Linear Algebra Home

## Linear Transformations

Definition. A function $$T: V \to W$$ from a vector space $$V$$ to a vector space $$W$$ (over the same field) is called a linear transformation if

1. $$T(\overrightarrow{u}+\overrightarrow{v})= T(\overrightarrow{u})+T(\overrightarrow{v})$$ for all $$\overrightarrow{u}, \overrightarrow{v} \in V$$ and

2. $$T(c\overrightarrow{v})=cT(\overrightarrow{v})$$ for for all $$\overrightarrow{v}\in V$$ and all scalars $$c$$.

In short, a function $$T: V \to W$$ is a linear transformation if it preserves the linearity among vectors:
$$T(c\overrightarrow{u}+d\overrightarrow{v})= cT(\overrightarrow{u})+dT(\overrightarrow{v})$$ for all $$\overrightarrow{u}, \overrightarrow{v} \in V$$ and all scalars $$c,d$$.

Definition. The set of all linear transformations from a vector space $$V$$ to a vector space $$W$$ (over the same field) is denoted by $$L(V,W)$$.

Example.

1. For an $$m\times n$$ matrix $$A$$, $$T:\mathbb R^n\to \mathbb R^m$$ defined by $$T(\overrightarrow{x})=A\overrightarrow{x}$$ is a linear transformation.

2. $$T:P_n\to P_{n-1}$$ defined by $$T(a_0+a_1t+a_2t^2+\cdots+a_{n}t^n)=a_1+2a_2t+\cdots+na_nt^{n-1}$$ is a linear transformation.

3. The trace function $$T:M_{n}(\mathbb R) \to \mathbb R$$ defined by $$T(A)=\operatorname{tr}(A)$$ is a linear transformation.

4. The right shift operator $$T:\mathbb R^{\infty}\to \mathbb R^{\infty}$$ defined by $$T(a_1,a_2,a_3,\ldots)=(0,a_1,a_2,a_3,\ldots)$$ is a linear transformation.

From the definition of a linear transformation we have the following properties.

Proposition. For a linear transformation $$T: V \to W$$,

1. $$T(\overrightarrow{0_V})=\overrightarrow{0_W}$$ and

2. for all $$\overrightarrow{v_1}, \ldots,\overrightarrow{v_k} \in V$$ and all scalars $$c_1,\ldots,c_k$$, $T(c_1\overrightarrow{v_1}+c_2\overrightarrow{v_2}+\cdots+c_k\overrightarrow{v_k}) = c_1 T(\overrightarrow{v_1})+c_2 T(\overrightarrow{v_2})+\cdots+c_k T(\overrightarrow{v_k}).$

Example. Consider the function $$T:\mathbb R^3 \to \mathbb R^3$$ defined by $$T(x_1,x_2,x_3)=(x_1,x_2,5)$$. Since $$T(0,0,0)=(0,0,5) \neq (0,0,0)$$, $$T$$ is not a linear transformation.

For any given linear transformation $$T: V \to W$$, the domain space is $$V$$ and the codomain space is $$W$$. We study a subspace of the domain space called Kernel or Null Space and a subspace of the codomain space called Image Space or Range.

Definition. The kernel or null space of a linear transformation $$T: V \to W$$, denoted by $$\ker (T)$$ or $$\ker T$$, is the following subspace of $$V$$: $\ker T= \{\overrightarrow{x} \in V \;|\; T(\overrightarrow{x})=\overrightarrow{0_W}\}.$ The nullity of $$T$$, denoted by $$\operatorname{nullity}(T)$$, is the dimension of $$\ker T$$, i.e., $\operatorname{nullity}(T)=\operatorname{dim}(\ker T).$
Example. For the linear transformation $$T:M_{n}(\mathbb R) \to M_{n}(\mathbb R)$$ defined by $$T(A)=A-A^T$$, $\ker T=\{A\in M_{n}(\mathbb R) \;|\; T(A)=A-A^T=O\}=\{A\in M_{n}(\mathbb R) \;|\; A^T=A\},$ the set of all $$n\times n$$ real symmetric matrices. Then $$\operatorname{nullity}(T)=\operatorname{dim}(\ker T)=n(n+1)/2.$$

