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4.8 Row Space, Column Space, and Null Space: Class Notes

Definition

A homogeneous linear system $A𝐱=\mathrm{𝟎}$ and an associated nonhomogeneous system $A𝐱=𝐛$ are corresponding linear systems.

#28

Find a general solution of the system, and use that solution to find a general solution of the associated homogeneous system and a particular solution of the given system.

Solution

Using Gauss-Jordan elimination, we find

$${\displaystyle \left[\begin{array}{ccccc}\hfill {\displaystyle 9}\hfill & \hfill {\displaystyle -3}\hfill & \hfill {\displaystyle 5}\hfill & \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle 4}\hfill \\ \hfill {\displaystyle 6}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 5}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 14}\hfill & \hfill {\displaystyle -8}\hfill \end{array}\right]⟶\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle -\frac{1}{3}}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle -\frac{13}{3}}\hfill & \hfill {\displaystyle \frac{13}{3}}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 9}\hfill & \hfill {\displaystyle -7}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]}$$

If the system were homogeneous (all constants on the right equal to 0), we would have ${\displaystyle x_1=\frac{1}{3}{x}_2+\frac{13}{3}{x}_4}$ and $x_3=-9x_4$. Letting $x_2=s$ and $x_4=t$, we get

$${\displaystyle \overrightarrow{x}=\left[\begin{array}{c}\frac{1}{3}\\ 1\\ 0\\ 0\end{array}\right]s+\left[\begin{array}{c}\frac{13}{3}\\ 0\\ -9\\ 1\end{array}\right]t.}$$

With the constants from $\overrightarrow{b}$, we get ${\displaystyle x_1=\frac{1}{3}{x}_2+\frac{13}{3}{x}_4+\frac{13}{3}}$ and $x_3=-9x_4-7$, so

$${\displaystyle \overrightarrow{x}=\left[\begin{array}{c}\frac{1}{3}\\ 1\\ 0\\ 0\end{array}\right]s+\left[\begin{array}{c}\frac{13}{3}\\ 0\\ -9\\ 1\end{array}\right]t+\left[\begin{array}{c}\frac{13}{3}\\ 0\\ -7\\ 0\end{array}\right].}$$

The first two terms are the general solution to the homogeneous system, and the last vector is a particular solution to the nonhomogeneous system, ${\overrightarrow{x}}_p$.

Definition

The solution set of a homogeneous linear system $A𝐱=\mathrm{𝟎}$ is called the general solution. A constant vector ${𝐱}_0$ in the solution set of a consistent linear system $A𝐱=𝐛$ is a particular solution.

Theorem 4.8.2

If ${𝐱}_0$ is any solution of a consistent linear system $A𝐱=𝐛$, and if $S=\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_k\}$ is a basis for the null space of $A$, then every solution of $A𝐱=𝐛$ can be expressed in the form $𝐱={𝐱}_0+c_1{𝐯}_1+c_2{𝐯}_2+\dots +c_k{𝐯}_k$. Conversely, for all choices of scalars $c_1,c_2,\dots ,c_k$, the vector $𝐱$ in this formula is a solution of $A𝐱=𝐛$.

Here ${𝐱}_0$ is a particular solution and $c_1{𝐯}_1+\dots +c_k{𝐯}_k$ is the general solution of the homogeneous system, as in #28.

Note that the solution set of $A\overrightarrow{x}=\overrightarrow{b}$ can be considered a translation of the solution space of $A\overrightarrow{x}=\overrightarrow{0}$ by the particular solution vector.

The vectors in the general solution of the homogeneous system are linearly independent and so form a basis for the null space of the homogeneous system. The general solution of the homogeneous system is denoted by ${\overrightarrow{x}}_h$.

Definition

For an $m\times n$ matrix

the vectors

in $R^n$ that are formed from the rows of $A$ are called the row vectors of $A$, and the vectors

in $R^m$ that are formed from the columns of $A$ are called the column vectors of $A$.

