3.5 Bases of a vector space

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Subspace spanned by a set

If $A$ is any subset of a vector space $V$ define the subspace spanned or generated by $A$, denoted $L(A)$ , as the set of allfinite linear combinations of elements of $A$,

$$L(A)=\left\{\sum _{i=1}^n{a}^i{v}_i|a^i\in \mathbb{K},v_i\in A,n=1,2,\dots \right\}.$$

The word ‘finite’ is emphasized here because no meaning can be attached to infinite sums until we have available the concept of‘limit’ (see Chapters 10 and 13). We may think of $L(A)$ as the intersection ofall subspaces of $V$ that contain $A$ – essentially, it is the ‘smallest’ vector subspace containing $A$. At first sight the notation whereby the indices on the coefficients $a^i$ of the linear combinations have been set in the superscript position may seem a little peculiar, but we will eventually see that judicious and systematic placements of indices can make many expressions much easier to manipulate.

Exercise

If $M$ and $N$ are subspaces of $V$ show that their sum $M+N$ is identical with the span of their union, $M+N=L(M\cup N)$

The vector space $V$ is said to be finite dimensional [7] if it can be spanned by a finite set, $V=L(A)$ , where $A=\{v_1,v_2,\dots ,v_n\}$ . Otherwise we say $V$ is infinite dimensional. When $V$ is finite dimensional its dimension, $\dim V$, is defined to be the smallest number $n$ such that $V$ is spanned by a set consisting ofjust $n$ vectors.

Example 3.17

${\mathbb{R}}^n$ is finite dimensional, since it can be generated by the set of‘unit vectors’,

$$A=\{e_1=(1,0,\dots ,0),e_2=(0,1,\dots ,0),\dots ,e_n=(0,0,\dots ,1)\}.$$

Since any vector $u$ can be written

$$u=(u_1,u_2,\dots ,u_n)=u_1{e}_1+u_2{e}_2+\cdots +u_n{e}_n,$$

these vectors span ${\mathbb{R}}^n$ , and $\dim {\mathbb{R}}^n\le n$ . We will see directly that $\dim {\mathbb{R}}^n=n,$ , as to be expected.

Example 3.18

${\mathbb{R}}^{\infty }$ is clearly infinite dimensional. It is not even possible to span this space with the set of vectors $A=\{e_1,e_2,\dots \}$ , where

$$e_1=(1,0,\dots ),e_2=(0,1,\dots ),\dots .$$

The reason is that any finite linear combination ofthese vectors will only give rise to vectors having at most a finite number of non-zero components. The set of all those vectors tha are finite linear combinations of vectors from $A$ does in fact form an infinite dimensiona subspace of ${\mathbb{R}}^{\infty }.$ , but it is certainly not the whole space. The space spanned by $A$ is precisely the subspace ${\hat{\mathbb{R}}}^{\infty }$ defined in Example 3.10.

Exercise

If $V$ is a vector space and $u\in V$ is any non-zero vector show that $\dim V=1$ if and only if every vector $v\in V$ is proportional to $u$; i.e., $v=au$ for some $a\in \mathbb{K}$

Exercise

Show that the set of functions on the real line, $\mathcal{F}(\mathbb{R})$ , is an infinite dimensional vector space.

Basis of a vector space

A set of vectors $A$ is said to be linearly independent, often written ‘l.i.’ , if every finite subset of vectors $\{v_1,v_2,\dots ,v_k\}\subseteq A$ has the property that

$$\sum _{i=1}^k{a}^i{v}_i=0⟹a^j=0\text{ for all }j=1,\dots ,k.$$

In other words, the zero vector 0 cannot be written as a non-trivial linear combination of these vectors. The zero vector can never be a member of a l.i. set since $a0=0$ for any $a\in \mathbb{K}$ . If $A$ is a finite set ofvectors, $A=\{v_1,v_2,...,v_n\}$ , it is sufficient to set $k=n$ in the above definition. A subset $E$ ofa vector space $V$ is called a basis ifit is linearly independen and spans the whole of $V$. A set of vectors is said to be linearly dependent if it is not l.i.

