## Linear Algebra and Its Applications, Exercise 2.4.18

Exercise 2.4.18. Given the vectors $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 5 \\ 0 \end{bmatrix}$

and the subspace $V$ that they span, find two matrices $A$ and $B$ such that $V = \mathcal R(A) = \mathcal N(B)$.

Answer: The easiest way to create a matrix whose row space is $V$ is to use the vectors above as the rows of the matrix: $\begin{bmatrix} 1&1&0 \\ 1&2&0 \\ 1&5&0 \end{bmatrix}$

However we can simplify things by doing Gaussian elimination on this matrix to obtain another matrix with the same row space $V$, first subtracting the first row from the second and third rows, and then subtracting four times the second row from the third: $\begin{bmatrix} 1&1&0 \\ 1&2&0 \\ 1&5&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&0 \\ 0&1&0 \\ 0&4&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix}$

We can then further simplify by subtracting the second row from the first; again, this does not change the row space: $\begin{bmatrix} 1&1&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix}$

Our final matrix is thus $A = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix}$

with $\mathcal R(A) = V$. Note that $V$ is the x-y plane in $\mathbf R^3$ and the coordinate vectors $e_1 = \begin{bmatrix} 1&0&0 \end{bmatrix}^T$ and $e_2 = \begin{bmatrix} 0&1&0 \end{bmatrix}^T$ are a basis for $V$.

We now want to find a second matrix $B$ whose nullspace is $V$. If $V = \mathcal N(B)$ then for any vector $v$ in $V$ we must have $Bv = 0$. In particular, we must have $Be_1 = Be_2 = 0$ for the coordinate vectors $e_1$ and $e_2$ that form a basis for $V$. The simplest way to define B is then as the 1 by 3 matrix $B = \begin{bmatrix} 0&0&1 \end{bmatrix}$

for which $Be_1 = \begin{bmatrix} 0&0&1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = 0$

and $Be_2 = \begin{bmatrix} 0&0&1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = 0$

For any solution $x = \begin{bmatrix} x_1&x_2&x_3 \end{bmatrix}^T$ to $Bx = 0$ we must then have $\begin{bmatrix} 0&0&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0$

From the above we see that $x_3 = 0$ and that $x_1$ and $x_2$ can take on any value. The nullspace $\mathcal N(B)$ therefore contains all vectors of the form $\begin{bmatrix} c_1 \\ c_2 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = c_1e_1 + c_2e_2$

But this is equivalent to all linear combinations of the coordinate vectors $e_1$ and $e_2$ that form a basis for $V$. We thus have $\mathcal N(B) = V$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books .

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