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Linear Algebra Gathering and Sharpening Tools

Ryan Higginbottom

Section 5.4 Rank and Nullity

In this section we will connect dimension with the subspaces associated with linear transformations (see Section 3.4).
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Subsection 5.4.1 Defining Rank and Nullity

We begin by defining the dimension of the image of a linear transformation. (Readers should recall that image and range have the same meaning, but we prefer to use image.)
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Definition 5.4.1.

Let \(T\) be a linear transformation. Then the rank of \(T\text{,}\) denoted \(\rank(T)\text{,}\) is the dimension of the image of \(T\text{:}\)
\begin{equation*} \rank(T) = \dim(\img(T))\text{.} \end{equation*}
The rank of a matrix \(A\) is the dimension of the column space of \(A\text{:}\)
\begin{equation*} \rank(A) = \dim(\col(A))\text{.} \end{equation*}
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It may seem strange to define the same word in two ways. However, since the image of \(T\) is exactly the column space of \(A\) when \(T\) is multiplication by \(A\text{,}\) these two definitions coincide.
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When \(A\) is an \(m\times n\) matrix over \(\ff\text{,}\) its rows are vectors in \(\ff^n\) and its columns are vectors in \(\ff^m\text{.}\) This is why the column space of \(A\) is a subspace of \(\ff^m\text{.}\) We can also examine the analogous space for the rows.
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Definition 5.4.2.

The set of all linear combinations of the rows of a matrix \(A\) is called the row space of \(A\text{.}\) We denote this by \(\row(A)\text{.}\)
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Note 5.4.3.

Since the rows of \(A\) are the columns of \(A^T\text{,}\) it is immediate that \(\row(A) = \col(A^T)\text{.}\)
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With the definition of the row space it is natural to wonder how the sizes of the row and column spaces compare to each other. The following results will help us settle this matter.
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Proof.

We will first show that \(\row(A) = \row(B)\) as sets. If \(A\) is reduced to \(B\text{,}\) then the rows of \(B\) are linear combinations of the rows of \(A\text{.}\) (The elementary row operations produce linear combinations of the original rows.) Therefore, any linear combination of the rows of \(B\) can be written as a linear combination of the rows of \(A\text{.}\) This proves that \(\row(B) \subseteq \row(A)\text{.}\) Since all row operations are reversible, we can use row operations to produce \(A\) from \(B\text{,}\) and this same argument shows that \(\row(A) \subseteq \row(B)\text{.}\) This proves that \(\row(A) = \row(B)\text{.}\)
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If the matrix \(B\) is in REF, the nonzero rows are linearly independent because no nonzero row is a linear combination of the rows below it. Here we are applying the Linear Dependence Lemma (Theorem 5.1.19) to the nonzero rows from bottom to top. Since the rows of \(B\) span \(\row(B)\) by definition, the fact that they are linearly independent means that they form a basis for \(\row(B)\text{.}\)
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Proof.

If we put \(A\) into REF, then Proposition 5.4.4 tells us that the number of pivots is \(\dim(\row(A))\) since that is the number of vectors in a basis of \(\row(A)\text{.}\) However, the number of pivots in a REF (or the RREF) of \(A\) is also \(\rank(A)\text{.}\) (See Exercise 5.3.4.8.) This proves that \({\rank(A) = \dim(\row(A))}\text{.}\)
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Note 5.4.6.

This theorem says that \(\rank(A) = \rank(A^T)\text{.}\) This theorem also answers the question about the relative sizes of \(\col(A)\) and \(\row(A)\)—they are the same!
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Example 5.4.7.

Consider the following matrix \(A \in M_{4,5}(\ff_5)\text{:}\)
\begin{equation*} A = \begin{bmatrix} 0 \amp 2 \amp 2 \amp 1 \amp 2 \\ 4 \amp 3 \amp 4 \amp 1 \amp 3 \\ 0 \amp 0 \amp 3 \amp 3 \amp 2 \\ 4 \amp 2 \amp 0 \amp 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
We will find a basis for \(\row(A)\text{.}\) Here is the RREF of \(A\text{:}\)
\begin{equation*} A \sim \begin{bmatrix} 1 \amp 0 \amp 0 \amp 4 \amp 4 \\ 0 \amp 1 \amp 0 \amp 2 \amp 2 \\ 0 \amp 0 \amp 1 \amp 1 \amp 4 \\ 0 \amp 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
We have used the RREF of a matrix in the past to find bases for the null space and column space of a matrix. Now, we will use it to find a basis for the row space. Proposition 5.4.4 tells us that the nonzero rows of this RREF are the basis we seek, therefore a basis for \(\row(A)\) is \(\{\bfv_1, \bfv_2, \bfv_3 \}\text{,}\) where
\begin{align*} \bfv_1 \amp = [1\; 0\; 0\; 4\; 4 ]\\ \bfv_2 \amp = [0\; 1\; 0\; 2\; 2 ]\\ \bfv_3 \amp = [0\; 0\; 1\; 1\; 4 ]\text{.} \end{align*}
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We have defined the dimension of the image of a linear transformation \(T\text{,}\) so we now turn to the kernel.
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Definition 5.4.8.