Example 3.19

The set of vectors $\{e_1=(1,0,\dots ,0),\dots ,e_n=(0,0,\dots 0,1)\}$ span ${\mathbb{K}}^n$ since every vector $v=(v_1,v_2,\dots ,v_n)$ can be expressed as a linear combination

$$v=v_1{e}_1+v_2{e}_2+\cdots +v_n{e}_n.$$

They are linearly independent, for if $v=0$ then we must have $v_1=v_2=\cdot \cdot \cdot v_n=0$ . Hence $e_1,\dots ,e_n$ is a basis of ${\mathbb{K}}^n$

Exercise

Show that the vectors $f_1=(1,0,0),f_2=(1,1,-1)\mathrm{and}{f}_3=(1,1,1)$ are l.i. and form a basis of ${\mathbb{R}}^3$

It is perhaps surprising to learn that even infinite dimensional vector space such as ${\mathbb{R}}^{\infty }$ always has a basis. Just try and construct a basis! The set $A=\{e_1=(1,0,0,\dots )$ $e_2=(0,1,0,\dots ),\dots \}$ clearly won’$t$ do, since any vector having an infinite number of non-zero components cannot be a finite linear combination of these vectors. We omit the proof as it is heavily dependent on Zorn’s lemma and such bases are only of limited use. For the rest of this section we only consider bases in finite dimensional spaces.

Theorem 3.3

Let $V$ be afinite dimensional vector space ofdimension $n$. A subset $E=$ $\{e_1,e_2,\dots ,e_n\}$ spans $V$ ifand only ifit is linearly independent

Proof

Only if: Assume $V=L(E)$ , so that every vector $v\in V$ is a linear combination

$$v=\sum _{i=1}^n{v}^i{e}_i.$$

The set $E$ is then linearly independent, for suppose there exists a vanishing linear combi nation

$$\sum _{i=1}^n{a}^i{e}_i=0(a^i\in \mathbb{K})$$

where, say, $a^1\ne 0$ . Replacing $e_1$ by $e_1=b^2{e}_2+\cdot \cdot \cdot +b^n{e}_n$ where $b^i=-a^i/a^1$ , we find

$$v=\sum _{j=2}^n{\bar{v}}^j{e}_j\text{ where }{\bar{v}}^j=v^j+b^j{v}^1.$$

Thus $E^{\prime }=E-\{e_1\}$ spans $V_{;}$ , contradicting the initial hypothesis that $V$ cannot be spanned by a set of fewer than $n$ vectors on account of $\dim V=n$

If: Assume $E$ is linearly independent. Our aim is to show that it spans all of $V$. Since $\dim V=n$ there must exist a set $F=\{f_1,f_2,\dots ,f_n\}$ of exactly $n$ vectors spanning $V$. By the above argument this set is l.i. Expand $e_1$ in terms of the vectors from $F$ ,

$$e_1=a^1{f}_1+a^2{f}_2+\cdots +a^n{f}_n,\text{\hspace{1em}(3.4)}$$

where, by a permutation of the vectors of the basis $F_{;}$ , we may assume that $a^1\ne 0$ . The set $F^{\prime }=\{e_1,f_2,...,f_n\}$ is a basis for $V$:

(a) $F^{\prime }$ is linearly independent, for if there were a vanishing linear combination

$$c{e}_1+c^2{f}_2+\cdots +c^n{f}_n=0,$$

then substituting Eq. (3.4) gives

$$c{a}^1{f}_1+(c{a}^2+c^2)f_2+\cdots +(c{a}^n+c^n)f_n=0.$$

By linear independence of $\{f_1,\dots ,f_n\}$ and $a^1\ne 0$ it follows that $c=0$ , and subsequently that $c^2=\cdots =c^n=0$

(b) The set $F^{\prime }$ spans the vector space $V$ since by Eq. (3.4)

$$f_1=\frac{1}{a^1}(e_1-a^2{f}_2-\cdots -a^n{f}_n),$$

and every $v\in V$ must be a linear combination of $\{e_1,f_2,\dots ,f_n\}$ since it is spanned by $\{f_1,\dots ,f_n\}$

