Important Results

THEOREM 3.3.7   Let $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n \}$ be a basis of a given vector space $V.$ If $\{{\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_m \}$ is a set of vectors from $V$ with $m > n$ then this set is linearly dependent.

Proof. Since we want to find whether the set $\{{\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_m \}$ is linearly independent or not, we consider the linear system

$$\alpha_1 {\mathbf w}_1 + \alpha_2 {\mathbf w}_2 + \cdots + \alpha_m {\mathbf w}_m = {\mathbf 0}$$ (3.3.1)

with $\alpha_1, \alpha_2, \ldots, \alpha_m$ as the $m$ unknowns. If the solution set of this linear system of equations has more than one solution, then this set will be linearly dependent.

As $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n \}$ is a basis of $V$ and ${\mathbf w}_i \in V,$ for each $i, \; 1 \leq i \leq m,$ there exist scalars $a_{ij}, \; 1 \leq i \leq n,\; 1 \leq j \leq m,$ such that

$${\mathbf w}_1 = a_{11} {\mathbf v}_1 + a_{21} {\mathbf v}_2 + \cdots + a_{n1} {\mathbf v}_n$$

$${\mathbf w}_2 = a_{12} {\mathbf v}_1 + a_{22} {\mathbf v}_2 + \cdots + a_{n2} {\mathbf v}_n$$

$$\vdots = \vdots$$

$${\mathbf w}_m = a_{1m} {\mathbf v}_1 + a_{2m} {\mathbf v}_2 + \cdots + a_{nm} {\mathbf v}_n.$$

The set of Equations (3.3.1) can be rewritten as

$$\alpha_1 \left( \sum\limits_{j=1}^n a_{j1} {\mathbf v}_j\right) +... ... \alpha_m \left( \sum\limits_{j=1}^n a_{jm} {\mathbf v}_j \right) = {\mathbf 0}$$

$${\mbox{i.e.,}} \left( \sum\limits_{i=1}^m \alpha_i a_{1i}\right) {\mathbf v}_1 +... ... \left( \sum\limits_{i=1}^m \alpha_i a_{ni}\right) {\mathbf v}_n = {\mathbf 0}.$$

Since the set $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n \}$ is linearly independent, we have

$$\sum\limits_{i=1}^m \alpha_i a_{1i}= \sum\limits_{i=1}^m \alpha_i a_{2i}= \cdots = \sum\limits_{i=1}^m \alpha_i a_{ni} = 0.$$

Therefore, finding $\alpha_i$ 's satisfying equation (3.3.1) reduces to solving the system of homogeneous equations $A \alpha = {\mathbf 0}$ where $\alpha^t =( \alpha_1, \alpha_2, \ldots, \alpha_m)$ and $A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cd... ... & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm} \end{bmatrix}.$ Since $n < m,$ i.e., THE NUMBER OF EQUATIONS is strictly less than THE NUMBER OF UNKNOWNS, Corollary 2.5.3 implies that the solution set consists of infinite number of elements. Therefore, the equation (3.3.1) has a solution with not all $\alpha_i, \; 1 \leq i \leq m,$ zero. Hence, the set $\{{\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_m \}$ is a linearly dependent set. height6pt width 6pt depth 0pt

Remark 3.3.8   Let $V$ be a vector subspace of ${\mathbb{R}}^n$ with spanning set $S.$ We give a method of finding a basis of $V$ from $S.$

1. Construct a matrix $A$ whose rows are the vectors in $S.$
2. Use only the elementary row operations $R_i(c)$ and $R_{ij}(c)$ to get the row-reduced form $B$ of $A$ (in fact we just need to make as many zero-rows as possible).
3. Let ${\cal B}$ be the set of vectors in $S$ corresponding to the non-zero rows of $B.$