Definition. The image space or range of a linear transformation $$T: V \to W$$, denoted by $$\operatorname{im} (T)$$ or $$\operatorname{im} T$$ or $$T(V)$$, is the following subspace of $$W$$: $\operatorname{im} T= \{T(\overrightarrow{x}) \;|\; \overrightarrow{x} \in V\}.$ The rank of $$T$$, denoted by $$\operatorname{rank}(T)$$, is the dimension of $$\operatorname{im} T$$, i.e., $\operatorname{rank}(T)=\operatorname{dim}(\operatorname{im} T).$
Example. For the linear transformation $$T:M_{n}(\mathbb R) \to M_{n}(\mathbb R)$$ defined by $$T(A)=A-A^T$$, $\operatorname{im} T=\{A-A^T \;|\; A\in M_{n}(\mathbb R) \},$ the set of all $$n\times n$$ real skew-symmetric matrices. Then $\operatorname{rank}(T)=\operatorname{dim}(\operatorname{im} T)=n(n-1)/2.$

Theorem.(Rank-Nullity Theorem) Let $$T: V \to W$$ be a linear transformation. If $$V$$ has finite dimension, then $\operatorname{rank}(T)+\operatorname{nullity}(T)=\operatorname{dim}(V).$

Let $$\operatorname{dim}(V)=n$$. Start with a basis $$\{\overrightarrow{v_1},\ldots,\overrightarrow{v_k}\}$$ of $$\ker T$$ and by the Extension Theorem, extend it to a basis $$\{\overrightarrow{v_1},\ldots,\overrightarrow{v_k},\overrightarrow{u_1},\ldots,\overrightarrow{u_{n-k}}\}$$ of $$V$$. Now show that $$\{T(\overrightarrow{u_1}),\ldots,T(\overrightarrow{u_{n-k}})\}$$ is a basis of $$\operatorname{im} T$$.

Example. For the linear transformation $$T:M_{n}(\mathbb R) \to M_{n}(\mathbb R)$$ defined by $$T(A)=A-A^T$$, $\operatorname{rank}(T)+\operatorname{nullity}(T)=\frac{n(n+1)}{2}+\frac{n(n-1)}{2}=n^2 =\operatorname{dim}(M_{n}(\mathbb R) ).$

Now we discuss two important types of linear transformation.

Definition. Let $$T: V \to W$$ be a linear transformation. $$T$$ is onto if each $$\overrightarrow{b}\in W$$ has a pre-image $$\overrightarrow{x}$$ in $$V$$ under $$T$$, i.e., $$T(\overrightarrow{x})=\overrightarrow{b}$$. $$T$$ is one-to-one if each $$\overrightarrow{b}\in \mathbb R^m$$ has at most one pre-image in $$V$$ under $$T$$.

Example.

1. The linear transformation $$T:\mathbb R^3 \to \mathbb R^2$$ defined by $$T(x_1,x_2,x_3)=(x_1,x_2)$$ is onto because each $$(x_1,x_2)\in \mathbb R^2$$ has a pre-image $$(x_1,x_2,0)\in \mathbb R^3$$ under $$T$$. But $$T$$ is not one-to-one because $$T(0,0,0)=T(0,0,1)=(0,0)$$, i.e., $$(0,0)$$ has two distinct pre-images $$(0,0,0)$$ and $$(0,0,1)$$ under $$T$$.

2. The linear transformation $$T:\mathbb R^2 \to \mathbb R^3$$ defined by $$T(x_1,x_2)=(x_1,x_2,0)$$ is one-to-one because $$T(x_1,x_2)=T(y_1,y_2) \implies (x_1,x_2,0)=(x_1,x_2,0) \implies (x_1,x_2)=(y_1,y_2)$$. But $$T$$ is not onto because $$(0,0,1)\in \mathbb R^3$$ has no pre-image $$(x_1,x_2)\in \mathbb R^2$$ under $$T$$.

3. The linear transformation $$T:\mathbb R^2 \to \mathbb R^2$$ defined by $$T(x_1,x_2)=(x_1+x_2,x_1-x_2)$$ is one-to-one and onto (exercise).

Theorem. Let $$T: V \to W$$ be a linear transformation. Then the following are equivalent.

1. $$T$$ is one-to-one.