If \(T\) is a linear transformation, then the nullity of \(T\) is the dimension of the kernel of \(T\text{.}\) If \(A\) is a matrix, then the nullity of \(A\) is the dimension of the null space of \(A\text{.}\)
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Note 5.4.9.

As we saw with rank, these two uses of “nullity” coincide for the situation when \(T\) is multiplication by \(A\text{.}\)
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Some other texts use the notation \(\nll(A)\) to indicate nullity instead of null space, as we have. We will not introduce any additional notation for the nullity, but we will use \(\dim(\ker(T))\) or \(\dim(\nll(A))\) as appropriate.
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Subsection 5.4.2 The Rank-Nullity Theorem

The following theorem brings together the rank and nullity of a matrix/linear transformation.
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Proof.

We will prove the result for matrices. The proof for linear transformations is a bit more technical. (The reader should note that the result for linear transformations implies the result for matrices!)
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If \(A \in M_{m,n}(\ff)\text{,}\) let \(B\) be the RREF of \(A\text{.}\) Then \(\rank(A)\) is the number of pivot columns in \(B\text{.}\) Further, \(\dim(\nll(A))\) is the number of non-pivot columns in \(B\text{.}\) (See Exercise 5.3.4.8.) Since each of the \(n\) columns of \(B\) must be either a pivot or a non-pivot column, and since \(A\) and \(B\) have the same number of columns, this proves the theorem.
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Example 5.4.11.

If \(A\) is a \(5\times 6\) matrix with a three-dimensional null space, this theorem tells us that the rank of \(A\) is \(6-3=3\text{.}\)
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Let us consider an additional scenario: Could a \(6\times 8\) matrix \(A\) have a one-dimensional null space? If such a matrix existed, it would have a rank of \(8-1=7\text{,}\) according to Theorem 5.4.10. But the largest rank that a \(6\times 8\) matrix can have is 6, since there cannot be more pivots than there are rows. So the answer is no, a \(6\times 8\) matrix cannot have a one-dimensional null space.
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When the dimensions of the domain and codomain of a linear transformation are equal, some properties of such a transformation coincide.
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Proof.

By Theorem 3.4.7, \(T\) is injective if and only if \(\kerr(T) = \{\bfo\}\text{.}\) In other words, \(T\) is injective if and only if \(\dim(\kerr(T)) = 0\text{.}\) By Theorem 5.4.10, this happens if and only if \(\rank(T) = \dim(V)\text{,}\) and if \(\dim(V) = \dim(W)\text{,}\) \(\rank(T)=\dim(W)\) if and only if \(T\) is injective. This proves that \(T\) is injective if and only if \(T\) is surjective. Since a bijective linear transformation is an isomorphism, our proof is complete.
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To close out this section, we present a long theorem with many equivalent statements. We will omit a proof, because the equivalence of most of these statements has been already established at various places in this text. (In other books, this theorem forms the central focus of the text. It is certainly important, but we have chosen a different emphasis.)
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This theorem ties together threads from almost every section we’ve covered, which is quite an achievement! The reader should note that this result only applies to square matrices.
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Reading Questions 5.4.3 Reading Questions

1.

Consider the following matrix \(A\text{:}\)
\begin{equation*} A = \begin{bmatrix} 2 \amp 0 \amp 2 \amp 4 \amp 0 \\ -1 \amp 1 \amp -4 \amp 6 \amp -7 \\ 6 \amp 3 \amp -3 \amp 2 \amp 13 \end{bmatrix} \text{.} \end{equation*}
Find a basis for \(\row(A)\text{.}\) Explain your answer.
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2.

Suppose that \(T: V \to W\) is a linear transformation and that \(\dim(V) = 4\) and \(\dim(W) = 5\text{.}\) What are the possible values for \(\dim(\kerr(T))\text{?}\) Explain.
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Exercises 5.4.4 Exercises

1.

Find the rank and nullity of each of the following matrices.
  1. \(A \in M_{3,4}(\rr)\text{,}\)
    \begin{equation*} A = \begin{bmatrix} 2 \amp -3 \amp 4 \amp 1 \\ -1 \amp 2 \amp -3 \amp 1 \\ 4 \amp -3 \amp 2 \amp 11 \end{bmatrix} \end{equation*}
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  2. \(A \in M_{4,5}(\ff_3)\text{,}\)
    \begin{equation*} A = \begin{bmatrix} 1 \amp 2 \amp 2 \amp 1 \amp 0 \\ 1 \amp 2 \amp 2 \amp 1 \amp 1 \\ 2 \amp 1 \amp 2 \amp 0 \amp 1 \\ 2 \amp 1 \amp 2 \amp 0 \amp 2 \end{bmatrix} \end{equation*}
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2.

