Linear Algebra and Its Applications, Exercise 2.3.2

Exercise 2.3.2. State whether the following sets of vectors are linearly independent or not

(a) $(1, 1, 2)$ , $(1, 2, 1)$ , and $(3, 1, 1)$

(b) given any four vectors $v_1$ , $v_2$ , $v_3$ , and $v_4$ , the vectors $v_1-v_2$ , $v_2-v_3$ , $v_3-v_4$ , and $v_4-v_1$

Answer: (a) We test for linear independence by doing elimination on the matrix with columns consisting of the three vectors in question:

$\begin{bmatrix} 1&1&3 \\ 1&2&1 \\ 2&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&3 \\ 0&1&-2 \\ 0&-1&-5 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&3 \\ 0&1&-2 \\ 0&0&-7 \end{bmatrix}$

Since the resulting matrix has three pivots and thus rank $r = 3 = n$ the columns (and thus the vectors in question) are linearly independent.

(b) If we simply add the four vectors (which corresponds to a linear combination with $c_1 = c_2 = c_3 = c_4 = 1$ ) we have

$(v_1-v_2) + (v_2-v_3) + (v_3-v_4) + (v_4-v_1)$

$= v_1 - v_2 + v_2 - v_3 + v_3 - v_4 + v_4 - v_1 = 0$

So the four vectors in question are always linearly dependent no matter whether the original four vectors are linearly independent or not.

(c) From theorem 2G on page 82 we see that any set of four vectors in $\mathbf{R}^3$ must be linearly dependent, so the four vectors in question are always linearly dependent for any values of $x$ , $y$ , and $z$ .

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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2 Responses to Linear Algebra and Its Applications, Exercise 2.3.2

Aswin j bhat says:

September 8, 2023 at 6:11 am

Part a of the question says (1,1,2) but sir you solved using (1,1,3)

- hecker says:
  
  September 8, 2023 at 2:27 pm
  
  Actually, the original matrix is correct. Note that the matrix has the original three vectors as its columns, not as its rows. I have redone the calculations and updated the post just to make sure I did this correctly.