## 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 .

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