32 Null Spaces
The null space of an  matrix
 matrix  can be thought of as the set of all matrices that can be multiplied by
 can be thought of as the set of all matrices that can be multiplied by  to get the zero matrix. That is, the null space of
 to get the zero matrix. That is, the null space of  , denoted as
, denoted as  , is the set of all solutions x of the homogeneous equation
, is the set of all solutions x of the homogeneous equation 
      ![Rendered by QuickLaTeX.com \[A\text{\textbf{x}=\textbf{0}}\]](https://iu.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-d89654e12920c07c5cd372958cc59597_l3.png)
according to Larson and Falvo [16].
Let’s now look at an example. This problem is one I completed as homework for Elementary Linear Algebra and comes from Larson and Falvo [16].
Find a basis for the null space of the matrix
      ![Rendered by QuickLaTeX.com \[A=\begin{bmatrix} 1&2&3\\ 0&1&0 \end{bmatrix}\]](https://iu.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-42a4d8480201689d494459b5ae5079fb_l3.png)
Solution
To find the null space, we need to find the solutions x to 
      ![Rendered by QuickLaTeX.com \[\begin{bmatrix} 1&2&3\\ 0&1&0 \end{bmatrix}\textbf{x}=\textbf{0},\]](https://iu.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-e6311563fb2eaff5d6b59606f4beb941_l3.png)
which can be rewritten as
      ![Rendered by QuickLaTeX.com \[\begin{bmatrix} 1&2&3\\ 0&1&0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}.\]](https://iu.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-7a31593c2fe0ec018252d76d4da38b8f_l3.png)
Using matrix multiplication, we have the following system of equations
      
Because the coefficient matrix of this system is already in row-echelon form, we do not need to perform any elementary row operations before solving this system.  So, we can let  , where
, where  is a scalar.   It follows from the second equation that
 is a scalar.   It follows from the second equation that  .  Finally, it follows from equation one that
.  Finally, it follows from equation one that  . Therefore,
. Therefore,
      ![Rendered by QuickLaTeX.com \begin{align*} \textbf{x}&=\left[      \begin{array}{@{}*{7}{r}@{}}              s\\              0 \\              -\displaystyle\frac{s}{3} \\           \end{array}      \right] \\ &=s\left[      \begin{array}{@{}*{7}{r}@{}}              1\\              0 \\              -\displaystyle\frac{1}{3} \\           \end{array}      \right] &&\text{where $s$ is a real number} \end{align*}](https://iu.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-acaf08ab437a3c8471c95cff5c06fcec_l3.png)
As we can see, every solution to
      ![Rendered by QuickLaTeX.com \[A\textbf{x}=\textbf{0}\]](https://iu.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-103f5194bbb25d61d90a8bfdadff02b5_l3.png)
can be written as a multiple of the vector
      ![Rendered by QuickLaTeX.com \[\left[      \begin{array}{@{}*{7}{r}@{}}              1\\              0 \\              -\displaystyle\frac{1}{3} \\           \end{array}      \right]\]](https://iu.pressbooks.pub/app/uploads/quicklatex/quicklatex.com-6be0a7135e4d29fc120c82460a7e5a51_l3.png)
so this vector forms a basis for the null space.
We have seen how matrices and other vectors work together regarding row, column, and null spaces along with their bases. We are now going to look at a topic that utilizes the ideas of square matrices, their determinants, and vectors known as eigenvectors.
