$ and $\{e_1, \ldots, e_n\}$ be the standard basis for $K^n$. For any $i \in \{1, \ldots, n\}$:

$$_{\mathcal{B}}[S \circ T]_{\mathcal{D}} \, e_i = {_{\mathcal{B}}[S \circ T]_{\mathcal{D}}} \, [v_i]_{\mathcal{B}}$$

$$=[(S\circ T)(v_{i})]_{\mathcal{D}}$$

$$= [S(T(v_i))]_{\mathcal{D}}$$

$={_{\mathcal{C}}[S]_{\mathcal{D}}}\,[T(v_{i})]_{\mathcal{C}}\quad$ $= {_{\mathcal{C}}[S]_{\mathcal{D}}} \, {_{\mathcal{B}}[T]_{\mathcal{C}}} \, [v_i]_{\mathcal{B}} \quad$

$$= {_{\mathcal{C}}[S]_{\mathcal{D}}} \, {_{\mathcal{B}}[T]_{\mathcal{C}}} \, e_i$$

Both matrices have the same columns, so they are equal.

When is a linear transformation an isomorphism (relating to the change of basis matrix), and how does this relate to its matrix?

With bases $\mathcal{B}$ and $\mathcal{C}$, let $T : V \to W$ be a linear transformation. Then $T$ is an isomorphism if and only if $_{\mathcal{B}}[T]_{\mathcal{C}}$ is invertible. In this case:

$$_{\mathcal{C}}[T^{-1}]_{\mathcal{B}} = \left({_{\mathcal{B}}[T]_{\mathcal{C}}}\right)^{-1}$$

Proof:

Forward direction: suppose $T$ is invertible.

Then $T^{-1}\circ T=I_V$ and $T\circ T^{-1}=I_W$. Using the matrix-of-composition theorem, ${}_{\mathcal C}[T^{-1}]_{\mathcal B}\,{}_{\mathcal B}[T]_{\mathcal C}=I$ and ${}_{\mathcal B}[T]_{\mathcal C}\,{}_{\mathcal C}[T^{-1}]_{\mathcal B}=I$ .

Hence ${}_{\mathcal B}[T]_{\mathcal C}$ is invertible, with inverse ${}_{\mathcal C}[T^{-1}]_{\mathcal B}$.

Conversely, suppose $A={}_{\mathcal B}[T]_{\mathcal C}$ is invertible.

If $v\in\ker T$, then $A[v]_{\mathcal B}=[T(v)]_{\mathcal C}=0$, so $[v]_{\mathcal B}=0$ and hence $v=0$.

Thus $T$ is injective. For any $w\in W$, choose $v\in V$ with $[v]_{\mathcal B}=A^{-1}[w]_{\mathcal C}$.

Then $[T(v)]_{\mathcal C}=A[v]_{\mathcal B}=[w]_{\mathcal C}$, so $T(v)=w$. Thus $T$ is surjective.

Therefore $T$ is invertible.

What is a linear operator, and what notation is used for its matrix?

A linear operator is a linear transformation from a vector space $V$ to itself: $T : V \to V$. When $V$ is finite-dimensional with basis $\mathcal{B}$, we write:

$$[T]_{\mathcal{B}} = {_{\mathcal{B}}[T]_{\mathcal{B}}}$$

Note that $[T]_{\mathcal{B}} \in M_n(K)$ where $n = \dim(V)$.

How do the matrices of a linear operator with respect to two different bases relate?

Let $V$ be a finite-dimensional vector space with bases $\mathcal{B}$ and $\mathcal{C}$, and let $T : V \to V$ be a linear transformation. Then:

$$[T]_{\mathcal{B}} = P_{\mathcal{B} \to \mathcal{C}}^{-1} \, [T]_{\mathcal{C}} \, P_{\mathcal{B} \to \mathcal{C}}$$

where $P_{B →C}$ is the change of basis matrix from B to C.

How is the matrix of the identity map related to the change of basis matrix?

Let $I_V:V\to V$ be the identity map, and let $\mathcal B=\{v_1,\dots,v_n\}$ and $\mathcal C$ be bases of $V$. The matrix ${}_{\mathcal B}[I_V]_{\mathcal C}$ has columns $[v_i]_{\mathcal C}$, so it is exactly the change of basis matrix from $\mathcal B$ to $\mathcal C$: ${}_{\mathcal B}[I_V]_{\mathcal C}=P_{\mathcal B\to\mathcal C}$. In particular, ${}_{\mathcal B}[I_V]_{\mathcal B}=I_n$.