## 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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