Continuing, $e_2$ must be a unique linear combination

$$e_2=b^1{e}_1+b^2{f}_2,+\cdots +b^n{f}_n.$$

Since, by hypothesis, $e_1$ and $e_2$ are linearly independent, at least one of the coefficients $b^2,\dots ,b^n$ must be non-zero, say $b^2\ne 0$ . Repeating the above argument we see that the set $F^{\prime \prime }=\{e_1,e_2,f_3,...,f_n\}$ is a basis for $V$. Continue the process $n$ times to prove that $E=F^{(n)}=\{e_1,e_2,...,e_n\}$ is a basis for $V$. -

Corollary 3.4

If $E=\{e_1,e_2,...,e_n\}$ is a basis ofthe vector space $V$ then $\dim V=n$

Proof

Suppose that $\dim V=m<n$ . The set of vectors $E^{\prime }=\{e_1,e_2,...,e_m\}$ is l.i., since it is a subset of the l.i. set $E$. Hence, by Theorem 3.3 it spans $V$, since it consists of exactly $m=\dim V$ vectors. But this is impossible since, for example, the vector $e_n$ cannot be a linear combination of the vectors in $E^{\prime }$ . Hence we must have $\dim V\ge n$ . However, by the definition of dimension it is impossible to have $\dim V>n;$ hence $\dim V=n$ -

Exercise

Show that if $A=\{v_1,v_2,\dots ,v_m\}$ is an l.i. set of vectors then $m\le n=\dim V$

Theorem 3.5

Let $V$ be a finite dimensional vector space, $n=\dim V$. If $\{e_1,\dots ,e_n\}$ is a basis of $V$ then each vector $v\in V$ has a unique decomposition

$$v=\sum _{i=1}^n{v}^i{e}_i,v^i\in \mathbb{K}.\text{\hspace{1em}(3.5)}$$

The $n$ scalars ${\mathbf{v}}^i\in \mathbb{K}$ are called the components ofthe vector $v$ with respect to this basis.

Proof

Since the $e_i$ span $V$ , every vector $v$ has a decomposition of the form Eq. (3.5). If there were a second such decomposition,

$$v=\sum _{i=1}^n{v}^i{e}_i=\sum _{i=1}^n{w}^i{e}_i$$

then

$$\sum _{i=1}^n(v^i-w^i)e_i=0.$$

Since the $e_i$ are linearly independent, each coefficient ofthis sum must vanish, $v^i-w^i=0$ Hence ${\mathbf{v}}^i={\mathbf{w}}^i$ , and the decomposition is unique.

Theorem 3.6

If $V$ and $W$ are finite dimensional then they are isomorphic if and only if they have the same dimension.

Proof

Suppose $V$ and $W$ have the same dimension $n$. Let $\{e_i\}$ be a basis of $V$ and $\{f_i\}$ a basis of $W$, where $i=1,2,\dots ,n$ . Set $T:V\to W$ to be the linear map defined by $T{e}_1=f_1,T{e}_2=f_2,...,T{e}_n=f_n$ . This map extends uniquely to all vectors in $V$ by linearity,

$$T\left(\sum _{i=1}^n{v}^i{e}_i\right)=\sum _{i=1}^n{v}^{i}T{e}_i=\sum _{i=1}^n{v}^i{f}_i$$

and is clearly one-to-one and onto. Thus $T$ is an isomorphism between $V$ and $W$

Conversely suppose $V$ and $W$ are isomorphic vector spaces, and let $T:V\to W$ be a linear map having inverse $T^{-1}:W\to V$. If $\{e_1,e_2,\dots ,e_n\}$ is a basis of $V$, we show that $\{f_i=T{e}_i\}$ is a basis of $W:$

(a) The vectors $\{f_i\}$ are linearly independent, for suppose there exist scalars $a^i\in \mathbb{K}$ such that

$$\sum _{i=1}^n{a}^i{f}_i=0.$$

Then,

$$0=T^{-1}\left(\sum _{i=1}^n{a}^i{f}_i\right)=\sum _{i=1}^n{a}^i{T}^{-1}{f}_i=\sum _{i=1}^n{a}^i{e}_i$$

and from the linear independence of $\left\{{\mathbf{e}}_i\right\}$ it follows that $a^1=a^2=\cdot \cdot \cdot =a^n=0$

