Vector Spaces

4.6 Rank

The Row Space

• If $A$ is an $m \times n$ matrix, each row of $A$ has $n$ entries and thus can be identified with a vector in $\mathbb{R}^n$.
• The set of all linear combinations of the row vectors is called the row space of $A$ and is denoted by Row $A$.
• Each row has $n$ entries, so Row $A$ is a subspace of $\mathbb{R}^n$.
• Since the rows of $A$ are identified with the columns of $A^T$, we could also write Col $A^T$ in place of Row $A$.

Theorem 13:

If two matrices $A$ and $B$ are row equivalent, then their row spaces are the same. If $B$ is in echelon form, the nonzero rows of B form a basis for the row space of A as well as for that of $B$.

Proof:

• If $B$ is obtained from $A$ by row operations, the rows of $B$ are linear combinations of the rows of $A$.
• It follows that any linear combination of the rows of $B$ is automatically a linear combination of the rows of $A$.
• Thus the row space of $B$ is contained in the row space of $A$.
• Since row operations are reversible, the same argument shows that the row space of $A$ is a subset of the row space of $B$.
• So the two row spaces are the same.
• If $B$ is in echelon form, its nonzero rows are linearly independent because no nonzero row is a linear combination of the nonzero rows below it. (Apply Theorem 4 to the nonzero rows of $B$ in reverse order, with the first row last).
• Thus the nonzero rows of $B$ form a basis of the (common) row space of $B$ and $A$.

Example 2:

Find bases for the row space, the column space, and the null space of the matrix

$$A=\begin{bmatrix} -2 & -5 & 8 & 0 & -17 \\ 1 & 3 & -5 & 1 & 5 \\ 3 & 11 & -19& 7 & 1 \\ 1 & 7 & -13& 5 & -3 \end{bmatrix}$$

Solution:

• To find bases for the row space and the column space, row reduce $A$ to an echelon form:

$$A \sim B = \begin{bmatrix} 1 & 3 & -5 & 1 & 5 \\ 0 & 1 & -2 & 2 & -7 \\ 0 & 0 & 0 & -4 & 20 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

• By Theorem 13, the first three rows of $B$ form a basis for the row space of $A$ (as well as for the row space of $B$).
• Thus Basis for Row $A$: $\left\{ (1,3,-5,1,5), (0,1,-2,2,-7), (0,0,0,-4,20)\right\}$.
• For the column space, observe from $B$ that the pivots are in columns 1, 2, and 4.

• Hence columns 1, 2, and 4 of $A$ (not $B$) form a basis for Col A: $$\text{Basis for Col }A:=\left\{ \begin{bmatrix} -2 \\ 1 \\ 3 \\1 \end{bmatrix}, \begin{bmatrix} -5 \\ 3 \\11 \\7 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 7 \\5 \end{bmatrix} \right\}$$

• Notice that any echelon form of $A$ provides (in its nonzero rows) a basis for Row $A$ and also identifies the pivot columns of $A$ for Col $A$.

• However, for Nul $A$, we need the reduced echelon form.
• Further row operations on $B$ yield

$$A \sim B \sim C= \begin{bmatrix} 1 & 0 & 1 & 0 & 1 \\ 0 & 1 & -2 & 0 & 3 \\ 0 & 0 & 0 & 1 & -5 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

• The equation $A\textbf{x}=\textbf{0}$ is equivalent to $C\textbf{x}=\textbf{0}$, that is,

$$\begin{align} x_1 & &+ x_3& &+ x_5 & = 0 \\ &x_2&-2x_3& &+3x_5 &= 0 \\ & & &x_4&-5x_5 &=0 \end{align}$$

• So $x_1=-x_3-x_5, x_2=2x_3-3x_5, x_4=5x_5$, with $x_3$ and $x_5$ free variables.
• The calculations show that

$$\text{Basis for Nul }A:\left\{ \begin{bmatrix} -1 \\ 2 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ -3 \\ 0 \\ 5 \\ 1 \end{bmatrix} \right\}$$

• Observe that, unlike the basis for Col $A$, the bases for Row $A$ and Nul $A$ have no simple connection with the entries in $A$ itself.

