Transformation Definition and Rank-Nullity Theorem

What is a linear map? What is a linear transformation? In this article, I cover the image and rank of a linear transformation and the rank-nullity theorem. In doing so, I work through several examples.

Introduction to Linear Maps

We begin with the definition of a linear transformation.

Definition. Let $V$ and $W$ be linear spaces. A function $T$ from $V$ to $W$ is called a linear map if $$T(f+g)=T(f)+T(g) \qquad \text{and}\qquad T(k f)=k T(f) $$ for all elements $f$ and $g$ of $V$ and for all scalars $k.$

The collection of all linear maps from $V$ to $W$ is denoted by $\mathcal{L}(V,W)$ and if $T$ is a linear map from $V$ to $W$ we denote this by $T\in \mathcal{L}(V,W).$

Definition. Let $T\in \mathcal{L}(V,W).$

(1) Then the kernel of $T$ is the subset of $V$ consisting of the vectors that $T$ maps to 0; that is $$ \ker(T)=\{v\in V \, | \, T v=0\}. $

(2) Then the image of $T$ is the subset of $W$ consisting of the vectors of the form $Tv$ for some $v\in V$; that is $$ \mathop{im}(T)=\{w\in W \, | \, \text{ there exists } v \in V \text{ such that } w=T v\}.$$

Example. The function from $C^{\infty}$ to $C^{\infty}$ defined by $T(f)=f”$ is a linear map. Find the kernel and image of $T.$

Example. The function from $C^{\infty}$ to $C^{\infty}$ defined by $T(f)=\int_0^1 f(x) \, dx $ is a linear map. Find the kernel and image of $T.$

Theorem. If $ T \in \mathcal{L}(V,W) $, then null $T$ is a subspace of $V.$

Proof. By definition, $\mathop{ker} T=\{v\in V \mid T v=0\}$ and so $\mathop{ker} T \subseteq W.$ Let $u, v\in \mathop{ker} T.$ Then, $T(u+v)=T u+ Tv=0+0=0$ which shows, $u+v\in \mathop{ker} T$, for all $u , v \in \mathop{ker} T.$ Let $k$ be a scalar and $u\in \mathop{ker} T.$ Then $T(k u)=k T u= k(0)=0$, which shows, $k u\in \mathop{ker} T$ for all scalars $k$ and all $u\in \mathop{ker} T.$ Since $T(0)=T(0+0)=T(0)+T(0)$, $T(0)=0$ in $W$ which shows, $0\in \mathop{ker} T.$ Therefore, $\mathop{ker} T$ is a subspace of $W.$

Definition. A linear map $T: V\rightarrow W$ is called injective whenever $u, v\in V$ and $T u=T v$, we have $u=v.$

Theorem. Let $T\in \mathcal{L}(V,W)$, then $T$ is injective if and only if null $T={0}.$

Proof. Suppose $T$ is injective. Since $T(0)=0$, $0\in \mathop{ker} T$ and so ${0}\subseteq \mathop{ker} T.$ Let $v\in \mathop{ker} T.$ Then $T(0)=0=T(v)$ yields $v=0$ because $T$ is injective. Thus, $\mathop{ker} T\subseteq {0}.$ Therefore, $\mathop{ker} T={0}.$ Conversely, assume $\mathop{ker} T={0}.$ Let $u, v\in V.$ If $Tu=Tv$, then $T(u-v)=Tu – T v=0$ which shows $u-v \in \mathop{ker} T.$ Thus $u=v$ and therefore, $T$ is injective.

Definition. For $T\in \mathcal{L}(V,W)$, the image of $T$, denoted by $\mathop{im}(T)$, is the subset of $W$ consisting of those vectors that are of the form $T v$ for some $v\in V.$

Definition. A linear map $T: V\rightarrow W$ is called surjective if its range equals $W.$

Theorem. If $T\in \mathcal{L}(V,W)$, then $\mathop{im} T$ is a subspace of $W.$

Proof. By definition, $\mathop{im} T=\{T v \mid v\in V\}\subseteq W.$ Let $w_1,w_2\in W \in \mathop{im} T.$ Then there exists $v_1,v_2\in V$ such that $w_1=Tv_1$ and $w_2=T v_2.$ By linearity of $T$, $w_1+w_2=Tv_1+Tv_2=T(v_1+v_2)$ which shows $w_1+w_2\in \mathop{im} T$ for all $w_1,w_2 \in \mathop{im} T.$ Let $w\in \mathop{im} T$ and let $k$ be a scalar. Then there exists $v\in V$ such that $w=T v.$ By linearity of $T$, $ kw =k Tv =T(k v)$ which shows $kw\in \mathop{im} T$ for all $w\in \mathop{im} T$ and for all scalars $k.$ Therefore, $\mathop{im} T$ is a subspace of $W.$