(b) To show that the vectors $\{f_i\}$ span $W$ let $w$ be any vector in $W$ and set $v=T^{-1}w\in V$

Since $\{e_i\}$ spans $V$ there exist scalars $v^i$ such that

$$v=\sum _{i=1}^n{v}^i{e}_i.$$

Applying the map $T$ to this equation results in

$$w=\sum _{i=1}^n{v}^{i}T{e}_i=\sum _{i=1}^n{v}^i{f}_i,$$

which shows that the set $\{f_1,\dots ,f_n\}$ spans $W.$

By Corollary 3.4 it follows that $\dim W=\dim V=n$ since both vector spaces have a basis consisting of $n$ vectors. -

Example 3.20

By Corollary 3.4 the space ${\mathbb{K}}^n$ is $n$-dimensional since, as shown in Example 3.19, the set $\{e_i=(0,0,\dots ,0,1,0,\dots ,0)\}$ is a basis. Using Theorem 3.6 every $n$-dimensional vector space $V$ over the field $\mathbb{K}$ is isomorphic to ${\mathbb{K}}^n$ , which may be though of as the archetypical $n$-dimensional vector space over $\mathbb{K}.$ . Every basis $\{f_1,f_2,\dots ,f_n\}$ of $V$ establishes an isomorphism $T:V\to {\mathbb{K}}^n$ defined by

$$Tv=\left(v^1,v^2,\dots ,v^n\right)\in {\mathbb{K}}^n\text{ where }v=\sum _{i=1}^n{v}^i{f}_i.$$

Example 3.20 may lead the reader to wonder why we bother at all with the abstract vector space machinery of Section 3.2, when all properties of a finite dimensional vector space $V$ could be referred to the space ${\mathbb{K}}^n$ by simply picking a basis. This would, however, have some unfortunate consequences. Firstly, there are infinitely many bases of the vector space $V$, each of which gives rise to a different isomorphism between $V$ and ${\mathbb{K}}^n$ . There is nothing natural in the correspondence between the two spaces, since there is no genera way of singling out a preferred basis for the vector space $V$. Furthermore, any vector space concept should ideally be given a basis-independent definition, else we are always faced with the task of showing that it is independent of the choice of basis. For these reasons we will persevere with the ‘invariant’ approach to vector space theory.

Matrix of a linear operator

Let $T:V\to V$ be a linear operator on a finite dimensional vector space $V$. Given a basis $\{e_1,e_2,\dots ,e_n\}$ of $V$ define the components $T_k^a$ of the linear operator $T$ with respect to this basis by setting

$$T{e}_j=\sum _{i=1}^n{T}_j^i{e}_i.\text{\hspace{1em}(3.6)}$$

By Theorem 3.5 the components $T_j^i$ are uniquely defined by these equations, and the square $n\times n$ matrix $\mathsf{T}=[T_j^i]$ is called the matrix of $T$ with respect to the basis $\{e_i\}$ . It is usua to take the superscript $i$ as the ‘first’ index, labelling rows, while the subscript $j$ labels the columns, and for this reason it is generally advisable to leave some horizontal spacing between these two indices. In Section 2.3 the components of a matrix were denoted by subscripted symbols such as $\mathsf{A}=[a_{ij}]$ , but in general vector spaces it is a good idea to display the components of a matrix representing a linear operator $T$ in this ‘mixed script notation.