Then the set ${\cal B}$ is a basis of $L(S)= V.$

EXAMPLE 3.3.9   Let $S= \{ (1,1,1,1), (1,1,-1,1), (1,1,0,1), (1, -1, 1,1)\}$ be a subset of ${\mathbb{R}}^4.$ Find a basis of $L(S).$
Solution: Here $A =\begin{bmatrix}1& 1& 1 & 1 \\ 1& 1 & -1 &1 \\ 1 & 1 & 0 &1\\ 1& -1 & 1 &1 \end{bmatrix}.$ Applying row-reduction to $A,$ we have
$\begin{bmatrix}1& 1& 1 &1 \\ 1& 1 & -1 &1 \\ 1 & 1 & 0 &1\\ 1& -1 & 1 &1 \end{... ...rix}1& 1& 1 &1\\ 0& 0 & 0 &0 \\ 0 & 0 & -1 &0 \\ 0 & -2 & 0 &0 \end{bmatrix}.$
Observe that the rows $1, 3$ and $4$ are non-zero. Hence, a basis of $L(S)$ consists of the first, third and fourth vectors of the set $S.$ Thus, ${\cal B}=\{(1,1,1,1), \; (1,1,0,1),\; (1, -1, 1,1) \}$ is a basis of $L(S).$

Observe that at the last step, in place of the elementary row operation $R_{32}(-2)$ , we can apply $R_{23}(-\frac{1}{2})$ to make the third row as the zero-row. In this case, we get $\{(1,1,1,1), \; (1,1,-1,1),\; (1, -1, 1,1) \}$ as a basis of $L(S)$ .

COROLLARY 3.3.10   Let $V$ be a finite dimensional vector space. Then any two bases of $V$ have the same number of vectors.

Proof. Let $\{{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n\}$ and $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_m \}$ be two bases of $V$ with $m > n.$ Then by the above theorem the set $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_m \}$ is linearly dependent if we take $\{{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n\}$ as the basis of $V.$ This contradicts the assumption that $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_m \}$ is also a basis of $V.$ Hence, we get $m = n.$ height6pt width 6pt depth 0pt

DEFINITION 3.3.11 (Dimension of a Vector Space)   The dimension of a finite dimensional vector space $V$ is the number of vectors in a basis of $V,$ denoted $\dim(V).$

Note that the Corollary 3.2.6 can be used to generate a basis of ANY NON-TRIVIAL FINITE DIMENSIONAL VECTOR SPACE.

EXAMPLE 3.3.12  

1. Consider the complex vector space ${\mathbb{C}}^2 ({\mathbb{C}}).$ Then,
   $$(a+ i b, c + i d) = (a+ i b) (1 , 0) + (c + i d) ( 0, 1).$$
   So, $\{(1, 0), (0,1) \}$ is a basis of ${\mathbb{C}}^2 ({\mathbb{C}})$ and thus $\dim (V) = 2.$
2. Consider the real vector space ${\mathbb{C}}^2 ({\mathbb{R}}).$ In this case, any vector
   $$(a + i b, c + i d) = a (1, 0) + b (i, 0) + c (0,1) + d (0, i).$$
   Hence, the set $\{(1,0), (i, 0), (0,1), (0, i)\}$ is a basis and $\dim (V) = 4.$

Remark 3.3.13   It is important to note that the dimension of a vector space may change if the underlying field (the set of scalars) is changed.

EXAMPLE 3.3.14   Let $V$ be the set of all functions $f: {\mathbb{R}}^n {\longrightarrow}{\mathbb{R}}$ with the property that $f({\mathbf x}+ {\mathbf y}) = f({\mathbf x}) + f({\mathbf y})$ and $f( \alpha {\mathbf x}) = \alpha f({\mathbf x}).$ For $f, g \in V,$ and $t \in {\mathbb{R}},$ define

$$(f \oplus g)({\mathbf x}) = f({\mathbf x}) + g({\mathbf x}) \;\; {\mbox{ and }}$$

$$(t \odot f) ({\mathbf x}) = f( t {\mathbf x}).$$

Then $V$ is a real vector space.

