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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. $ {\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).$
    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

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


Solution: By Definition $ {\cal R}(T) = L ( T(1,0,0), T(0,1,0), T(0,0,1) ).$ We therefore have
$\displaystyle {\cal R}(T)$ $\displaystyle =$ $\displaystyle L \bigl( (1,0,1,2), (-1,1,0,-5), (1,-1,0,5) \bigr)$  
  $\displaystyle =$ $\displaystyle L \bigl( (1,0,1,2), (1,-1,0,5) \bigr)$  
  $\displaystyle =$ $\displaystyle \{ {\alpha}(1,0,1,2) + \beta (1,-1,0,5) \; : {\alpha}, \beta \in {\mathbb{R}}\}$  
  $\displaystyle =$ $\displaystyle \{ ({\alpha}+ \beta, -\beta, {\alpha}, 2 {\alpha}+ 5 \beta) \; : {\alpha}, \beta \in {\mathbb{R}}\}$  
  $\displaystyle =$ $\displaystyle \{(x,y,z,w) \in {\mathbb{R}}^4 \; : x +y - z = 0, 5 y - 2 z + w = 0 \}.$  

Also, by definition
$\displaystyle {\cal N}(T)$ $\displaystyle =$ $\displaystyle \{ (x,y,z) \in {\mathbb{R}}^3 \; : T(x,y,z) = {\mathbf 0}\}$  
  $\displaystyle =$ $\displaystyle \{ (x,y,z) \in {\mathbb{R}}^3 \; : (x-y+z, y-z,x, 2x-5y+5z) = {\mathbf 0}\}$  
  $\displaystyle =$ $\displaystyle \{ (x,y,z) \in {\mathbb{R}}^3 \; : x-y+ z = 0, y-z = 0,$  
    $\displaystyle \hspace{1.75in} x = 0, 2x - 5y + 5z=0 \}$  
  $\displaystyle =$ $\displaystyle \{ (x,y,z) \in {\mathbb{R}}^3 \; : y-z = 0, x = 0 \}$  
  $\displaystyle =$ $\displaystyle \{ (x,y,z) \in {\mathbb{R}}^3 \; : y=z, x = 0 \}$  
  $\displaystyle =$ $\displaystyle \{ (0,y,y) \in {\mathbb{R}}^3 \; : y \;\; {\mbox{arbitrary}} \}$  
  $\displaystyle =$ $\displaystyle L ( (0,1,1) )$  

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

    $\displaystyle T\bigl( (1,0) \bigr)= (1,0,0), \; T\bigl( (0,1) \bigr) = (1, 0, 0).$

    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

    $\displaystyle T: {\mathbb{R}}^3 \longrightarrow {\mathbb{R}}^2 \; {\mbox{ such that }} \; T(1,-1,1) = (1,2),\;\; {\mbox{ and }} \;\; T(-1,1,2) = (1,0)?$

  4. Recall the vector space $ {{\cal P}}_n ({\mathbb{R}}).$ Define a linear transformation

    $\displaystyle D: {{\cal P}}_n({\mathbb{R}}) {\longrightarrow}{{\cal P}}_n({\mathbb{R}})$

    by

    $\displaystyle D(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n) = a_1 + 2 a_2 x + \cdots + n a_n x^{n-1}.$

    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

    $\displaystyle T(1,0,0) = (0,0,1),\;\; T(1,1,0) = (1,1,1)\; {\mbox{ and }} \; T(1,1,1) = (1,1,0).\;\;$

    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

    $\displaystyle T( (3,4) ) = (0,1), \; T( (-1,1) ) = (2,3).$

    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.

    $\displaystyle A \longrightarrow B_1 \longrightarrow B_1 \longrightarrow B_2 \cdots \longrightarrow B_{k-1} \longrightarrow B_k \longrightarrow B.$

    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

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

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
$\displaystyle {\cal R}(T)$ $\displaystyle =$ $\displaystyle L ( T(u_1), T(u_2), \ldots, T(u_n) )$  
  $\displaystyle =$ $\displaystyle L ( {\mathbf 0}, \ldots, {\mathbf 0}, T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) )$  
  $\displaystyle =$ $\displaystyle L ( T(u_{r+1}), T(u_{r+2}), \ldots, T(u_n) ).$  

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

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

That is,

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

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

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

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

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

That is,

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

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

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

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,

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

height6pt width 6pt depth 0pt

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

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

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

    $\displaystyle T: V \longrightarrow {\mathbb{C}}^k \;\; {\mbox{ by }} \;\; T\bigl( P(z) \bigr) = \bigl( P(z_1), P(z_2), \ldots, P(z_k) \bigr).$

    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,

      $\displaystyle {\mathbf x}= T_A({\mathbf y}) = (T_A \circ T_A)({\mathbf y}) = T_A \bigl( T_A({\mathbf y}) \bigr) = T_A ( {\mathbf x}) = {\mathbf 0}.$

    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).$

next up previous contents
Next: Similarity of Matrices Up: Linear Transformations Previous: Matrix of a linear   Contents
A K Lal 2007-09-12