Theorem. (Rank-Nullity Theorem) If $V$ is finite-dimensional and $T\in \mathcal{L}(V,W)$, then range $T$ is finite-dimensional and $\mathop{dim} V= \mathop{dim} (\mathop{im} T) + \mathop{dim} (\mathop{ker} T).$

Proof. Since $V$ is a finite-dimensional and $\mathop{ker} T$ is a subspace of $V$, $\mathop{ker} T$ is finite-dimensional and so let $(u_1,\ldots,u_n)$ be a basis of $\mathop{ker} T.$ Since $(u_1,\ldots,u_n)$ is linearly independent in $V$, it can be extended to a basis of $V$, say $\mathcal{B}=(u_1,\ldots,u_n,v_1,\ldots,v_m).$ It suffices to show $(T v_1,\ldots, T v_m)$ is a basis of $\mathop{im} T$, for then $\mathop{im} T$ is finite-dimensional is proven and $\mathop{dim} V=n+m=\mathop{dim} \mathop{ker} T+\mathop{dim} \mathop{im} T$ holds as well.

Let $w\in \mathop{im} T.$ Then there exists $v\in V$ and scalars $a_1,\ldots,a_n, b_1,\ldots,b_m$ such that $$ w=Tv=T(a_1 u_1+\cdots + a_n u_n+b_1 v_1+\cdots + b_m v_m)=b_1 T v_1+\cdots + b_m T v_m. $$ Thus $(T v_1,\ldots,T v_m)$ spans $\mathop{im} T.$ Suppose $c_1,\ldots,c_m$ are scalars such that $c T v_1+\cdots + c_m T v_m=0.$ Then there exist scalars $d_1,\ldots,d_n$ such that $$c_1 v_1+\cdots + c_m v_m=d_1 u_1+\cdots +d_n u_n. $$ Since $\mathcal{B}$ is a basis of $V$ and $$c_1 v_1+\cdots c_m v_m+(-d_1) u_1 + \cdots +(-d_n)u_n=0$$ it follows $$c_1=\cdots = c_m = d_1 = \cdots = \cdots =d_n=0.$$ In particular, $c_1=\cdots = c_m=0$ shows $T v_1,\ldots, T v_m$ are linearly independent. Therefore, $(T v_1,\ldots, T v_m)$ is a basis for $\mathop{im} T.$

Corollary. If $V$ and $W$ are finite-dimensional vector spaces with $T\in \mathcal{L}(V,W)$, then $\mathop{dim} V > \mathop{dim} W \implies T$ is not injective, and $\mathop{dim} V < \mathop{dim} W \implies T$ is not surjective.

Proof. By the Rank-Nullity Theorem, $$\mathop{dim} V=\mathop{dim} \mathop{ker} T+ \mathop{dim} \mathop{im} T \leq \mathop{dim} \mathop{ker} T +\mathop{dim} W.$$ Since $\mathop{dim} V > \mathop{dim} W $, it follows $$0<\mathop{dim} V-\mathop{dim} W\leq \mathop{dim} \mathop{ker} T. $$ Thus, $\mathop{ker} T \neq {0}$ and so $T$ is not injective.

For the second statement, again by the Rank-Nullity Theorem, $\mathop{dim} V=\mathop{dim} \mathop{ker} T+ \mathop{dim} \mathop{im} T$ and so $\mathop{dim} \mathop{im} T\leq \mathop{dim} V.$ Since $\mathop{dim} V < \mathop{dim} W $, $0<\mathop{dim} W-\mathop{dim} V\leq \mathop{dim} W-\mathop{dim} \mathop{im} T$ and so there exists a nonzero element $w\in W$ with $w\in \mathop{im} T.$ Therefore, $T$ is not surjective.

Example. Suppose that $T$ is a linear map from $V$ to $\mathbb{F}.$ Prove that if $u \in V$ is not in $\mathop{ker} T$, then $V=\mathop{ker} T \, \oplus \{a u:a\in F\}.$ Let $U=\{a u \mid a\in \mathbb{F}\}.$ The following arguments show $V=\mathop{ker} T +U$ and $\mathop{ker} T \cap U={0}$, respectively. Let $v\in V$ with $Tv=b.$ Since $Tu\neq 0$ there is a $u_1 \in U$ such that $T u_1=b.$ Then we can write $v=u_1+(v-u_1)$, and $v-u_1\in \mathop{ker} T.$ This gives $V=\mathop{ker} T+ U.$ Let $v\in \mathop{ker} T \cap U$, then there exists $a \in \mathbb{F}$ such that $v=a u$ and so $T(v)=a T u=0$ since $T v\in \mathop{ker} T.$ Thus $a=0$ and so $v=0$ meaning $\mathop{ker} T \cap U\subseteq {0}.$ Since $0 \in \mathop{ker} T \cap U$, it follows $\mathop{ker} T \cap U= {0}.$