For $1 \leq i \leq n,$ consider the functions

$${\mathbf e_i}({\mathbf x}) = {\mathbf e_i} \bigl( (x_1, x_2, \ldots, x_n) \bigr) = x_i.$$

Then it can be easily verified that the set $\{ {\mathbf e_1}, {\mathbf e_2}, \ldots, {\mathbf e_n} \}$ is a basis of $V$ and hence $\dim (V) = n.$

The next theorem follows directly from Corollary 3.2.6 and Theorem 3.3.7. Hence, the proof is omitted.

THEOREM 3.3.15   Let $S$ be a linearly independent subset of a finite dimensional vector space $V.$ Then the set $S$ can be extended to form a basis of $V.$

Theorem 3.3.15 is equivalent to the following statement:
Let $V$ be a vector space of dimension $n.$ Suppose, we have found a linearly independent set
[4] $S = \{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_r\} \subset V.$ Then there exist vectors ${\mathbf v}_{r+1}, \ldots, {\mathbf v}_n$ in $V$ such that $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n \}$ is a basis of $V.$

COROLLARY 3.3.16   Let $V$ be a vector space of dimension $n.$ Then any set of $n$ linearly independent vectors forms a basis of $V.$ Also, every set of $m$ vectors, $m > n,$ is linearly dependent.

COROLLARY 3.3.17   Let $V$ be a vector space of dimension $n.$ Then any set of vectors $S = \{{\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n \}$ such that $L(S) = V$ forms a basis of $V$ .

EXAMPLE 3.3.18   Let $V = \{(v,w,x,y,z) \in {\mathbb{R}}^5\; : v + x - 3y + z = 0\}$ and $W = \{(v,w,x,y,z) \in {\mathbb{R}}^5\; : w -x - z = 0, v = y\}$ be two subspaces of ${\mathbb{R}}^5.$ Find bases of $V$ and $W$ containing a basis of $V \cap W.$
Solution: Let us find a basis of $V \cap W.$ The solution set of the linear equations

$$v + x - 3y + z = 0, \;\; w -x - z = 0 \;\; {\mbox{ and }} \;\; v = y$$

is given by

$$(v,w,x,y,z)^t = (y, 2y, x, y, 2y-x)^t = y (1,2,0,1,2)^t + x (0,0,1,0,-1)^t.$$

Thus, a basis of $V \cap W$ is

$$\{ (1,2,0,1,2), (0,0,1,0,-1)\}.$$

To find a basis of $W$ containing a basis of $V \cap W,$ we can proceed as follows:

1. Find a basis of $W.$
2. Take the basis of $V \cap W$ found above as the first two vectors and that of $W$ as the next set of vectors.

   Now use Remark 3.3.8 to get the required basis.

Heuristically, we can also find the basis in the following way:
A vector of $W$ has the form $(y,x+z,x,y,z)$ for $x,y,z \in {\mathbb{R}}.$ Substituting $y=1,x=1,$ and $z = 0$ in $(y,x+z,x,y,z)$ gives us the vector $(1,1,1,1,0) \in W.$ It can be easily verified that a basis of $W$ is

$$\{ (1,2,0,1,2), (0,0,1,0,-1), (1,1,1,1,0) \}.$$

Similarly, a vector of $V$ has the form $(v,w,x,y,3y-v-x)$ for $v,w, x,y\in {\mathbb{R}}.$ Substituting $v=0,w=1,x=0$ and $y=0,$ gives a vector $(0,1,0,0,0) \in V.$ Also, substituting $v=0,w=1,x=1$ and $y=1,$ gives another vector $(0,1,1,1,2) \in V.$ So, a basis of $V$ can be taken as

$$\{ (1,2,0,1,2), (0,0,1,0,-1), (0,1,0,0,0), (0,1,1,1,2)\}.$$

Recall that for two vector subspaces $M$ and $N$ of a vector space $V({\mathbb{F}}),$ the vector subspace $M + N$ (see Exercise 3.1.18.7) is defined by

$$M+N = \{ {\mathbf u}+ {\mathbf v}: {\mathbf u}\in M, \; {\mathbf v}\in N \}.$$

With this definition, we have the following very important theorem (for a proof, see Appendix 14.4.1).

