Rank-Nullity Theorem

DEFINITION 4.3.1 (Range and Null Space)   Let $V, W$ be finite dimensional vector spaces over the same set of scalars and $T: V {\longrightarrow}W$ be a linear transformation. We define

1. ${\cal R}(T) = \{ T({\mathbf x}) : {\mathbf x}\in V \},$ and
2. ${\cal N}(T) = \{{\mathbf x}\in V : T({\mathbf x}) = {\mathbf 0}\}.$

We now prove some results associated with the above definitions.

PROPOSITION 4.3.2   Let $V$ and $W$ be finite dimensional vector spaces and let $T: V {\longrightarrow}W$ be a linear transformation. Suppose that $({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n)$ is an ordered basis of $V.$ Then

1. 1. ${\cal R}(T)$ is a subspace of $W.$
   2. ${\cal R}(T) = L( T({\mathbf v}_1), T({\mathbf v}_2), \ldots, T({\mathbf v}_n) ).$
   3. $\dim({\cal R}(T)) \leq \dim(W).$
2. 1. ${\cal N}(T)$ is a subspace of $V.$
   2. $\dim({\cal N}(T)) \leq \dim(V).$
3. $T$ is one-one $\Longleftrightarrow \; \; {\cal N}(T) = \{{\mathbf 0}\}$ is the zero subspace of $V \Longleftrightarrow \; \;$ $\{ T(u_i): 1 \leq i \leq n \}$ is a basis of ${\cal R}(T).$
4. $\dim ({\cal R}(T)) = \dim (V)$ if and only if ${\cal N}(T) = \{{\mathbf 0}\}.$

Proof. The results about ${\cal R}(T)$ and ${\cal N}(T)$ can be easily proved. We thus leave the proof for the readers.
We now assume that $T$ is one-one. We need to show that ${\cal N}(T) = \{{\mathbf 0}\}.$
Let ${\mathbf u}\in {\cal N}(T).$ Then by definition, $\; T({\mathbf u}) = {\mathbf 0}.$ Also for any linear transformation (see Proposition 4.1.3), $T({\mathbf 0}) = {\mathbf 0}.$ Thus $T({\mathbf u}) = T({\mathbf 0}).$ So, $T$ is one-one implies ${\mathbf u}= {\mathbf 0}.$ That is, ${\cal N}(T) = \{{\mathbf 0}\}.$

Let ${\cal N}(T) = \{{\mathbf 0}\}.$ We need to show that $T$ is one-one. So, let us assume that for some ${\mathbf u}, {\mathbf v}\in V, \; T({\mathbf u}) = T({\mathbf v}).$ Then, by linearity of $T, \; T({\mathbf u}- {\mathbf v}) = {\mathbf 0}.$ This implies, ${\mathbf u}- {\mathbf v}\in {\cal N}(T) = \{{\mathbf 0}\}.$ This in turn implies ${\mathbf u}= {\mathbf v}.$ Hence, $T$ is one-one.

The other parts can be similarly proved. height6pt width 6pt depth 0pt

Remark 4.3.3  

1. The space ${\cal R}(T)$ is called the RANGE SPACE of $T$ and ${\cal N}(T)$ is called the NULL SPACE of $T.$
2. We write $\rho(T) = \dim ({\cal R}(T))$ and $\nu(T) = \dim ({\cal N}(T)).$
3. $\rho(T)$ is called the rank of the linear transformation $T$ and $\nu(T)$ is called the nullity of $T.$

EXAMPLE 4.3.4   Determine the range and null space of the linear transformation

$$T: {\mathbb{R}}^3 {\longrightarrow}{\mathbb{R}}^4 \;\; {\mbox{ with }} \;\; T(x,y,z) = (x-y+z, y-z,x, 2x - 5y +5z).$$