Example. Suppose that $T\in \mathcal{L}(U,W)$ is injective and $(v_1,\ldots,v_n)$ is linearly independent in $V.$ Prove that $(T v_1,\ldots,T v_n)$ is linearly independent in $W.$

Suppose $a_1 T v_1+ \cdots a_n T v_n=0$ in $W$ where $a_1,\ldots,a_n \in \mathbb{F}.$ Then by linearity of $T$, $T(a_1 v_1 + \cdots + a_n v_n)=0.$ Since $T(0)$ and $T$ is injective, $a_1 v_1+\cdots + a_n v_n=0.$ Since $(v_1,\ldots,v_n)$ is linearly independent $a_1=\cdots = a_n =0.$ Therefore, $(T v_1,\ldots, T v_n)$ is linearly independent.

Example. Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\mathop{dim} V=1$ and $T\in \mathcal{L}(V,V)$, then there exists $a\in \mathbb{F}$ such that $T v=a v$ for all $v\in V.$

Let ${w}$ be a basis of $V$, and let $v\in V.$ Then there exists $c\in \mathbb{F}$ such that $v=c w.$ Applying $T$ yields, $$ T v=T(c w)=cTw=c(a w)=(ca)w=a(cw)=a v $$ where $Tw=aw$ since $Tw\in V$ and ${w}$ is a basis of $V.$

Example. Suppose that $V$ is finite-dimensional. Prove that any linear map on a subspace of $V$ can be extended to a linear map on $V.$ In other words, show that if $U$ is a subspace of $V$ and $S\in \mathcal{L}(U,W)$, then there exists $T\in \mathcal{L}(V,W)$ such that $Tu=Su$ for all $u\in U.$

Let $(u_1,\ldots,u_n)$ be a basis of $U$ and extend this basis of $U$ to a basis of $V$, say $$(u_1,\ldots,u_n, v_1,\ldots,v_m).$$ Define $T$ as the linear extension of $S$, as follows $$ T(u_i)=S(u_i) \text{ for } 1\leq i \leq n \hspace{.5cm} \text{ and } \hspace{.5cm} T(v_j)=v_j \text{ for } 1\leq j \leq m. $$ Then for all $u\in U$, \begin{align} T(u) & =T(a_1 u_1+\cdots + a_n u_n) =a_1 T u_1+\cdots a_n T u_n \\ & =a_1 S u_1+\cdots +a_n S u_n =S(a_1u_1+\cdots a_n u_n) =S(u) \end{align} where $a_1,\ldots,a_n\in \mathbb{F}.$ By definition of $T$, $T\in \mathcal{L}(V,W).

Example. Prove that if $S_1,\ldots,S_n$ are injective linear maps such that $S_1 \cdots S_n$ makes sense, then $S_1 \cdots S_n$ is injective.

Suppose $v$ and $w$ are any vectors and $(S_1\cdots S_n)v=(S_1\cdots S_n)w.$ Then by definition of composition, $$S_1(S_2\cdots S_n)v=S_1(S_2\cdots S_n)w,$$ and since $S_1$ is injective $$S_2(S_3\cdots S_n)v=S_2(S_3\cdots S_n)w.$$ Since $S_2,\ldots,S_n$ are all injective $v=w$ as desired showing $S_1\cdots S_n$ is injective.

Example. Prove that if $(v_1,\ldots,v_n)$ spans $V$ and $T\in \mathcal{L}(V,W)$ is surjective, then $(T v_1,\ldots,T v_n)$ spans $W.$

Let $w\in W.$ Then, since $T$ is surjective, there exists $v\in V$ such that $T(v)=w.$ Since $(v_1,\ldots,v_n)$ spans $V$ there exists scalars $a_1,\ldots,a_n$ such that $v=a_1 v_1+ \dots a_n v_n$ and by linearity of $T$, $T(v)=a_1 T v_1+ \cdots a_n T v_n=w.$ Therefore, $(T v_1,\ldots,T v_n)$ spans $W$ because every element $w$ in $W$ is a linear combination of the $T v$’s.

Example. Suppose that $V$ is finite-dimensional and that $T\in \mathcal{L}(V,W).$ Prove that there exists a subspace $U$ of $V$ such that $U\cap \mathop{ker} T={0}$ and range $T = \{T u : u\in U\}.$