THEOREM 3.3.19   Let $V ({\mathbb{F}})$ be a finite dimensional vector space and let $M$ and $N$ be two subspaces of $V.$ Then

$$\dim (M) + \dim(N) = \dim (M + N) + \dim (M \cap N).$$ (3.3.2)

EXERCISE 3.3.20  

1. Find a basis of the vector space ${\cal P}_n({\mathbb{R}}).$ Also, find $\dim ( {\cal P}_n ({\mathbb{R}}) ).$ What can you say about the dimension of ${\cal P}({\mathbb{R}})?$
2. Consider the real vector space, $C([0, 2 \pi]),$ of all real valued continuous functions. For each $n$ consider the vector ${\mathbf e}_n$ defined by ${\mathbf e}_n(x) = \sin (n x).$ Prove that the collection of vectors $\{ {\mathbf e}_n : 1 \leq n < \infty \}$ is a linearly independent set.
   [Hint: On the contrary, assume that the set is linearly dependent. Then we have a finite set of vectors, say $\{{\mathbf e}_{k_1}, {\mathbf e}_{k_2}, \ldots, {\mathbf e}_{k_\ell}\}$ that are linearly dependent. That is, there exist scalars ${\alpha}_i \in {\mathbb{R}}$ for $1 \leq i \leq \ell$ not all zero such that
   $${\alpha}_1 \sin(k_1 x) + {\alpha}_2 \sin (k_2 x) + \cdots + {\alpha}_\ell \sin(k_\ell x) = {\mathbf 0}\;\;$$    for all $$\;\; x \in [0, 2 \pi].$$
   Now for different values of $m$ integrate the function
   $$\int_0^{2 \pi} \sin(m x) \left( {\alpha}_1 \sin(k_1 x) + {\alpha}_2 \sin (k_2 x) + \cdots + {\alpha}_\ell \sin(k_\ell x) \right)$$    dx$$$$
   to get the required result.]
3. Show that the set $\{(1,0,0), (1,1,0), (1,1,1) \}$ is a basis of ${\mathbb{C}}^3 ({\mathbb{C}}).$ Is it a basis of ${\mathbb{C}}^3 ({\mathbb{R}})$ also?
4. Let $W= \{(x,y,z,w) \in {\mathbb{R}}^4 : x + y - z + w = 0 \}$ be a subspace of ${\mathbb{R}}^4.$ Find its basis and dimension.
5. Let $V = \{(x,y,z,w) \in {\mathbb{R}}^4 : x + y - z + w = 0, x + y + z + w = 0 \}$ and $W = \{(x,y,z,w)\in {\mathbb{R}}^4 : x - y - z + w = 0, x + 2 y -w = 0 \}$ be two subspaces of ${\mathbb{R}}^4.$ Find bases and dimensions of $V,$ $W,$ $V \cap W$ and $V + W.$
6. Let $V$ be the set of all real symmetric $n \times n$ matrices. Find its basis and dimension. What if $V$ is the complex vector space of all $n \times n$ Hermitian matrices?
7. If $M$ and $N$ are $4$ -dimensional subspaces of a vector space $V$ of dimension $7$ then show that $M$ and $N$ have at least one vector in common other than the zero vector.
8. Let $P=L\{(1,0,0), (1,1,0) \}$ and $Q=L\{(1,1,1)\}$ be vector subspaces of ${\mathbb{R}}^3.$ Show that $P+Q = {\mathbb{R}}^3$ and $P \cap Q = \{{\mathbf 0}\}.$ If ${\mathbf u}\in {\mathbb{R}}^3,$ determine ${\mathbf u}_P, {\mathbf u}_Q$ such that ${\mathbf u}= {\mathbf u}_P + {\mathbf u}_Q$ where ${\mathbf u}_P \in P$ and ${\mathbf u}_Q \in Q.$ Is it necessary that ${\mathbf u}_P$ and ${\mathbf u}_Q$ are unique?
