Linear Algebra and Its Applications, Exercise 2.4.8

Exercise 2.4.8. Is it possible for the row space and nullspace of a matrix to both contain the vector $\begin{bmatrix} 1&1&1 \end{bmatrix}^T$? If not, why not?

Answer: Suppose $A$ is an $m$ by 3 matrix and $\begin{bmatrix} 1&1&1 \end{bmatrix}^T$ is in the nullspace $\mathcal N(A)$. Then we must have

$A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{21}&a_{22}&a_{23} \\ \vdots&\vdots&\vdots \\ a_{m1}&a_{m2}&a_{m3} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 0$

which means that $a_{i1} + a_{i2} + a_{i3} = 0$ for $1 \le i \le m$.

Now suppose that $\begin{bmatrix} 1&1&1 \end{bmatrix}^T$ is also in the row space $\mathcal R(A^T)$. Then $\begin{bmatrix} 1&1&1 \end{bmatrix}^T$ can be expressed as a linear combination of the rows of $A$ so that

$c_1 \begin{bmatrix} a_{11} \\ a_{12} \\ a_{13} \end{bmatrix} + c_2 \begin{bmatrix} a_{21} \\ a_{22} \\ a_{23} \end{bmatrix} + \cdots + c_m \begin{bmatrix} a_{m1} \\ a_{m2} \\ a_{m3} \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$

for some set of coefficients $c_1, c_2, \dotsc, c_m$.

We now add the equations on both sides of the system above. For the right side we obtain $1 + 1 + 1 = 3$. For the left side we have

$(c_1a_{11} + c_2a_{21} + \cdots + c_ma_{m1})$

$\qquad + (c_1a_{12} + c_2a_{22} + \cdots + c_ma_{m2})$

$\qquad + (c_1a_{13} + c_2a_{23} + \cdots + c_ma_{m3})$

$= c_1(a_{11} + a_{12} + a_{13}) + c_2(a_{21} + a_{22} + a_{23})$

$\qquad + \cdots + c_m(a_{m1} + a_{m2} + a_{m3})$

But recall from above that we have $a_{i1} + a_{i2} + a_{i3} = 0$ for $1 \le i \le m$ (as a consequence of $\begin{bmatrix} 1&1&1 \end{bmatrix}^T$ being in the nullspace of $A$) so that the equation above reduces to

$c_1 \cdot 0 + c_2 \cdot 0 + \cdots + c_m \cdot 0 = 0$

Since the left side of the above system of equations sums to 0 and the right side sums to 3 we have a contradiction and conclude that the vector $\begin{bmatrix} 1&1&1 \end{bmatrix}^T$ cannot be in the row space of $A$.

So if the vector $\begin{bmatrix} 1&1&1 \end{bmatrix}^T$ is in the nullspace of $A$ then it cannot also be in the row space of $A$.

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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