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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