Solution: By Definition ${\cal R}(T) = L ( T(1,0,0), T(0,1,0), T(0,0,1) ).$ We therefore have

$${\cal R}(T) = L \bigl( (1,0,1,2), (-1,1,0,-5), (1,-1,0,5) \bigr)$$

$$= L \bigl( (1,0,1,2), (1,-1,0,5) \bigr)$$

$$= \{ {\alpha}(1,0,1,2) + \beta (1,-1,0,5) \; : {\alpha}, \beta \in {\mathbb{R}}\}$$

$$= \{ ({\alpha}+ \beta, -\beta, {\alpha}, 2 {\alpha}+ 5 \beta) \; : {\alpha}, \beta \in {\mathbb{R}}\}$$

$$= \{(x,y,z,w) \in {\mathbb{R}}^4 \; : x +y - z = 0, 5 y - 2 z + w = 0 \}.$$

Also, by definition

$${\cal N}(T) = \{ (x,y,z) \in {\mathbb{R}}^3 \; : T(x,y,z) = {\mathbf 0}\}$$

$$= \{ (x,y,z) \in {\mathbb{R}}^3 \; : (x-y+z, y-z,x, 2x-5y+5z) = {\mathbf 0}\}$$

$$= \{ (x,y,z) \in {\mathbb{R}}^3 \; : x-y+ z = 0, y-z = 0,$$

$$\hspace{1.75in} x = 0, 2x - 5y + 5z=0 \}$$

$$= \{ (x,y,z) \in {\mathbb{R}}^3 \; : y-z = 0, x = 0 \}$$

$$= \{ (x,y,z) \in {\mathbb{R}}^3 \; : y=z, x = 0 \}$$

$$= \{ (0,y,y) \in {\mathbb{R}}^3 \; : y \;\; {\mbox{arbitrary}} \}$$

$$= L ( (0,1,1) )$$

EXERCISE 4.3.5  

1. Let $T: V {\longrightarrow}W$ be a linear transformation and let $\{T({\mathbf v}_1), T({\mathbf v}_2), \ldots, T({\mathbf v}_n)\}$ be linearly independent in ${\cal R}(T).$ Prove that $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n\} \subset V$ is linearly independent.
2. Let $T: {\mathbb{R}}^2 {\longrightarrow}{\mathbb{R}}^3$ be defined by
   $$T\bigl( (1,0) \bigr)= (1,0,0), \; T\bigl( (0,1) \bigr) = (1, 0, 0).$$
   Then the vectors $(1,0)$ and $(0,1)$ are linearly independent whereas $T\bigl( (1,0) \bigr)$ and $T\bigl( (0,1) \bigr)$ are linearly dependent.
3. Is there a linear transformation
   $$T: {\mathbb{R}}^3 \longrightarrow {\mathbb{R}}^2 \; {\mbox{ such that }} \; T(1,-1,1) = (1,2),\;\; {\mbox{ and }} \;\; T(-1,1,2) = (1,0)?$$

4. Recall the vector space ${{\cal P}}_n ({\mathbb{R}}).$ Define a linear transformation
   $$D: {{\cal P}}_n({\mathbb{R}}) {\longrightarrow}{{\cal P}}_n({\mathbb{R}})$$
   by
   $$D(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n) = a_1 + 2 a_2 x + \cdots + n a_n x^{n-1}.$$
   Describe the null space and range space of $D.$ Note that the range space is contained in the space ${{\cal P}}_{n-1}({\mathbb{R}}).$
5. Let $T : {\mathbb{R}}^3 \longrightarrow {\mathbb{R}}^3$ be defined by
   $$T(1,0,0) = (0,0,1),\;\; T(1,1,0) = (1,1,1)\; {\mbox{ and }} \; T(1,1,1) = (1,1,0).\;\;$$
   1. Find $T(x,y,z)$ for $x,y,z \in {\mathbb{R}},$
   2. Find ${\cal R}(T)$ and ${\cal N}(T).$ Also calculate $\rho(T)$ and $\nu(T).$
   3. Show that $T^3 = T$ and find the matrix of the linear transformation with respect to the standard basis.
6. Let $T: {\mathbb{R}}^2 \longrightarrow {\mathbb{R}}^2$ be a linear transformation with
   $$T( (3,4) ) = (0,1), \; T( (-1,1) ) = (2,3).$$
   Find the matrix representation $T[{\cal B}, {\cal B}]$ of $T$ with respect to the ordered basis ${\cal B}= \bigl( (1,0), (1,1) \bigr)$ of ${\mathbb{R}}^2.$
7. Determine a linear transformation $T : {\mathbb{R}}^3 \longrightarrow {\mathbb{R}}^3$ whose range space is $L \{ (1,2,0), (0,1,1), (1,3,1) \}.$
8. Let $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n \}$ be a basis of a vector space $V ({\mathbb{F}})$ . If $W({\mathbb{F}})$ is a vector space and ${\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_n \in W$ then prove that there exists a unique linear transformation $T: V {\longrightarrow}W$ such that $T({\mathbf v}_i) = {\mathbf w}_i$ for all $i = 1, 2, \ldots, n$ .
9. Suppose the following chain of matrices is given.
   $$A \longrightarrow B_1 \longrightarrow B_1 \longrightarrow B_2 \cdots \longrightarrow B_{k-1} \longrightarrow B_k \longrightarrow B.$$
   If row space of $B$ is in the row space of $B_k$ and the row space of $B_l$ is in the row space of $B_{l-1}$ for $2 \leq l \leq k$ then show that the row space of $B$ is in the row space of $A.$