If $v={\sum }_{k=1}^n{v}^k{e}_k$ is an arbitrary vector of $V$ then its image vector $\begin{array}{r}\tilde{\mathbf{v}}=\mathbf{T}\mathbf{v}={\sum }_{j=1}^m{w}^j{\mathbf{e}}_j\end{array}$ is given by

$$\tilde{v}=Tv=T\left(\sum _{j=1}^n{v}^j{e}_j\right)=\sum _{j=1}^n{v}^{j}T{e}_j=\sum _{j=1}^n\sum _{i=1}^n{v}^j{T}_j^i{e}_i,$$

and the components of $\tilde{v}$ are given by

$${\tilde{v}}^i=(Tv{)}^i=\sum _{j=1}^n{T}_j^i{v}^j.\text{\hspace{1em}(3.7)}$$

If we write the components of $v$ and $\tilde{v}$ as column vectors or $n\times 1$ matrices, $v$ and $\tilde{\mathbf{v}},$

$$\mathbf{v}=\left(\begin{array}{c}{v}^1\\ v^2\\ ⋮\\ v^n\end{array}\right),\tilde{\mathbf{v}}=\left(\begin{array}{c}{\tilde{v}}^1\\ {\tilde{v}}^2\\ ⋮\\ {\tilde{v}}^n\end{array}\right),$$

then Eq. (3.7) is the componentwise representation of the matrix equation

$$\tilde{\mathbf{v}}=\mathsf{T}\mathbf{v}.\text{\hspace{1em}(3.8)}$$

The matrix of the composition of two operators $ST\equiv S\circ T$ is given by

$$\begin{array}{l}ST(e_i)={\sum }_{j=1}^{n}S(T_i^j{e}_j)\\ ={\sum }_{j=1}^n{T}_i^{j}S{e}_j\\ ={\sum }_{j=1}^n{\sum }_{k=1}^n{T}_i^j{S}_j^k{e}_k\\ ={\sum }_{k=1}^n(ST{)}_i^k{e}_k\end{array}$$

where

$$(ST{)}_i^k=\sum _{j=1}^n{S}_j^k{T}_i^j.\text{\hspace{1em}(3.9)}$$

This can be recognized as the componentwise formula for the matrix product $ST$.

Example 3.21

Care should be taken when reading off the components of the matrix $T$ from Eq. (3.6) as it is very easy to come up mistakenly with the ‘transpose’ array. For example, if a transformation $T$ of a three-dimensional vector space is defined by its effect on a basis

$e_1$, $e_2$, $e_3$,

$$\begin{array}{l}T{e}_1=e_1-e_2+e_3\\ T{e}_2=e_1-e_3\\ T{e}_3=e_2+2e_3,\end{array}$$

then its matrix with respect to this basis is

$$\mathsf{T}=\left(\begin{array}{ccc}1& 1& 0\\ -1& 0& 1\\ 1& -1& 2\end{array}\right).$$

The result of applying $T$ to a vector $\mathbf{u}=x{\mathbf{e}}_1+y{\mathbf{e}}_2+z{\mathbf{e}}_3$ is

$$Tu=xT{e}_1+yT{e}_2+zT{e}_3=(x+y)e_1+(-x+z)e_2+(x-y+2z)e_3,$$

which can also be obtained by multiplying the matrix $T$ and the column vector $\mathbf{u}=$ $(x,y,z{)}^T$

$$\mathsf{Tu}=\left(\begin{array}{ccc}1& 1& 0\\ -1& 0& 1\\ 1& -1& 2\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}x+y\\ -x+z\\ x-y+2z\end{array}\right).$$

If $S$ is a transformation given by

$$\begin{array}{l}S{e}_1=e_1+2e_3\\ S{e}_2=e_2\\ S{e}_3=e_1-e_2\end{array}$$

whose matrix with respect to this basis is

$$\mathsf{S}=\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\\ 2& 0& 0\end{array}\right),$$

the product of these two transformations is found from

$$\begin{array}{ll}ST{e}_1=S{e}_1-S{e}_2+S{e}_3& =2e_1-2e_2+2e_3\\ ST{e}_2=S{e}_1-S{e}_3& =e_2+2e_3\\ ST{e}_3=S{e}_2+2S{e}_3& =2e_1-e_2.\end{array}$$

Thus the matrix of $ST$ is the matrix product of $S$ and $T$,

$$\mathsf{ST}=\left(\begin{array}{ccc}2& 0& 2\\ -2& 1& -1\\ 2& 2& 0\end{array}\right)=\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\\ 2& 0& 0\end{array}\right)\left(\begin{array}{ccc}1& 1& 0\\ -1& 0& 1\\ 1& -1& 2\end{array}\right).$$

Exercise

In Example 3.21 compute $TS$ by calculating $T(S{e}_i)$ and also by evaluating the matrix product of $T$ and $S$.