Definition

If $A$ is an $m\times n$ matrix, then the subspace of $R^n$ spanned by the row vectors of $A$ is denoted by $\mathrm{r}\mathrm{o}\mathrm{w}(A)$ and is called the row space of $A$, and the subspace of $R^m$ spanned by the column vectors of $A$ is denoted by $\mathrm{c}\mathrm{o}\mathrm{l}(A)$ and is called the column space of $A$. The solution space of the homogeneous system of equations $A𝐱=\mathrm{𝟎}$, which is a subspace of $R^n$, is denoted by $\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}(A)$ and is called the null space of $A$.

Theorem 4.8.1

A system of linear equations $A𝐱=𝐛$ is consistent if and only if $𝐛$ is in the column space of $A$.

Theorem 4.8.3

a) Row equivalent matrices have the same row space.
b) Row equivalent matrices have the same null space.

Theorem 4.8.4

If a matrix $R$ is in row echelon form, then the row vectors with the leading 1’s (the nonzero row vectors) form a basis for the row space of $R$, and the column vectors with the leading 1’s of the row vectors form a basis for the column space of $R$.

In exercises 9 and 10, find bases for the null space and row space of $A$.

#9

Solution

Subtract 2 times row 1 from row 2:

$$\left[\begin{array}{ccc}2& 0& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

A basis for the row space is $(2,0,-1)$ (from the reduced matrix).

A basis for the null space comes from solving the associated system. The only equation is $2x_1-x_3=0$, so $x_3=2x_1$, and $x_2$ is a free variable. With $x_1=s$ and $x_2=t$:

$$\overrightarrow{x}=\left[\begin{array}{c}1\\ 0\\ 2\end{array}\right]s+\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]t$$

#10

Solution

$R_2\to R_2-2R_1$:

$$(2,1,3,0)+(-2,-8,-10,-4)=(0,-7,-7,-4)$$

$R_3\to R_3+R_1$:

$$(-1,3,2,2)+(1,4,5,2)=(0,7,7,4)$$

The new rows 2 and 3 are negatives of each other, so one of them becomes a zero row:

$$\left[\begin{array}{cccc}1& 4& 5& 2\\ 0& 7& 7& 4\\ 0& 0& 0& 0\end{array}\right]$$

The pivot rows are rows 1 and 2. To read off the null space we need $\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{f}(A)$ (or something close to it), so clear the 4 above the second pivot without introducing fractions: $R_1\to 7R_1-4R_2$:

$$(7,28,35,14)-(0,28,28,16)=(7,0,7,-2)$$

$$\left[\begin{array}{cccc}7& 0& 7& -2\\ 0& 7& 7& 4\\ 0& 0& 0& 0\end{array}\right]$$

Row space basis: $\{(7,0,7,-2),(0,7,7,4)\}$.

Dividing each row by 7, ${\displaystyle x_1=-x_3+\frac{2}{7}{x}_4}$ and ${\displaystyle x_2=-x_3-\frac{4}{7}{x}_4}$. With $x_3=s$ and $x_4=t$:

$${\displaystyle \overrightarrow{x}=\left[\begin{array}{c}-1\\ -1\\ 1\\ 0\end{array}\right]s+\left[\begin{array}{c}\frac{2}{7}\\ -\frac{4}{7}\\ 0\\ 1\end{array}\right]t}$$

Scaling each vector (by $-1$ and by 7) to clear signs and fractions, a basis for the null space is

$$\{\left[\begin{array}{c}1\\ 1\\ -1\\ 0\end{array}\right],\left[\begin{array}{c}2\\ -4\\ 0\\ 7\end{array}\right]\}.$$

Theorem 4.8.5

If $A$ and $B$ are row equivalent matrices, then:
a) A given set of column vectors of $A$ is linearly independent if and only if the corresponding column vectors of $B$ are linearly independent.
b) A given set of column vectors of $A$ forms a basis for the column space of $A$ if and only if the corresponding column vectors of $B$ form a basis for the column space of $B$.

