## Linear Algebra and Its Applications, Exercise 2.4.20

Exercise 2.4.20. For each of the following properties find a matrix with that property. If no such matrix exists, explain why that is the case.

a) The column space of the matrix contains the vectors

$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

and the row space contains the vectors

$\begin{bmatrix} 1 \\ 1 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

b) The column space has as a basis the vector

$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$

and the nullspace has as a basis the vector

$\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$

c) The column space is all of $\mathbf R^4$ and the row space is all of $\mathbf R^3$.

Answer: a) The easiest way to create a matrix $A$ with the specified vectors in the column space $\mathcal R(A)$ is to use the vectors in question as the only columns in the matrix:

$A = \begin{bmatrix} 1&0 \\ 0&0 \\ 0&1 \end{bmatrix}$

Note that adding the first row of $A$ to the third row of $A$ produces the vector $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ so that vector is in the row space $\mathcal R(A^T)$. Similarly adding the first  row of $A$ to twice the third row of $A$ produces the vector $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ so that vector is also in the row space $\mathcal R(A^T)$.

Thus the matrix $A$ specified above has the required property.

b) Since the basis vector for the nullspace is in $\mathbf R^3$ the number of columns of the matrix must be $n = 3$ (since each vector in the nullspace must be a solution to $Ax = 0$ and the elements of $x$ multiply the columns of $A$).

However for an $m$ by $n$ matrix the dimension of the  nullspace is $n - r$ where $r$ is the rank of the matrix. In this case the column space and nullspace each have only a single vector as a basis, so the dimension of the column space and nullspace are each 1. This implies that the rank $r = 1$ and the number of columns $n = 2$ for a matrix with the specified property.

So we must have both $n = 3$ and $n = 2$ which is a contradiction. Thus there is no matrix with the required property.

c) If the column space is $\mathbf R^4$ then its dimension is 4, and if the row space is $\mathbf R^3$ then its dimension is 3. However the dimensions of the row and columns spaces should both be equal to the rank $r$ of the matrix and therefore should be the same. This is a contradiction, so no matrix exists with the required property.

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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