Exercise

If $V$ is a finite dimensional vector space, $\dim V=n$ , over the field $\mathbb{K}=\mathbb{R}or\mathbb{C},$ , show that the group $GL(V)$ oflinear transformations of $V$ is isomorphic to the matrix group ofinvertible $n\times n$ matrices, $GL(n,\mathbb{K})$

Basis extension theorem

While specific bases should not be used in general definitions of vector space concepts if at all possible, there are specific instances when the singling out of a basis can prove of great benefit. The following theorem is often useful, in that it allows us to extend any l.i. set to a basis. In particular, it implies that if $v\in V$ is any non-zero vector, one can always find a basis such that $e_1=v$

Theorem 3.7

Let $A=\{v_1,v_2,\dots ,v_m\}$ be any l.i. subset of $V$, where $m\le n=\dim V$ Then there exists a basis $E=\{e_1,e_2,...,e_n\}$ of $V$ such that $e_1=v_1,e_2=v_2,...,e_m=$ $v_m$

Proof

If $m=n$ then by Theorem 3.3 the set $E$ is a basis of $V$ and there is nothing to show. Assuming $m<n$ , we set $e_1=v_1,e_2=v_2,\dots ,e_m=v_m$ . By Corollary 3.4 the set $A$ cannot span $V$ since it consists of fewer than $n$ elements, and there must exist a vector $e_{m+1}\in V$ that is not a linear combination of $e_1,\dots ,e_m$ . The set $A^{\prime }=\{e_1,e_2,\dots ,e_{m+1}\}$ is l.i., for if

$$a^1{e}_1+\cdots +a^m{e}_m+a^{m+1}{e}_{m+1}=0,$$

then we must have ${\mathbf{a}}^{m+1}=0$ , else $e_{m+1}$ would be a linear combination of $e_1,\dots ,e_m$ . The linear independence of $e_1,\dots ,e_m$ then implies that $a^1=a^2=\cdot \cdot \cdot =a^m=0.\mathrm{If}m+1<n$ continue adding vectors that are linearly independent of those going before, until we arrive at a set $E=A^{(n-m)}$ , which is l.i. and has $n$ elements. This set must be a basis and the process can be continued no further. -

The following examples illustrate how useful this theorem can be in applications.

Example 3.22

Let $W$ be a $k$-dimensional vector subspace ofa vector space $V$ ofdimension $n$. We will demonstrate that the dimension ofthe factor space $V/W_{;}$ , known as the codimen sion of $W$, is $n-k.$ . By Theorem 3.7 it is possible to find a basis $\{e_1,e_2,\dots ,e_n\}$ of $V$ such that the first $k$ vectors $e_1,\dots ,e_k$ are a basis of $W$. Then $\{e_{k+1}+W,e_{k+2}+W,\dots ,e_n+W\}$ forms a basis for $V/W$ since every coset $v+W$ can be written

$$v+W=v^{k+1}\left(e_{k+1}+W\right)+v^{k+2}\left(e_{k+2}+W\right)+\cdots +v^n\left(e_n+W\right)$$

where $v={\sum }_{i=1}^n{v}^i{e}_i$ is the unique expansion given by Theorem 3.5. These cosets therefore span $V/W$ . They are also l.i., for if

$$0+W=a^{k+1}\left(e_{k+1}+W\right)+a^{k+2}\left(e_{k+2}+W\right)+\cdots +a^n\left(e_n+W\right)$$

then $a^{k+1}{e}_{k+1}+a^{k+2}{e}_{k+2}+\cdot \cdot \cdot +a^n{e}_n\in W$ , which implies that there exist $b^1,\dots ,b^k$ such that

$$a^{k+1}{e}_{k+1}+a^{k+2}{e}_{k+2}+\cdots +a^n{e}_n=b^1{e}_1+\cdots +b^k{e}_k.$$