9. Let $W_1$ be a $k$ -dimensional subspace of an $n$ -dimensional vector space $V ({\mathbb{F}})$ where $k \geq 1.$ Prove that there exists an $(n-k)$ -dimensional subspace $W_2$ of $V$ such that $W_1 \cap W_2 = \{ {\mathbf 0}\}$ and $W_1 + W_2 = V.$
10. Let $P$ and $Q$ be subspaces of ${\mathbb{R}}^n$ such that $P + Q = {\mathbb{R}}^n$ and $P \cap Q = \{{\mathbf 0}\}.$ Then show that each ${\mathbf u}\in {\mathbb{R}}^n$ can be uniquely expressed as ${\mathbf u}= {\mathbf u}_P + {\mathbf u}_Q$ where ${\mathbf u}_P \in P$ and ${\mathbf u}_Q \in Q.$
11. Let $P=L\{(1,-1,0), (1,1,0) \}$ and $Q=L\{(1,1,1), (1,2,1)\}$ be vector subspaces of ${\mathbb{R}}^3.$ Show that $P+Q = {\mathbb{R}}^3$ and $P \cap Q \neq \{{\mathbf 0}\}.$ Show that there exists a vector ${\mathbf u}\in {\mathbb{R}}^3$ such that ${\mathbf u}$ cannot be written uniquely in the form ${\mathbf u}= {\mathbf u}_P + {\mathbf u}_Q$ where ${\mathbf u}_P \in P$ and ${\mathbf u}_Q \in Q.$
12. Let $W_1$ and $W_2$ be two subspaces of a vector space $V$ such that $W_1 \subset W_2$ . Show that $W_1 = W_2$ if and only if $\dim(W_1) = \dim(W_2)$ .
13. Let $W_1$ and $W_2$ be two subspaces of a vector space $V$ . If $\dim(W_1) + \dim(W_2) > \dim(V)$ , then prove that $W_1 \cap W_2$ contains a non-zero vector.
14. Recall the vector space ${\cal P}_4({\mathbb{R}}).$ Is the set,
    $$W = \{p(x) \in {\cal P}_4({\mathbb{R}}) \; : \; p(-1) = p(1) = 0 \}$$
    a subspace of ${\cal P}_4({\mathbb{R}})?$ If yes, find its dimension.
15. Let $V$ be the set of all $2 \times 2$ matrices with complex entries and $a_{11} + a_{22} = 0.$ Show that $V$ is a real vector space. Find its basis. Also let $W = \{A \in V : a_{21} = {\overline{- a_{12}}} \}.$ Show $W$ is a vector subspace of $V,$ and find its dimension.
16. Let $A = \begin{bmatrix}1 & 2 & 1 & 3 & 2\\ 0 & 2 & 2 & 2 & 4\\ 2 & -2 & 4 & 0 & 8\\ 4 & 2 & 5 & 6 & 10 \end{bmatrix},$ and $B =\begin{bmatrix}2 & 4 & 0 & 6\\ -1 & 0 & -2 &5 \\ -3 & -5 & 1 & -4 \\ -1 & -1 & 1 & 2 \end{bmatrix}$ be two matrices. For $A$ and $B$ find the following:
    1. their row-reduced echelon forms.
    2. the matrices $P_1$ and $P_2$ such that $P_1 A$ and $P_2 B$ are in row-reduced form.
    3. a basis each for the row spaces of $A$ and $B.$
    4. a basis each for the range spaces of $A$ and $B.$
    5. bases of the null spaces of $A$ and $B.$
    6. the dimensions of all the vector subspaces so obtained.
17. Let $M(n,{\mathbb{R}})$ denote the space of all $n \times n$ real matrices. For the sets given below, check that they are subspaces of $M(n,{\mathbb{R}})$ and also find their dimension.
    1. ${\mathit sl}(n,{\mathbb{R}}) = \{ A \in M(n,{\mathbb{R}}) \; : \; {\mbox{tr}}(A) = 0\},$ where recall that ${\mbox{tr }}(A)$ stands for trace of $A.$
    2. $S(n,{\mathbb{R}}) = \{ A \in M(n,{\mathbb{R}}) \; : \; A = A^t\}.$
    3. $A(n,{\mathbb{R}}) = \{ A \in M(n,{\mathbb{R}}) \; : \; A+A^t = {\mathbf 0}\}.$