Since $V$ is finite-dimensional so is $\mathop{ker} T.$ Let $(v_1,\ldots,v_n)$ be a basis of $\mathop{ker} T$ and extend this basis to a basis of $V$, namely let $$\mathcal{B}=(v_1,\ldots,v_n,u_1,\ldots ,u_m)$$ be a basis of $V.$ Let $U=\mathop{span} (u_1,\ldots,u_m)$, we will show $U\cap \mathop{ker} T=\{0\}$ and range $T = \{T u : u\in U\}.$ Clearly, ${0}\subseteq U\cap \mathop{ker} T$ since both $U$ and $\mathop{ker} T$ are subspaces. Let $v\in U\cap \mathop{ker} T.$ Then $v\in U$ implies there exists $a_1,\ldots,a_m$ such that $v=a_1 u_1+\cdots +a_m u_m.$ Then $v\in \mathop{ker} T$ implies there exists $b_1,\ldots,b_n$ such that $v=b_1 v_1+\cdots +b_n v_n.$ Then $a_1 u_1+\cdots +a_m u_m+(-b_1)v_1+\cdots +(-b_n)v_n=0$ and since $\mathcal{B}$ is a basis of $V$ the $u$’s and $v$’s are linearly independent and so $a_1=\cdots =a_m=b_1=\cdots =b_n=0$ meaning $v=0$; and so $\mathop{ker} T = {0}.$ Let $v\in V.$ Then $$ T(v)=a_1T v_1+\cdots +a_n T v_n+b_1 T u_1+\cdots + b_m T u_m=b_1 Tu_1+\cdots +b_m T u_m $$ showing $\{T v : v\in T\}\subseteq \{T v : v\in U\}$ conversely is obvious, since $U\subseteq V.$

Example. Prove that if there exists a linear map on $V$ whose null space and range are both finite-dimensional, then $V$ is finite-dimensional.

Example. Suppose that $V$ and $W$ are finite-dimensional and $T\in \mathcal{L}(V,W).$ Prove that there exists a surjective linear map from $V$ onto $W$ if and only if $\mathop{dim} W \leq \mathop{dim} V.$

Suppose $T$ is a linear map of $V$ onto $W$, then $\mathop{dim} V=\mathop{dim} \mathop{ker} T+\mathop{dim} \mathop{im} T.$ Since $\mathop{dim} \mathop{ker} T \geq 0$, $\mathop{dim} V\geq \mathop{dim} \mathop{im} T=\mathop{dim} W$, since $T$ is onto. Conversely, assume $m=\mathop{dim} W\leq \mathop{dim} V=n$ with bases of $W$ say $(w_1,\ldots,w_m)$ and of $V$ say $(v_1,\ldots,v_n).$ Define $T$ to be the linear extension of: $$ \begin{cases} T(v_i)=w_i & \text{ if } 1\leq i\leq m \\ T(v_i)=0 & \text{ if } i>m \end{cases} $$

Then $T$ is surjective since: if $w\in W$ then there exists $a_i\in \mathbb{F}$ such that $$ w=a_1 w_1+\cdots + a_m w_m=a_1 T(v_1)+\cdots + a_m T(v_m)=T(a_1 v_1+\cdots + a_m v_m)$$ showing every element in $W$ is in $\mathop{im} T$.

Example. Suppose that $W$ is finite-dimensional and $T\in \mathcal{L}(V,W).$ Prove that $T$ is injective if and only if there exists $S\in \mathcal{L}(W,V)$ such that $S T$ is the identity map on $V.$

Suppose $S\in\mathcal{L}(W,V)$ and $ST$ is the identity on $V.$ Then $\mathop{ker} T\subseteq \mathop{ker} ST={0}$ and so $\mathop{ker} T={0}.$ Therefore, $T$ is injective. Conversely, suppose $T$ is injective. So we can define $S_1\in\mathcal{L}(W,V)$ by the following: $S_1(w)=v$ where $v=T^{-1}(w)$ (since $T$ is injective). So $S_1:\mathop{im} T \rightarrow V$ and since $W$ is finite-dimensional $S_1$ can be extended to $S: W\rightarrow V$ and by definition of $S_1$, if $v\in V$, then $ST(v)=S(Tv)=T^{-1}(Tv)=v.$

Example. Suppose that $V$ is finite-dimensional and $T\in \mathcal{L}(V,W).$ Prove that $T$ is surjective if and only if there exists $S\in \mathcal{L}(W,V)$ such that $T S$ is the identity map on $W.$

We will present two proofs.

Suppose $S\in \mathcal{L}(W,V)$ and $TS=I_W.$ Let $w\in W.$ Then $v=S(w)\in V$ is such that $T(v)=TS(w)=w$ and therefore $T$ is surjective. Conversely, suppose $T$ is surjective. Since $V$ is a finite-dimensional vector space and $T$ is a linear map, $W$ is finite-dimensional. Let $(v_1,\ldots,v_n)$ be a basis for $V.$ Since $T$ is surjective, $(T v_1,\ldots,T v_n)$ spans $W.$ Also since $T$ is surjective $\mathop{dim} W\leq \mathop{dim} V=n$ Any spanning set reduced to a basis say $(T v_1,\ldots,T v_m)$ is a basis of $W$ where $m\leq n.$ Define $S$ as the linear extension of $S(T v_i)=v_i$ for each $1\leq i \leq m.$ Then, for all $w\in W$, $$ S(w)=S(a_1 T v_1+\cdots +a_m T v_m)=a_1 ST v_1+\cdots + a_m ST v_m=a_1 v_1+\cdots + a_m v_m=w$$ for scalars $a_1,\ldots,a_m\in \mathbb{F}$, and so $TS=I_W.$