We now state and prove the rank-nullity Theorem. This result also follows from Proposition 4.3.2.

THEOREM 4.3.6 (Rank Nullity Theorem)   Let $T: V {\longrightarrow}W$ be a linear transformation and $V$ be a finite dimensional vector space. Then

$$\dim ( {\cal R}(T)) + \dim ({\cal N}(T)) = \dim (V),$$

or equivalently $\rho(T) + \nu (T) = \dim(V).$

Proof. Let $\dim(V) = n$ and $\dim ({\cal N}(T)) = r.$ Suppose $\{u_1, u_2, \ldots, u_r \}$ is a basis of ${\cal N}(T).$ Since $\{u_1, u_2, \ldots, u_r \}$ is a linearly independent set in $V,$ we can extend it to form a basis of $V$ (see Corollary 3.3.15). So, there exist vectors $\{u_{r+1}, u_{r+2}, \ldots, u_n \}$ such that $\{u_1, \ldots, u_r, u_{r+1}, \ldots, u_n \}$ is a basis of $V.$ Therefore, by Proposition 4.3.2

$${\cal R}(T) = L ( T(u_1), T(u_2), \ldots, T(u_n) )$$

$$= L ( {\mathbf 0}, \ldots, {\mathbf 0}, T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) )$$

$$= L ( T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) ).$$

We now prove that the set $\{ T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) \}$ is linearly independent. Suppose the set is not linearly independent. Then, there exists scalars, $\alpha_{r+1}, \alpha_{r+2}, \ldots, \alpha_n,$ not all zero such that

$$\alpha_{r+1} T(u_{r+1}) + \alpha_{r+2} T(u_{r+2}) + \cdots + \alpha_n T(u_n) = {\mathbf 0}.$$

That is,

$$T ( \alpha_{r+1} u_{r+1} + \alpha_{r+2} u_{r+2} + \cdots + \alpha_n u_n )= {\mathbf 0}.$$

So, by definition of ${\cal N}(T),$

$$\alpha_{r+1} u_{r+1} + \alpha_{r+2} u_{r+2} + \cdots + \alpha_n u_n \in {\cal N}(T) = L ( u_1, \ldots, u_r).$$

Hence, there exists scalars $\alpha_i, \; 1 \leq i \leq r$ such that

$$\alpha_{r+1} u_{r+1} + \alpha_{r+2} u_{r+2} + \cdots + \alpha_n u_n = \alpha_{1} u_{1} + \alpha_{2} u_{2} + \cdots + \alpha_r u_r.$$

That is,

$$\alpha_1 u_1 + + \cdots + \alpha_r u_r - \alpha_{r+1} u_{r+1} - \cdots - \alpha_n u_n = {\mathbf 0}.$$

But the set $\{ u_1, u_2, \ldots, u_n \}$ is a basis of $V$ and so linearly independent. Thus by definition of linear independence

$$\alpha_i = 0 \; {\mbox{ for all }} \; i, \; 1 \leq i \leq n.$$

In other words, we have shown that $\{ T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) \}$ is a basis of ${\cal R}(T).$ Hence,

$$\dim({\cal R}(T)) + \dim({\cal N}(T)) = (n-r) + r = n = \dim(V).$$

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Using the Rank-nullity theorem, we give a short proof of the following result.

