## Linear Algebra and Its Applications, Review Exercise 2.17

Review exercise 2.17. Suppose that $x$ is a vector in $\mathbb{R}^n$ and that $x^Ty = 0$ for all $y$. Show that $x$ must be the zero vector.

Answer: For $x = (x_1, x_2, \ldots, x_n)$ and $y = (y_1, y_2, \ldots, y_n)$ we have

$x^Ty = \sum_{j=1}^n x_jy_j$

Since  $x^Ty = 0$ for all $y$ we must have $x^Te_i = 0$ for each of the elementary vectors $e_1$, $e_2$, through $e_n$. By the definition of the elementary vectors we have $e_{i_j}= 1$ if $i = j$ and $e_{i_j} = 0$ otherwise.

So for all $1 \le i \le n$ we have

$0 = x^Te_i = \sum_{j=1}^n x_je_{i_j} = x_ie_{i_i} = x_i \cdot 1 = x_i$

Since $x_i = 0$ for all $1 \le i \le n$ we have $x = 0$.

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.

This entry was posted in linear algebra. Bookmark the permalink.