Suppose $T$ is surjective. There exists a subspace $U$ of $V$ such that $U\cap \mathop{ker} T={0}$ and $\mathop{im} T={T u : u\in U}.$ Define $T_1: U\rightarrow W$ by $T_1 u=T u$ for $u\in U.$ Notice $T_1$ is injective and surjective and so $T_1$ has an inverse. Define $S=T_1^{-1}$ we have $TS w= T_1 T_1^{-1}w=w$ for all $w\in W.$

Definition. A linear map $T\in \mathcal{L}(V,W)$ is called invertible if there exists a linear map $S\in \mathcal{L}(W,V)$ such that $S T$ equals the identity map on $V$ and $TS$ equal the identity map on $W.$

Given $T\in\mathcal{L}(V,W).$ A linear map $S\in\mathcal{L}(W,V)$ satisfying $S T=I$ and $T S=I$ is called an inverse of $T.$ Two vector spaces are called isomorphic if there is an invertible linear map from one vector space onto the other one.

Theorem. If $\mathcal{B}=(f_1,\ldots,f_n)$ is a basis of a linear space $V$, then the coordinate transformation $L_\mathcal{B}(f)=[f]_\mathcal{B}$ from $V$ to $\mathbb{R}^n$ is an isomorphism. Thus any linear space $V$ is isomorphic to $\mathbb{R}^n.$

An invertible linear transformation $T$ is called an isomorphism. Recall the rank-nullity theorem states that if $V$ and $W$ are finite-dimensional vector spaces and $T$ is a linear transformation from $V$ to $W$ then $\mathop{dim} V=\mathop{dim} \ker T+\mathop{dim} \mathop{im} T$ where $\mathop{dim} \ker T$ is called the nullity and $\mathop{dim} \mathop{im} T$ is called the rank.

Theorem. If $V$ and $W$ are finite-dimensional linear spaces, then

(1) $T$ is an isomorphism from $V$ to $W$ if and only if $\ker(T)= {0}$ and $\mathop{im} (T)=W$,

(2) if $V$ is isomorphic to $W$, then $\mathop{dim} (V)=\mathop{dim} (W)$,

(3) if $\mathop{dim} (V)=\mathop{dim} (W)$ and $\ker(T)={0}$, then $T$ is an isomorphism, and

(4) if $\mathop{dim} (V)=\mathop{dim} (W)$ and $\mathop{im}(T)=W$, then $T$ is an isomorphism.

Proof. The proof of each part follows.

Suppose $T$ is an isomorphism from $V$ to $W.$ This means there exists an invertible linear transformation $T^{-1}$ from $W$ to $V$ such that $T(T^{-1})=I_W$ and $T^{-1}(T)=I_V$, where $I_W$ and $I_V$ are the identity transformations on $W$ and $V$ respectively. We will use $T^{-1}$ to show $\ker T={0}$ and $\mathop{im} T= W.$ Since $\ker T$ is a subspace, ${0}\subset \ker T.$ Conversely, suppose $f\in \ker T.$ Then $T(f)=0$ and so $T^{-1}(T(f))=f=0$ so that $\ker T\subseteq {0}.$ Thus $\ker T = {0}.$ Since $\mathop{im} T$ is a subspace of $W$, $\mathop{im} T\subseteq W.$ To show conversely, let $w\in W.$ Then $T^{-1}(w)\in V$ and so $T(T^{-1}w)=w$ shows $w\in \mathop{im} T.$ Thus $\mathop{im} T=W.$

Conversely, suppose $\ker T={0}$ and $\mathop{im} T=W.$ We will show $T$ is an isomorphism, which means we must show $T(f)=g$ has a unique solution $f$ for each $g\in W.$ Now since $\mathop{im} T=W$, $T(f)=g$ has at least one solution, say $f.$ Suppose $T(f)=g$ has two solutions, say $T(f_1)=g=T(f_2).$ Then $0=T(f_1)-T(f_2)=T(f_1-f_2)$ shows $f_1-f_2\in \ker T.$ Since $\ker T={0}$, $f_1=f_2$ must follow, which means $T(f)=g$ must have only one solution, so that $T^{-1}$ exists, and therefore, $T$ is an isomorphism.