#15

Find a basis for the subspace of $R^4$ that is spanned by the given vectors.
$(1,1,0,0)$, $(0,0,1,1)$, $(-2,0,2,2)$, $(0,-3,0,3)$

Solution

Put the vectors in the columns of a matrix and row reduce:

$$\left[\begin{array}{cccc}1& 0& -2& 0\\ 1& 0& 0& -3\\ 0& 1& 2& 0\\ 0& 1& 2& 3\end{array}\right]⟶\left[\begin{array}{cccc}1& 0& -2& 0\\ 0& 1& 2& 0\\ 0& 0& -2& 3\\ 0& 0& 0& -3\end{array}\right]$$

$${\displaystyle ⟶\left[\begin{array}{cccc}1& 0& 0& -3\\ 0& 1& 0& 3\\ 0& 0& 1& -\frac{3}{2}\\ 0& 0& 0& 1\end{array}\right]⟶\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]}$$

Since all 4 columns have leading 1’s, the given set is the desired basis (by Theorem 4.8.5).

#17

Find a subset of the given vectors that forms a basis for the space spanned by those vectors, and then express each vector that is not in the basis as a linear combination of the basis vectors.
${𝐯}_1=(1,-1,5,2)$, ${𝐯}_2=(-2,3,1,0)$, ${𝐯}_3=(4,-5,9,4)$, ${𝐯}_4=(0,4,2,-3)$, ${𝐯}_5=(-7,18,2,-8)$

Solution

Put the vectors in the columns of a matrix. Row echelon form gives the leading 1’s, in columns 1, 2, and 4:

$$\left[\begin{array}{ccccc}1& -2& 4& 0& -7\\ -1& 3& -5& 4& 18\\ 5& 1& 9& 2& 2\\ 2& 0& 4& -3& -8\end{array}\right]⟶\left[\begin{array}{ccccc}1& 0& 2& 8& 15\\ 0& 1& -1& 4& 11\\ 0& 0& 0& 1& 2\\ 0& 0& 0& 0& 0\end{array}\right]$$

Basis: $\{{\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_4\}$.

Continue to reduced row echelon form, and let the columns of this matrix be denoted by ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2,{\overrightarrow{w}}_3,{\overrightarrow{w}}_4,{\overrightarrow{w}}_5$:

$$⟶\left[\begin{array}{ccccc}1& 0& 2& 0& -1\\ 0& 1& -1& 0& 3\\ 0& 0& 0& 1& 2\\ 0& 0& 0& 0& 0\end{array}\right]$$

Reduced row echelon form gives the dependency relationships between the ${\overrightarrow{w}}_i$. It turns out these are preserved by elementary row operations. We therefore find

$${\overrightarrow{w}}_3=2{\overrightarrow{w}}_1-{\overrightarrow{w}}_2\text{and}{\overrightarrow{w}}_5=-{\overrightarrow{w}}_1+3{\overrightarrow{w}}_2+2{\overrightarrow{w}}_4,$$

so

$${\overrightarrow{v}}_3=2{\overrightarrow{v}}_1-{\overrightarrow{v}}_2\text{and}{\overrightarrow{v}}_5=-{\overrightarrow{v}}_1+3{\overrightarrow{v}}_2+2{\overrightarrow{v}}_4.$$

Check:

$$2(1,-1,5,2)-(-2,3,1,0)=(4,-5,9,4)={\overrightarrow{v}}_3$$

$$-(1,-1,5,2)+3(-2,3,1,0)+2(0,4,2,-3)=(-7,18,2,-8)={\overrightarrow{v}}_5$$

Definition

An equation that expresses a column vector of a matrix $A$ that does not contain a pivot as a linear combination of column vectors that contain pivots is a dependency equation.

#18

Find a basis for the row space of $A$ that consists entirely of row vectors of $A$.

Solution

The rows of $A$ are the columns of $A^T$, so row reduce $A^T$:

$$A^T=\left[\begin{array}{ccc}1& 2& -1\\ 4& 1& 3\\ 5& 3& 2\\ 2& 0& 2\end{array}\right]⟶\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

The pivots are in columns ${\overrightarrow{c}}_1$ and ${\overrightarrow{c}}_2$ of $A^T$, so the corresponding rows of $A$ are