Before going to the next section, we prove that for any matrix $A$ of order $m \times n$

$${\mbox{Row rank}}(A) = {\mbox{Column rank}}(A).$$

PROPOSITION 3.3.21   Let $A$ be an $m \times n$ real matrix. Then

$${\mbox{Row rank}}(A) = {\mbox{Column rank}}(A).$$

Proof. Let $R_1, R_2, \ldots, R_m$ be the rows of $A$ and $C_1, C_2, \ldots, C_n$ be the columns of $A.$ Note that ${\mbox{Row rank}}(A)= r,$ means that

$$\dim \bigl( L(R_1, R_2, \ldots, R_m) \bigr) = r.$$

Hence, there exists vectors

$${\mathbf u}_1= (u_{11}, \ldots, u_{1n}), {\mathbf u}_2=(u_{21}, \ldots, u_{2n}), \ldots, {\mathbf u}_r=(u_{r1}, \ldots, u_{rn}) \in {\mathbb{R}}^n$$

with

$$R_i \in L ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_r) \in {\mathbb{R}}^n, \; {\mbox{ for all }} \; i, 1 \leq i \leq m.$$

Therefore, there exist real numbers ${\alpha}_{ij}, \; 1 \leq i \leq m, \; 1 \leq j \leq r$ such that

$$R_1 = {\alpha}_{11} {\mathbf u}_1 + {\alpha}_{12}{\mathbf u}_2 + ... ..., \sum_{i=1}^r {\alpha}_{1i}u_{i2}, \ldots, \sum_{i=1}^r {\alpha}_{1i}u_{in} ),$$

$$R_2 = {\alpha}_{21} {\mathbf u}_1 + {\alpha}_{22}{\mathbf u}_2 + ... ..., \sum_{i=1}^r {\alpha}_{2i}u_{i2}, \ldots, \sum_{i=1}^r {\alpha}_{2i}u_{in} ),$$

and so on, till

$$R_m = {\alpha}_{m1} {\mathbf u}_1 + \cdots + {\alpha}_{mr} {\math... ..., \sum_{i=1}^r {\alpha}_{mi}u_{i2}, \ldots, \sum_{i=1}^r {\alpha}_{mi}u_{in} ).$$

So,

$$C_1 = \begin{bmatrix}\sum\limits_{i=1}^r {\alpha}_{1i}u_{i1} \\ ... ...atrix} {\alpha}_{1r} \\ {\alpha}_{2r} \\ \vdots \\ {\alpha}_{mr} \end{bmatrix}.$$

In general, for $1 \leq j \leq n,$ we have

$$C_j = \begin{bmatrix}\sum\limits_{i=1}^r {\alpha}_{1i}u_{ij} \\ ... ...atrix} {\alpha}_{1r} \\ {\alpha}_{2r} \\ \vdots \\ {\alpha}_{mr} \end{bmatrix}.$$

Therefore, we observe that the columns $C_1, C_2, \ldots, C_n$ are linear combination of the $r$ vectors

$$({\alpha}_{11}, {\alpha}_{21}, \ldots, {\alpha}_{m1})^t, ({\alpha... ...lpha}_{m2})^t, \ldots, ({\alpha}_{1r}, {\alpha}_{2r}, \ldots, {\alpha}_{mr})^t.$$

Therefore,

$${\mbox{Column rank}}(A) = \dim\bigl( L(C_1, C_2, \ldots, C_n) \bigr) = \leq r = {\mbox{Row rank}}(A).$$

A similar argument gives

$${\mbox{Row rank}}(A) \leq {\mbox{Column rank}}(A).$$

Thus, we have the required result. height6pt width 6pt depth 0pt