If $V$ is isomorphic to $W$, then there exists a linear transformation $T$ such that $\ker T={0}$ and $\mathop{im} T=W.$ By the rank-nullity theorem, $\mathop{dim} V=\mathop{dim} \ker T + \mathop{dim} \mathop{im} T=0+\mathop{dim} W.$

By the previous part, $T$ is an isomorphism when $\ker ={0}$ and $\mathop{im} T=W.$ We are assuming $\ker T={0}$ so it remains to show $\mathop{im} T=W.$ By the rank-nullity theorem $\mathop{dim} V=\mathop{dim} \ker T+\mathop{dim} \mathop{im} T=\mathop{dim} \ker T+\mathop{dim} W=\mathop{dim} W.$ Thus, $\ker T=0$ and so $\ker T={0}.$ Therefore, $T$ is an isomorphism.

By the previous part, $T$ is an isomorphism when $\ker ={0}$ and $\mathop{im} T=W.$ We are assuming $\mathop{im} T=W$ so it remains to show $\ker T={o}.$ By the rank-nullity theorem $$\mathop{dim} V=\mathop{dim} \ker T+\mathop{dim} \mathop{im} T=\mathop{dim} \mathop{im} T=\mathop{dim} W.$$ Therefore, $\mathop{im} T=W$ and so $T$ is an isomorphism.

Example. Show that the transformation $T(A)=S^{-1}AS$, from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$ where $S=\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$ is an isomorphism.

Notice that $\mathop{dim} \mathbb{R}^{2\times 2}=\mathop{dim} \mathbb{R}^{2\times 2}.$ Can we determine an invertible linear transformation $T^{-1}$? Checking the linearity conditions,$$
T(A_1+A_2)=S^{-1}(A_1+A_2)S=S^{-1}A_1S+S^{-1}A_2S=T(A_1)+T(A_2) $$
and $$ T(k A)=S^{-1}(k A)S=kS^{-1}A S=k T(A).$$

Now we know $T$ is a linear transformation. Is it invertible? We need to solve the equation, $B=S^{-1} A S$ for input $A.$ We can do this because $S$ is an invertible matrix, so $S B=S(S^{-1}AS)=AS$ and multiplying by $S^{-1}$ on the right $S BS^{-1}=A$ so that $T$ is invertible, and the linear transformation is $T^{-1}(B)=S B S^{-1}.$

Example. Is the transformation $L( f)=\begin{bmatrix}f(1) \\ f(2) \\ f(3)\end{bmatrix}$ from $\mathcal{P}_3$ to $\mathbb{R}^3$ an isomorphism?

Since $\mathop{dim} \mathcal{P}_3=4$ and $\mathop{dim} \mathbb{R}^3=3$, the spaces $\mathcal{P}_3$ and $\mathbb{R}^3$ fail to be isomorphic, so that $L$ is not an isomorphism.

Example. Is the transformation $L( f)=\begin{bmatrix}f(1) \\ f(2) \\ f(3)\end{bmatrix}$ from $\mathcal{P}_2$ to $\mathbb{R}^3$ an isomorphism?

Notice $\mathop{dim} \mathcal{P}_2=3$ and $\mathop{dim} \mathbb{R}^3=3$ so the dimensions of the domain and the target space have the same dimension. Checking the linearity conditions, $$
L(f_1+f_2)=\begin{bmatrix}(f_1+f_2)(1) \\ (f_1+f_2)(2) \\ (f_1+f_2 (3)\end{bmatrix}=\begin{bmatrix} f_1(1) \\ f_1(2) \\ f_1(3) \end{bmatrix}+\begin{bmatrix} f_2(1) \\ f_2(2) \\ (f_2(3) \end{bmatrix} =L(f_1)+L(f_2) $$ and $$ T(k f)=\begin{bmatrix} (k f)(1) \\ (k f)(2) \\ (kf)(3)\end{bmatrix}=k\begin{bmatrix}f(1) \\ f(2) \\ f(3)\end{bmatrix}=k T(f).$$ The kernel of $T$ consists of all polynomials $f(x)$ in $\mathcal{P}_2$ such that $$T(f(x))=\begin{bmatrix}f(1) \\ f(2) \\ f(3)\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}.$$ Since a nonzero polynomial in $\mathcal{P}_2$ has at most two zeros, the zero polynomial is the only solution, so that $\ker T={0}.$ Therefore, $T$ is an isomorphism.

Example. Determine whether the transformation $$T(M)=M\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}M$$ from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^{2\times 2}$ is an isomorphism.

Notice that $\mathop{dim} \mathbb{R}^{2\times 2}=\mathop{dim} \mathbb{R}^{2\times 2}.$ Can we determine an invertible linear transformation $T^{-1}$? Checking the linearity conditions,

$$ \begin{array}{rl} T(M+N) &=(M+N)\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}(M+N) \\ &=M\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}M + N\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}N =T(M)+T(N) \end{array} $$

$$ \begin{array}{rl} T(k M) & =(kM)\begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}(Mk) \\ & =k \left ( M\begin{bmatrix} 2 & 0 \ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix}M \right ) = k T(M)\end{array} $$

The kernel of $T$ consists of $M=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ such that

$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} -a & 0 \\ -2c & -d \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}. $$

Thus $a=c=d=0.$ However, when $M=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ then $T(M)=0$ so that $\ker T \neq {0}$, and so $T$ is not an isomorphism.

