Linear Algebra and Its Applications, Review Exercise 2.14

Review exercise 2.14. Do the three vectors (1, 1, 3), (2, 3, 6), and (1, 4, 3) form a basis for the vector space \mathbb{R}^3?

Answer: In order for the three vectors to be a basis for \mathbb{R}^3 they must be linearly independent. We can test this by doing elimination on the following matrix with the three vectors as the columns:

\begin{bmatrix} 1&2&1 \\ 1&3&4 \\ 3&6&3 \end{bmatrix}

Subtracting 1 times the first row from the second row we have

\begin{bmatrix} 1&2&1 \\ 1&3&4 \\ 3&6&3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&1 \\ 0&1&3 \\ 3&6&3 \end{bmatrix}

and subtracting 3 times the first row from the third row we have

\begin{bmatrix} 1&2&1 \\ 0&1&3 \\ 3&6&3 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&1 \\ 0&1&3 \\ 0&0&0 \end{bmatrix}

Since there are only two pivots, the rank of the vector’s column space is r = 2. This means that the three columns are not linearly independent and thus the original three vectors cannot be a basis for \mathbb{R}^3.

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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