2. $$\ker T=\{\overrightarrow{0_V}\}$$.

3. $$\operatorname{nullity}(T)=0$$.

Example. The linear transformation $$T:\mathbb R^2 \to \mathbb R^3$$ defined by $$T(x_1,x_2)=(x_1,x_2,0)$$ has the standard matrix $$A=[T(\overrightarrow{e_1})\: T(\overrightarrow{e_2}) ] =\left[\begin{array}{rr} 1&0\\ 0&1\\ 0&0 \end{array} \right]$$. Note that the columns of $$A$$ are linearly independent , $$\ker T=\operatorname{NS}(A)=\{\overrightarrow{0_2}\}$$, and $$\operatorname{nullity}(T)=\operatorname{nullity}(A)=0$$. Thus $$T$$ (i.e., $$\overrightarrow{x} \mapsto A\overrightarrow{x}$$) is one-to-one.

Theorem. Let $$T: V \to W$$ be a linear transformation. Then the following are equivalent.

1. $$T$$ is onto.

2. $$\operatorname{im} T=W$$.

3. $$\operatorname{rank}(T)=\operatorname{dim}(W)$$.

Example. The linear transformation $$T:\mathbb R^3 \to \mathbb R^2$$ defined by $$T(x_1,x_2,x_3)=(x_1,x_2)$$ has the standard matrix $$A=[T(\overrightarrow{e_1})\: T(\overrightarrow{e_3}) \: T(\overrightarrow{e_2})] =\left[\begin{array}{rrr} 1&0&0\\ 0&1&0 \end{array} \right]$$. Note that each row of $$A$$ has a pivot position, $$\operatorname{im} T=\operatorname{CS}\left(A\right)=\mathbb R^2$$, and $$\operatorname{rank}(T)=\operatorname{rank}(A)=2=\dim(\mathbb R^2)$$. Thus $$T$$ (i.e., $$\overrightarrow{x} \mapsto A\overrightarrow{x}$$) is onto.

Definition. A linear transformation $$T: V \to W$$ is an isomorphism if it is one-to-one and onto. When $$T: V \to W$$ is an isomorphism, $$V$$ and $$W$$ are called isomorphic, denoted by $$V\cong W$$.

Example.

1. Define $$T:P_{n}\to \mathbb R^{n+1}$$ by $$T(a_0+a_1t+\cdots+a_nt^n)=[a_0,a_1,\ldots,a_n]^T$$. Verify that $$\ker T=\{\overrightarrow{0}\}$$ which implies $$T:P_n\to \mathbb R^{n+1}$$ is one-to-one. Also $$T:P_{n}\to \mathbb R^{n+1}$$ is onto since each $$[a_0\;a_1\;\cdots\;a_n]^T\in \mathbb R^{n+1}$$ has a pre-image $$a_0+a_1t+\cdots+a_nt^n \in P_n$$ under $$T$$. Thus $$T:P_{n}\to \mathbb R^{n+1}$$ is an isomorphism and consequently $$P_n$$ and $$\mathbb R^{n+1}$$ are isomorphic.

2. The left shift operator $$T:\mathbb R^{\infty}\to \mathbb R^{\infty}$$ defined by $$T(a_1,a_2,a_3,\ldots)=(a_2,a_3,a_4,\ldots)$$ is a linear transformation. Since each $$(a_2,a_3,a_4,\ldots) \in R^{\infty}$$ has a pre-image $$(0,a_2,a_3,\ldots) \in R^{\infty}$$ under $$T$$, $$T$$ is onto equivalently $$\operatorname{im} T=R^{\infty}$$. Verify that $\ker T=\{(a_1,0,0,\ldots) \;|\; a_1\in R \}=\operatorname{Span}\{(1,0,0,\ldots) \}.$ Thus $$T$$ is not one-to-one and hence not an isomorphism. Note that ontoness of $$T$$ does not imply that $$T$$ is one-to-one which may happen when domain and codomain spaces are infinite dimensional.

Theorem. Let $$T: V \to W$$ be a linear transformation. If $$V$$ and $$W$$ have the same finite dimension, then the following are equivalent.

1. $$T$$ is an isomorphism.

2. $$T$$ is one-to-one.

3. $$\ker T=\{\overrightarrow{0_V}\}$$.