Example. Determine whether the transformation $T(f(t))=\begin{bmatrix} f(1) \\ f'(2) \\ f(3) \end{bmatrix}$ from $\mathcal{P}_2$ to $\mathbb{R}^3$ is a linear transformation. Then determine whether $T$ is an isomorphism and if not then find its image and kernel.

This transformation is linear since $$ T(f(t)+g(t)) = \begin{bmatrix} f(1)+g(1) \\ f'(2)+g'(2) \\ f(3)+g(3) \end{bmatrix} = \begin{bmatrix} f(1) \\ f'(2) \\ f(3)\end{bmatrix} + \begin{bmatrix} g(1) \\ g'(2) \\ g(3)\end{bmatrix}= T(f(t))+T(g(t)) $$

and

$$ T(k f(t))=\begin{bmatrix} kf(1) \\ k f'(2) \\ k f(3) \end{bmatrix}= k \begin{bmatrix} f(1) \\ f'(2) \\ f(3) \end{bmatrix} = k T(f(t)).$$

Since $T(t-1)(t-3))=T(t^2-4t+3)= 0$ shows $\ker T \neq {0}$, T is not an isomorphism.

Theorem. A linear map is invertible if and only if it is injective and surjective.

Proof. Suppose $T\in \mathcal{L}(V,W)$ is invertible with inverse $T.$ Let $u,v\in V.$ If $Tu=Tv$ then $$u=(T^{-1})u)=T^{-1}(Tu)=T^{-1}(Tv)=(T^{-1}T)(v)=v $$ and so $T$ is injective. If $w\in W$, then $v=T^{-1} w\in V$ with $T v=T(T^{-1}w)=w$ shows $T$ is surjective. Assume $T$ is injective and surjective. For each $w\in W$ assign $T(v)=w.$ Such $S(w)=v$ exists because $T$ is surjective and is unique since $T$ is injective. Then $T(v)=w$ shows $ST(v)=S(w)=v$ so that $ST$ is the identity on $V.$ Also, $TS(w)=T(Sw)=Tv=w$ shows $TS$ is the identity on $W.$ Thus $S$ and $T$ are inverses.

Now $S$ is linear since:

  • if $w_1,w_2\in W$, then there exists a unique $v_1$ and $v_2$ such that $Tv_1=w_1$ and $Tv_2=w_2$, $S(w_1)=v_1$, and $S(w_2)=v_2.$ By linearity of $T$, $S(w_1+w_2)=S(T v_1+Tv_2)=(ST)(v_1+v_2)=v_1+v_2=S(w_1)+S(w_2).$
  • If $w\in W$ and $k\in \mathbb{F}$ then there exists a unique $v\in V$ such that $Tv=w$ and $S(w)=v.$ By linearity of $T$, $S(kw)=S(k Tv)=S(T k v)=k v=kS(w).$

Therefore $S$ is linear and is the inverse of $T.$

Consider the transformation $$ L \left( \, \begin{bmatrix} a & b & c & d \end{bmatrix} \, \right) = \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} $$ from $\mathbb{R}^{2\times 2}$ to $\mathbb{R}^4.$ Note that $L$ is the coordinate transformation $L_\mathcal{B}$ with respect to the basis

$$ \mathcal{B}= \left ( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right ) $$

Being a coordinate transformation, $L$ is both linear and invertible. Therefore $L$ is an isomorphism.

Theorem. Let $T$ be the linear transformation from $V$ to $V$ and let $B$ be the matrix of $T$ with respect to a basis $\mathcal{B}=(f_1,\ldots,f_n)$ of $V.$ Then $$ B= \begin{bmatrix}[T(f_1)]_{\mathcal{B}} & \cdots & [T(f_n)]_{\mathcal{B}}] \end{bmatrix} $$ In other words, the columns of $B$ are the $\mathcal{B}$-coordinate vectors of the transformation of the basis elements $f_1,\ldots,f_n$ of $V.$

Theorem. Two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

Proof. Suppose $V$ and $W$ are finite-dimensional vector spaces that are isomorphic. If $\mathop{dim} V > \mathop{dim} W$, then no linear map between $T$ and $W$ can be injective. Thus $V$ and $W$ are not isomorphic contrary to hypothesis. If $\mathop{dim} V < \mathop{dim} W$, then no linear map between $V$ and $W$ can be surjective. Thus $V$ and $W$ are not isomorphic contrary to hypothesis. Therefore, $\mathop{dim} V =\mathop{dim} W.$

Suppose $V$ and $W$ are vector spaces with $\mathop{dim} V=\mathop{dim} W=n.$ Let $(v_1,\ldots,v_n)$ ba a basis of $V$ and let $(w_1,\ldots,w_n)$ be a basis of $W.$ Define a linear map $T$ by $T(a_1 v_1+\cdots +a_n v_n)=a_1 w_1+\cdots +a_n w_n.$ It’s an easy exercise to show $T$ is linear, injective, and surjective. Thus $T$ is an isomorphism as required.