By the linear independence of $e_1,\dots ,e_n$ we have that $a^{k+1}=a^{k+2}=\cdots =a^n=0$ . The desired result now follows,

$$\mathrm{codim}W\equiv \dim (V/W)=n-k=\dim V-\dim W.$$

Example 3.23

Let $A:V\to V$ be a linear operator on a finite dimensional vector space $V$. Define its rank $\rho (A)$ to be the dimension of its image $\mathrm{im}A$, and its nullity $\nu (A)$ to be the dimension of its kernel $\ker A$,

$$\rho (A)=\dim \mathrm{im}A,\nu (A)=\dim \ker A.$$

By Theorem 3.7 there exists a basis $\{e_1,e_2,\dots ,e_n\}$ of $V$ such that the first $\nu$ vectors $e_1,\dots ,e_{\nu }$ form a basis of $\ker A$ such that $A{e}_1=A{e}_2=\cdot \cdot \cdot =A{e}_{\nu }=0$ . For any vector $\begin{array}{r}u={\sum }_{i=1}^n{u}^i{e}_i\end{array}$

$$Au=\sum _{i=\nu +1}^n{u}^{i}A{e}_i,$$

and $\mathrm{im}A=L(\{A{e}_{\nu +1},\dots ,A{e}_n\})$ ). Furthermore the vectors $A{e}_{\nu +1},\dots ,A{e}_n$ are l.i., for if there were a non-trivial linear combination

$$\sum _{i=\nu +1}^n{b}^{i}A{e}_i=A\left(\sum _{i=\nu +1}^n{b}^i{e}_i\right)=0$$

then ${\sum }_{i=\nu +1}^n{b}^i{e}_i\in \ker A$ , which is only possible if all $b^i=0$ . Hence $\dim \mathrm{im}A=n-\dim \ker A$, so that

$$\rho (A)=n-\nu (A)\text{ where }n=\dim V.\text{\hspace{1em}(3.10)}$$

Problems

Problem 3.7

Show that the vectors $(1,x)$ and $(1,y)$ in ${\mathbb{R}}^2$ are linearly dependent iff $x=y$ . In ${\mathbb{R}}^3$ , show that the vectors $(1,x,x^2),(1,y,y^2)$ and $(1,z,z^2)$ are linearly dependent if $x=y\mathrm{or}y=z$ or $x=z$

Generalize these statements to $(n+1)$ dimensions.

Problem 3.8

Let $V$ and $W$ be any vector spaces, which are possibly infinite dimensional, and $T:V\to W$ a linear map. Show that if $M$ is a l.i. subset of $W$, then $T^{-1}(M)=\{v\in V|Tv\in M\}$ is a linearly independent subset of $V$.

Problem 3.9

Let $V$ and $W$ be finite dimensional vector spaces of dimensions $n$ and $m$ respectively, and $T:V\to W$ a linear map. Given a basis $\{e_1,e_2,\dots ,e_n\}$ of $V$ and a basis $\{f_1,f_2,\dots ,f_m\}$ of $W,$ , show that the equations

$$T{e}_k=\sum _{a=1}^m{T}_k^a{f}_a(k=1,2,\dots ,n)$$

serve to uniquely define the $m\times n$ matrix of components $\mathsf{T}=[T_k^a]$ of the linear map $T$ with respec to these bases.

If $v={\sum }_{k=1}^n{v}^k{e}_k$ is an arbitrary vector of $V$ show that the components ofits image vector $w=T$ $v$ are given by

$$w^a=(Tv{)}^a=\sum _{k=1}^n{T}_k^a{v}^k.$$

Write this as a matrix equation.

Problem 3.10

Let $V$ be a four-dimensional vector space and $T:V\to V$ a linear operator whose effect on a basis $e_1,\dots ,e_4$ is

$$\begin{array}{l}T{e}_1=2e_1-e_4\\ T{e}_2=-2e_1+e_4\\ T{e}_3=-2e_1+e_4\\ T{e}_4=e_1.\end{array}$$

Find a basis for $\ker T$ and $\mathrm{im}T$ and calculate the rank and nullity of $T$.