Definition : Rank

The rank of $A$ is the dimension of the column space of $A$. Since Row $A$ is the same as Col $A^T$, the dimension of row space of $A$ is the rank of $A^T$. The dimension of the null space is sometimes called the nullity of $A$.

뒤에 나오지만, rank A는 Col A와 Row A의 dimension이다. 즉, dim Col A = dim Row A 임.

Theorem 14: Rank Theorem

The dimensions of the column space and the row space of an $m \times n$ matrix A are equal. This common dimension, the rank of $A$, also equals the number of pivot positions in $A$ and satisfies the equation

$$\text{rank }A + \text{dim Nul }A = n$$

Proof:

• By Theorem 6, rank $A$ is the number of pivot columns in $A$.
• Equivalently, rank $A$ is the number of pivot positions in an echelon form $B$ of $A$.
• Since $B$ has a nonzero row for each pivot, and since these rows form a basis for the row space of $A$, the rank of $A$ is also the dimension of the row space.
• The dimension of Nul $A$ equals the number of free variables in the equation $A\textbf{x}=\textbf{0}$.
• Expressed another way, the dimension of Nul $A$ is the number of columns of $A$ that are not pivot columns. (It is the number of these columns, not the columns themselves, that is related to Nul $A$).
• Obviously,

$$\left[ \text{# of pivot columns} \right] + \left[ \text{# of nonpivot columns} \right] = \left[ \text{# of columns} \right]$$

• This proves the theorem.

Example 3:

• a. If $A$ is a $7 \times 9$ matrix with a two-dimensional null space, what is the rank of $A$?
• b. Could a $6 \times 9$ matrix have a two-dimensional null space?

Solution:

a.

• Since $A$ has 9 columns, (rank $A$)+2=9, and hence rank $A$=7.

b.

• No.
• If a $6 \times 9$ matrix, call it $B$, has a two-dimensional null space, it would have to have rank 7, by the Rank Theorem.
• But the columns of $B$ are vectors in $\mathbb{R}^6$, and so the dimension of Col $B$ cannot exceed 6; that is, rank $B$ cannot exceed 6.

Theorem: The Invertible Matrix Theorem (continued)

Let $A$ be an $n \times n$ matrix. Then the following statements are each equivalent to the statement that $A$ is an invertible matrix.

• 13.The columns of $A$ form a basis of $\mathbb{R}^n$.
• 14.Col $A$ = $\mathbb{R}^n$.
• 15.Dim Col $A$ = $n$.
• 16.rank $A$=$n$
• 17.Nul $A$=$\left\{ \textbf{0} \right\}$
• 18.Dim Nul $A$ = 0

Proof:

• Statement (13) is logically equivalent to statements (5) and (8) regarding linear independence and spanning.

• The other five statements are linked to the earlier ones of the theorem by the following chain of almost trivial implications: $$(7) \rightarrow (14) \rightarrow (15) \rightarrow (16) \rightarrow (18) \rightarrow (17) \rightarrow (4)$$

• Statement (7), which says that the equation $A\textbf{x}=\textbf{b}$ has at least one solution for each $b$ in $\mathbb{R}^n$, implies (14), because Col $A$ is precisely the set of all $b$ such that the equation $A\textbf{x}=\textbf{b}$ is consistent.

• The implications $(14) \rightarrow (15) \rightarrow (16)$ follow from the definitions of dimension and rank.
• If the rank of $A$ is $n$, the number of columns of $A$, then dim Nul $A$=0, by the Rank Theorem, and so Nul $A$=$\left\{ \textbf{0} \right\}$.
• Thus $(16) \rightarrow (18) \rightarrow (17)$.
• Also, (17) implies that the equation $A\textbf{x}=\textbf{0}$ has only the trivial solution, which is statement (4).
• Since statements (4) and (7) are already known to be equivalent to the statement that $A$ is invertible, the proof is complete.