Theorem. If $A$ and $B$ are matrices of a linear transformation $T$ with respect to different bases, then $A$ and $B$ are similar matrices.

Theorem. Suppose $V$ and $W$ are finite-dimensional vector spaces, then $\mathcal{M}$ is an invertible linear map between $\mathcal{L}(V,W)$ and $Mat(m,n,\mathbb{F})$

Proof. Suppose $(v_1,\ldots,v_n)$ and $w_1,\ldots,w_m)$ are bases of $V$ and $W$ respectively. For each $T\in \mathcal{L}(V,W)$ we assign, $$ \mathcal{M}(T)=\begin{bmatrix}a_{11} & & a_{1n} \\ \vdots & \cdots & \vdots \\ a_{m1} & & a_{mn} \end{bmatrix}$$ where $T v_k=a_{1k} w_1+\cdots + a_{mk} w_m.$ Since $(w_1,\ldots,w_m)$ is a basis of $W$, This assignment is well-defined. The following arguments show $\mathcal{M}$ is linear, injective, and surjective, respectively.

  • If $S,T\in \mathcal{L}(V,W)$, then $\mathcal{M}(S+T)=\mathcal{M}(S)+\mathcal{M}(T)$ follows by the definition of matrix addition. If $S\in \mathcal{L}(V,W)$ and $k\mathbb{F}$, then $\mathcal{M}(kS)=k\mathcal{M}(S)$ follows by the definition of scalar multiplication of matrices.
  • If $T\in \mathcal{L}(V,W)$ with $\mathcal{M}(T)=0$, then for entries in $[a_{ij}]$ are zero showing $T v_k=0$ for each $k.$ Since $(v_1,\ldots,v_n)$ is a basis of $V$, $T$ must be the zero map, and so $\mathop{ker} \mathcal{M}={0}.$ Therefore, $\mathcal{M}$ is injective.
  • If $A \in Mat(m,n, \mathbb{F})$ then $A$ is an $m\times n$ matrix with entries $[a_{ij}].$ We define $Tv_k=a_{1k} w_1+\cdots a_{mk} w_m$ for each $k.$ Then, by the definition of the matrix of a linear map, $\mathcal{M}(T)=A$ and so $\mathcal{M}$ is surjective.

Theorem. If $V$ and $W$ are finite-dimensional, then $\mathcal{L}(V,W)$ is finite-dimensional and $\mathop{dim} \mathcal{L}(V,W) =(\mathop{dim} V) (\mathop{dim} W).$

Proof. If $V$ and $W$ are finite-dimensional vector spaces then $\mathcal{L}(V,W)$ is isomorphic to $Mat(m,n,\mathbb{F})$ where $n=\mathop{dim} V$ and $m=\mathop{dim} W.$ If follows $$ \mathop{dim} \mathcal{L}(V,W) =\mathop{dim} Mat(mn,\mathbb{F})= m n=(\mathop{dim} W)(\mathop{dim} V). $$ since $\mathop{dim} Mat(m,n,\mathbb{F})= m n.$

Theorem. If $V$ is finite-dimensional and $T\in \mathcal{L}(V)$, the the following are equivalent: $T$ is invertible, $T$ is injective, and $T$ is surjective.

Proof. The proof of each part follows.

  • Suppose $T$ is invertible, then by definition $T$ is injective.
  • Suppose $T$ is injective. Thus $\mathop{ker} T={0}.$ Since $V$ is finite-dimensional, we can apply the Rank-Nullity Theorem so $\mathop{dim} V=\mathop{dim} \mathop{im} T.$ Since $\mathop{im} T$ is a subspace of $V$ with the same dimension as $V$, $\mathop{im} T=V.$ Therefore, $T$ is surjective.
  • Suppose $T$ is surjective. Then $T(V)=V$ and so $\mathop{dim} V=\mathop{dim} \mathop{im} T.$ Since $V$ is finite-dimensional, $\mathop{dim} V=\mathop{dim} \mathop{ker} T+\mathop{dim} \mathop{im}$ T by the Rank-Nullity Theorem. Thus, $\mathop{dim} \mathop{ker} T =0$ and so $\mathop{ker} T={0}.$ Therefore, $T$ is injective, and since $T$ is injective and surjective, $T$ is invertible.

David A. Smith at Dave4Math

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.

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