4. $$\operatorname{nullity}(T)=0$$.

5. $$T$$ is onto.

6. $$\operatorname{im} T=W$$.

7. $$\operatorname{rank}(T)=\operatorname{dim}(W)$$.

By the Rank-Nullity Theorem, $\operatorname{nullity}(T)=0 \iff \operatorname{rank}(T)=\operatorname{dim}(V)=\operatorname{dim}(W).$

Theorem. If $$V$$ and $$W$$ are isomorphic via an isomorphism $$T:V\to W$$, then $$V$$ and $$W$$ have similar linear algebraic properties such as follows.

1. $$H$$ is a subspace of $$V$$ if and only if $$T(H)$$ is a subspace of $$W$$.

2. $$\{\overrightarrow{v_1},\ldots,\overrightarrow{v_n}\}$$ is linearly independent in $$V$$ if and only if $$\{T(\overrightarrow{v_1}),\ldots,T(\overrightarrow{v_n})\}$$ is linearly independent in $$W$$.

3. $$\{\overrightarrow{v_1},\ldots,\overrightarrow{v_n}\}$$ spans $$V$$ if and only if $$\{T(\overrightarrow{v_1}),\ldots,T(\overrightarrow{v_n})\}$$ spans $$W$$.

4. $$\{\overrightarrow{v_1},\ldots,\overrightarrow{v_n}\}$$ is a basis of $$V$$ if and only if $$\{T(\overrightarrow{v_1}),\ldots,T(\overrightarrow{v_n})\}$$ is a basis of $$W$$.

5. $$\operatorname{dim}(V)=\operatorname{dim}(W)$$.

Exercise.

Remark. The preceding theorem is true even when $$V$$ and $$W$$ have infinite dimensions.

Example. Consider the following three polynomials of $$P_2$$: $\overrightarrow{p_1}(t)=1+t^2, \overrightarrow{p_2}(t)=-1+2t-t^2, \mbox{and }\overrightarrow{p_3}(t)=-1+4t.$ Show that $$\{\overrightarrow{p_1},\;\overrightarrow{p_2},\;\overrightarrow{p_3}\}$$ is a basis of $$P_2$$.

Solution. First recall that $$T:P_{2}\to \mathbb R^3$$ defined by $$T(a_0+a_1t+a_2t^2)=[a_0,a_1,a_2]^T$$ is an isomorphism. \begin{align*} T(\overrightarrow{p_1})&=T(1+t^2)=\left[\begin{array}{r}1\\0\\1\end{array}\right]\\ T(\overrightarrow{p_2})&=T(-1+2t-t^2)=\left[\begin{array}{r}-1\\2\\-1\end{array}\right]\\ T(\overrightarrow{p_3})&=T(-1+4t)=\left[\begin{array}{r}-1\\4\\0\end{array}\right] \end{align*} Now $$A=[T(\overrightarrow{p_1})\;T(\overrightarrow{p_2})\;T(\overrightarrow{p_3})] =\left[\begin{array}{rrr}1&-1&-1\\0&2&4\\1&-1&0\end{array}\right]\xrightarrow{-R1+R3} \left[\begin{array}{rrr}\boxed{1}&-1&-1\\0&\boxed{2}&4\\0&0&\boxed{1}\end{array}\right].$$ Since $$3\times 3$$ matrix $$A$$ has 3 pivot positions, by the IMT, the columns of $$A$$ are linearly independent and span $$\mathbb R^3$$. Thus $$\{T(\overrightarrow{p_1}),T(\overrightarrow{p_2}),T(\overrightarrow{p_3})\}$$ is a basis of $$\mathbb R^3$$. Since $$T:P_{2}\to \mathbb R^3$$ is an isomorphism, $$\{\overrightarrow{p_1},\;\overrightarrow{p_2},\;\overrightarrow{p_3}\}$$ is a basis of $$P_2$$.