COROLLARY 4.3.7   Let $T: V {\longrightarrow}V$ be a linear transformation on a finite dimensional vector space $V.$ Then

$$T {\mbox{ is one-one }} \Longleftrightarrow T {\mbox{ is onto}} \Longleftrightarrow T {\mbox{ is invertible}}.$$

Proof. By Proposition 4.3.2, $T$ is one-one if and only if ${\cal N}(T) = \{{\mathbf 0}\}.$ By the rank-nullity Theorem 4.3.6 ${\cal N}(T) = \{{\mathbf 0}\}$ is equivalent to the condition $\dim ( {\cal R}(T)) = \dim (V).$ Or equivalently $T$ is onto.

By definition, $T$ is invertible if $T$ is one-one and onto. But we have shown that $T$ is one-one if and only if $T$ is onto. Thus, we have the last equivalent condition. height6pt width 6pt depth 0pt

Remark 4.3.8   Let $V$ be a finite dimensional vector space and let $T: V {\longrightarrow}V$ be a linear transformation. If either $T$ is one-one or $T$ is onto, then $T$ is invertible.

The following are some of the consequences of the rank-nullity theorem. The proof is left as an exercise for the reader.

COROLLARY 4.3.9   The following are equivalent for an $m \times n$ real matrix $A.$

1. ${\mbox{ Rank }}(A) = k.$
2. There exist exactly $k$ rows of $A$ that are linearly independent.
3. There exist exactly $k$ columns of $A$ that are linearly independent.
4. There is a $k \times k$ submatrix of $A$ with non-zero determinant and every $(k + 1) \times (k+1)$ submatrix of $A$ has zero determinant.
5. The dimension of the range space of $A$ is $k.$
6. There is a subset of ${\mathbb{R}}^m$ consisting of exactly $k$ linearly independent vectors ${\mathbf b}_1, {\mathbf b}_2, \ldots, {\mathbf b}_k$ such that the system $A {\mathbf x}= {\mathbf b}_i$ for $1 \leq i \leq k$ is consistent.
7. The dimension of the null space of $A = n - k.$

EXERCISE 4.3.10  

1. Let $T: V {\longrightarrow}W$ be a linear transformation.
   1. If $V$ is finite dimensional then show that the null space and the range space of $T$ are also finite dimensional.
   2. If $V$ and $W$ are both finite dimensional then show that
      1. if $\dim(V) < \dim(W)$ then $T$ is onto.
      2. if $\dim(V) > \dim(W)$ then $T$ is not one-one.
2. Let $A$ be an $m \times n$ real matrix. Then
   1. if $n > m,$ then the system $A {\mathbf x}= {\mathbf 0}$ has infinitely many solutions,
   2. if $n < m,$ then there exists a non-zero vector ${\mathbf b}=(b_1, b_2, \ldots, b_m)^t$ such that the system $A {\mathbf x}= {\mathbf b}$ does not have any solution.
3. Let $V$ be a vector space of dimension $n$ and let ${\cal B}= ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n)$ be an ordered basis of $V$ . Suppose ${\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_n \in V$ and let $[{\mathbf w}_i]_{{\cal B}} = [ a_{1i}, \; a_{2i}, \ldots, a_{ni} \;]^t$ . Put $A=[a_{ij}]$ . Then prove that ${\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_n$ is a basis of $V$ if and only if the matrix $A$ is invertible.
4. Let $A$ be an $m \times n$ matrix. Prove that
   ${\mbox{ Row Rank }}(A) = {\mbox{ Column Rank }}(A).$
   [Hint: Define $T_A: {\mathbb{R}}^n {\longrightarrow}{\mathbb{R}}^m$ by $T_A({\mathbf v}) = A {\mathbf v}$ for all ${\mathbf v}\in {\mathbb{R}}^n.$ Let ${\mbox{ Row Rank }}(A) = r.$ Use Theorem 2.5.1 to show, $A {\mathbf x}= {\mathbf 0}$ has $n-r$ linearly independent solutions. This implies,
   $\nu(T_A) = \dim ( \{ {\mathbf v}\in {\mathbb{R}}^n: T_A({\mathbf v}) = {\mathb... ... \dim ( \{ {\mathbf v}\in {\mathbb{R}}^n: A {\mathbf v}= {\mathbf 0}\}) = n-r.$
   Now observe that ${\cal R}(T_A)$ is the linear span of columns of $A$ and use the rank-nullity Theorem 4.3.6 to get the required result.]
5. Prove Theorem 2.5.1.
   [Hint: Consider the linear system of equation $A {\mathbf x}= {\mathbf b}$ with the orders of $A, {\mathbf x}$ and ${\mathbf b},$ respectively as $m \times n, n \times 1$ and $m \times 1.$ Define a linear transformation $T : {\mathbb{R}}^n {\longrightarrow} {\mathbb{R}}^m$ by $T({\mathbf v}) = A {\mathbf v}.$ First observe that if the solution exists then ${\mathbf b}$ is a linear combination of the columns of $A$ and the linear span of the columns of $A$ give us ${\cal R}(T).$ Note that $\rho(A) = {\mbox{ column rank}} (A) = \dim ({\cal R}(T)) = \ell {\mbox{(say)}}.$ Then for part $i)$ one can proceed as follows.
   $i)$ Let $C_{i_1}, C_{i_2}, \ldots, C_{i_\ell}$ be the linearly independent columns of $A.$ Then ${\mbox{rank}} (A) < {\mbox{rank}} ([A \; b])$ implies that $\{ C_{i_1}, C_{i_2}, \ldots, C_{i_\ell}, {\mathbf b}\}$ is linearly independent. Hence ${\mathbf b}\not\in L(C_{i_1}, C_{i_2}, \ldots, C_{i_\ell}).$ Hence, the system doesn't have any solution.