Let \(D: P_n \to P_{n-1}\) be the differentiation linear transformation. Calculate \(\rank(D)\) and \(\dim(\ker(D))\text{.}\)
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3.

  1. If a \(5\times 3\) matrix \(A\) has rank 3, find \(\dim(\nll(A))\text{,}\) \(\dim(\row(A))\text{,}\) and \(\rank(A^T)\text{.}\)
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  2. If the null space of a \(7\times 6\) matrix \(A\) is 5-dimensional, what is the dimension of the column space of \(A\text{?}\)
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  3. If \(A\) is a \(7\times 9\) matrix, what is the smallest possible dimension of \(\nll(A)\text{?}\)
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4.

Suppose a nonhomogeneous linear system of nine equations and ten variables has a solution for all possible constants on the right side of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Explain.
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Answer.
This is not possible. A \(9\times 10\) linear system has a \(9\times 10\) coefficient matrix which we will call \(A\text{.}\) This situation can be reinterpreted as a linear transformation from \(\ff^{10} \to \ff^9\) which is multiplication by \(A\text{.}\) The information given in the problem tells us that this transformation is surjective, meaning that the rank of \(A\) is as large as possible, which is 9 in this case. Therefore, the Rank-Nullity Theorem says that \(\dim(\nll(A)) = 1\text{.}\) This means that every nonzero vector in the null space of \(A\) is a scalar multiple of a single basis vector. Therefore, the associated homogeneous linear system only has nonzero solutions which are multiples of each other.
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5.

Suppose \(A \in M_{m,n}(\ff)\) and \(\bfb \in \ff^m\text{.}\) What has to be true about the two numbers \(\rank([A\; \bfb ])\) and \(\rank(A)\) in order for the equation \(A\bfx = \bfb\) to be consistent? Explain.
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Writing Exercises

6.
If \(A\) and \(B\) are matrices, prove that \(\rank(AB) \le \min\{\rank(A), \rank(B) \}\text{.}\)
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7.
Suppose that \(T \in L(V,W)\) and that \(V\) and \(W\) are both finite-dimensional.
  1. Prove that \(T\) is surjective if and only if \(\rank(T) = \dim(W)\text{.}\)
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  2. Prove that \(T\) is injective if and only if \(\rank(T) = \dim(V)\text{.}\)
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Answer.
  1. If \(T\) is surjective, then \(\img(T) = W\text{,}\) which means that \(\dim(\img(T)) = \dim(W)\text{.}\) This implies that \(\rank(T) = \dim(W)\text{.}\) On the other hand, if \(\rank(T) = \dim(W)\text{,}\) then \(\dim(\img(T)) = \dim(W)\text{.}\) Since \(\img(T)\) is a subspace of \(W\text{,}\) Theorem 5.3.18 implies that \(\img(T) = W\text{,}\) which means that \(T\) is surjective.
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  2. We will be able to argue both implications at once. A linear transformation \(T\) is injective if and only if \(\ker(T) = \{\bfo\}\text{.}\) This occurs if and only if \(\dim(\ker(T)) = 0\text{.}\) The Rank-Nullity Theorem then says that \(\dim(\ker(T)) = 0\) if and only if \(\rank(T) = \dim(V)\text{.}\) This proves the result.
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8.
Suppose that \(A\bfx = \bfb\) is a \(6\times 6\) linear system which is consistent but which does not have a unique solution. Prove that there must be a vector \(\mathbf{c} \in \ff^6\) such that the system \(A\bfx = \mathbf{c}\) is inconsistent.
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Answer.
The given information means there is no pivot in the last column of the augmented matrix but that there is at least one non-pivot column among the first six columns. The coefficient matrix \(A\) then has at least one non-pivot column, meaning that \(\rank(A) \le 5\text{.}\) But this means that the linear transformation \(T:\ff^6 \to \ff^6\) which is multiplication by \(A\) cannot be surjective, since the rank of \(A\) is less than six. Therefore, we can find a vector \(\mathbf{c} \in \ff^6\) such that \(A\bfx = \mathbf{c}\) is inconsistent.
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9.
Prove that if \(T \in L(V,W)\text{,}\) then \(\rank(T) \le \min \{\dim(V), \dim(W) \}\text{.}\)
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10.
Prove that if \(A \in M_{m,n}(\ff)\text{,}\) then \(\rank(A) \le \min \{m,n \}\text{.}\)
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Answer.
We know that \(\rank(A)\) is the number of pivots in the RREF of \(A\text{,}\) but the number of pivots must be less than both the number of columns (\(n\)) and the number of rows (\(m\)) of \(A\text{.}\) Then if \(\rank(A) \le m\) and \(\rank(A) \le n\text{,}\) we have \(\rank(A) \le \min \{m,n \}\text{.}\)
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11.
Let \(T \in L(V,W)\text{.}\)
  1. Prove that if \(\dim(V) > \dim(W)\text{,}\) then \(T\) cannot be injective.
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  2. Prove that if \(\dim(W) > \dim(V)\text{,}\) then \(T\) cannot be surjective.
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