Definition. Suppose $$B=\left(\overrightarrow{b_1},\ldots, \overrightarrow{b_n}\right)$$ is an ordered basis of a real vector space $$V$$. Then any vector $$\overrightarrow{x}\in V$$ can be written as $$\overrightarrow{x}=c_1\overrightarrow{b_1}+c_2\overrightarrow{b_2}+\cdots+c_n\overrightarrow{b_n}$$ for some unique scalars $$c_1,c_2,\ldots,c_n$$. The coordinate vector of $$\overrightarrow{x}$$ relative to $$B$$ or the $$B$$-coordinate of $$\overrightarrow{x}$$, denoted by $$[\overrightarrow{x}]_B$$, is $[x]_B=[c_1b_1+c_2b_2+\cdots+c_nb_n]_B=\left[\begin{array}{c}c_1\\c_2\\ \vdots \\c_n\end{array}\right].$

Remark. $$[\;\;]_B:V\to \mathbb R^n$$ is an isomorphism.

Theorem. If $$V$$ is a real vector space of dimension $$n$$, then $$V$$ is isomorphic to $$\mathbb R^n$$.

Let $$B$$ be an ordered basis of $$V$$. Define the coordinate map $$T:V\to \mathbb R^n$$ by $$T(\overrightarrow{x})=[\overrightarrow{x}]_B$$. It can be verified that $$T$$ is an isomorphism. Thus $$V\cong \mathbb R^n$$.

Definition. Let $$V$$ and $$W$$ be real vector spaces with ordered bases $$B=\left(\overrightarrow{b_1},\ldots, \overrightarrow{b_n}\right)$$ and $$C=\left(\overrightarrow{c_1},\ldots, \overrightarrow{c_m}\right)$$ respectively. Let $$T:V\to W$$ be a linear transformation. The matrix of $$T$$ from $$B$$ to $$C$$, denoted by $$[T]_{C\leftarrow B}$$ or $$_{C}[T]_B$$, is the following $$m\times n$$ matrix: $_{C}[T]_B=\left[ [T(\overrightarrow{b_1})]_C \cdots [T(\overrightarrow{b_n})]_C\right].$ Note that for all $$\overrightarrow{x}\in V$$, $[T(\overrightarrow{x})]_C=_{C}[T]_B [\overrightarrow{x}]_B.$

The following diagram shows the "equivalence" of a linear transformation $$T:V\to W$$ and the corresponding matrix transformation $$_{C}[T]_B: \mathbb R^n\to \mathbb R^m$$.

Example. $$P_n$$ and $$P_{n-1}$$ are real vector spaces with ordered bases $$B=\left(1,x,\ldots,x^n\right)$$ and $$C=\left(1,x,\ldots,x^{n-1}\right)$$ respectively. For the linear transformation $$T:P_n\to P_{n-1}$$ defined by $$T(a_0+a_1x+a_2x^2+\cdots+a_{n}x^n)=a_1+2a_2x+\cdots+na_nx^{n-1}$$, we have $_{C}[T]_B =\left[\begin{array}{ccccc}0&1&0&\cdots&0\\ 0&0&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&n \end{array}\right].$ The columns of $$_{C}[T]_B$$ are calculated as follows: \begin{align*} \text{Column 1} &:\; [[T(1)]_C=[0]_C=[0\cdot 1+0 x+0x^2+\cdots +0x^{n-1}]_C=[0,0,0\ldots,0]^T\\ \text{Column 2} &:\; [[T(x)]_C=[1]_C=[1\cdot 1+0 x+0x^2+\cdots +0x^{n-1}]_C=[1,0,0,\ldots,0]^T\\ \text{Column 3} &:\; [[T(x^2)]_C=[2x]_C=[0\cdot 1+2 x+0x^2+\cdots +0x^{n-1}]_C=[0,2,0,\ldots,0]^T\\ \vdots &\\ \text{Column n+1} &:\; [[T(x^n)]_C=[nx^{n-1}]_C=[0\cdot 1+0 x+\cdots +0x^{n-2}+nx^{n-1}]_C=[0,0,\ldots,0,n]^T\\ \end{align*}

The following theorem shows the relation between two matrices of a linear transformation corresponding to different bases.

Theorem. Let $$V$$ be a real vector space with ordered bases $$B$$ and $$B'$$. Let $$W$$ be a real vector space with ordered bases $$C$$ and $$C'$$. For a linear transformation $$T:V\to W$$, $_{C'}[T]_{B'} =_{C'}[I]_C\; {}_{C}[T]_B\; {}_{B}[I]_{B'} .$

Last edited