# 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 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.