## 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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