   On similar lines prove the other two parts.]

6. Let $T, S: V {\longrightarrow}V$ be linear transformations with $\dim (V) = n.$
   1. Show that ${\cal R}(T+S) \subset {\cal R}(T) + {\cal R}(S).$ Deduce that $\rho(T+S) \leq \rho(T) + \rho(S).$
      Hint: For two subspaces $M , N$ of a vector space $V,$ recall the definition of the vector subspace $M+N.$
   2. Use the above and the rank-nullity Theorem 4.3.6 to prove $\nu(T+S) \geq \nu(T) + \nu(S) - n.$
7. Let $V$ be the complex vector space of all complex polynomials of degree at most $n$ . Given $k$ distinct complex numbers $z_1, z_2, \ldots, z_k,$ we define a linear transformation
   $$T: V \longrightarrow {\mathbb{C}}^k \;\; {\mbox{ by }} \;\; T\bigl( P(z) \bigr) = \bigl( P(z_1), P(z_2), \ldots, P(z_k) \bigr).$$
   For each $k \ge 1$ , determine the dimension of the range space of $T.$
8. Let $A$ be an $n \times n$ real matrix with $A^2 = A.$ Consider the linear transformation $T_A: {\mathbb{R}}^n \longrightarrow {\mathbb{R}}^n,$ defined by $T_A({\mathbf v}) = A {\mathbf v}$ for all ${\mathbf v}\in {\mathbb{R}}^n.$ Prove that
   1. $T_A \circ T_A = T_A$ (use the condition $A^2 = A$ ).
   2. ${\cal N}(T_A) \cap {\cal R}(T_A) = \{{\mathbf 0}\}.$
      Hint: Let ${\mathbf x}\in {\cal N}(T_A) \cap {\cal R}(T_A).$ This implies $T_A({\mathbf x}) = {\mathbf 0}$ and ${\mathbf x}= T_A({\mathbf y})$ for some ${\mathbf y}\in {\mathbb{R}}^n.$ So,
      $${\mathbf x}= T_A({\mathbf y}) = (T_A \circ T_A)({\mathbf y}) = T_A \bigl( T_A({\mathbf y}) \bigr) = T_A ( {\mathbf x}) = {\mathbf 0}.$$

   3. ${\mathbb{R}}^n = {\cal N}(T_A) + {\cal R}(T_A).$
      Hint: Let $\{{\mathbf v}_1, \ldots, {\mathbf v}_k\}$ be a basis of ${\cal N}(T_A).$ Extend it to get a basis $\{{\mathbf v}_1, \ldots, {\mathbf v}_k, {\mathbf v}_{k+1}, \ldots, {\mathbf v}_n\}$ of ${\mathbb{R}}^n.$ Then by Rank-nullity Theorem 4.3.6, $\{T_A({\mathbf v}_{k+1}), \ldots, T_A({\mathbf v}_n)\}$ is a basis of ${\cal R}(T_A).$