# Determining which vector is linearly dependent in a set

I'm confused with a particular problem when given something like this to solve:

I am told to consider sets of vectors in $$\mathbb R^3$$ and $$\mathbb R^4$$ respectively. I am then told to construct a subset of this set that will span the subspace $$U = Span \ (A)$$, so basically I have to snub out any linearly dependent vectors in these two sets:

I'll merely look at one of the examples. The solution to this involved bringing the vectors together to create a matrix with the columns as the vectors, and then row-reduce to echelon form. Once doing so, analyzing the system of equations included something I find concerning. In my own solution, I wrote the column vectors as row vectors, and row-reduced. I'll compare our row reduced matrices.

$$\begin{bmatrix} 2 & -1 & 2 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix}$$

Which lead me to believe that the fourth vector was linearly dependent, as the row referring to it is a zero row. Thus, I stated that the basis that will span $$U$$ is $$B = \{(2,-1,2), (0,3,4), (0,0,-1)\}$$. However, my lecturer's answer was much different. Instead, they did what I stated in that they just kept the column vectors as columns for their matrix and then converted it to echelon form:

$$\begin{bmatrix} 1 & -1 & -1 & 1\\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & 1 & 7 \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \gamma \\ \delta\\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$

Where $$\alpha, \beta, \gamma, \delta \in F$$. This led to the conclusion:

$$\alpha - \beta - \gamma + \delta = 0$$ $$\beta + (2/3) \ \gamma + (2/3) \ \delta = 0$$ $$\gamma - 7 \ \delta= 0$$

The lecturer used $$\delta = 1$$ and was able to show that, by determining the other constants with this statement, a $$\{(4,-1,3)\}$$ was able to be expressed as a linear combination of the vectors preceding it.

My questions are two-fold:

a) Why was my method, and echelon form matrix, not valid for this purpose?

b) Is any initialization of the free parameter $$\delta$$ viable? If so, I did not succeed in showing this for $$\delta = 3$$, as I could not represent the fourth vector as a linear combination of vectors like the lecturer did. If only $$\delta = 1$$ is the only viable option to get the correct answer, how is one supposed to know what to set $$\delta$$ to to arrive at the correct answer in this case? Just for clarity, I mean "correct answer" in the context of it being the correct initialization of $$\delta$$ such that the fourth vector can be shown to be linearly dependent.

• Comment only since I haven't read your whole answer. If a list of vectors is dependent there's no way to single out one of them as "the" dependent vector. Whenever you have a linear combination that's $0$ any vector with a nonzero coefficient is linearly dependent on the others. – Ethan Bolker Oct 1 '18 at 19:58
• Short answer to a): Your method found a basis for the span, but you can’t identify which vectors of the original set are redundant. – amd Oct 1 '18 at 20:01
• @amd why can’t I? – sangstar Oct 1 '18 at 20:07
• The zero rows are always at the bottom of the matrix and the process might reorder rows. There’s no direct correspondence between the rows of the echelon form and the rows of the original matrix as there is for the columns. Try it with $\small\begin{bmatrix}0&1&1\\0&-1&-1\\1&0&0\end{bmatrix}$. Your conclusion would be that $(1,0,0)$ is a linear combination of the other two rows, which is patently impossible. – amd Oct 1 '18 at 20:10

If you perform Gaussian elimination on the matrix with the given vectors as columns, then the columns with no leading $$1$$ can be removed.
\begin{align} \begin{bmatrix} 2 & 1 & 0 & 4 \\ -1 & 1 & 1 & -1 \\ 2 & 3 & 1 & 3 \end{bmatrix} &\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 3/2 & 1 & 1 \\ 0 & 2 & 1 & -1 \end{bmatrix} && \begin{aligned} R_1&\gets\tfrac{1}{2}R_1 \\ R_2&\gets R_2+R_1 \\ R_3&\gets R_3-2R_1\end{aligned} \\&\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & -1/3 & -7/3 \end{bmatrix} && \begin{aligned} R_2&\gets\tfrac{2}{3}R_2 \\ R_3&\gets R_3-2R_2\end{aligned} \\&\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & 1 & 7 \end{bmatrix} && R_3\gets -3R_3 \end{align} At this point we see that the fourth vector can be removed. If we also compute the RREF \begin{align} \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 2/3 & 2/3 \\ 0 & 0 & 1 & 7 \end{bmatrix} &\to \begin{bmatrix} 1 & 1/2 & 0 & 2 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 5 \end{bmatrix} && R_2\gets -\frac{2}{3}R_3 \\&\to \begin{bmatrix} 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & -4 \\ 0 & 0 & 1 & 7 \end{bmatrix} && R_1\gets -\frac{1}{2}R_2 \end{align} We see that $$\begin{bmatrix} 4 \\ -1 \\ 3 \end{bmatrix}= 4\, \begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix}\, -4\, \begin{bmatrix} 1 \\ 1 \\ 3 \end{bmatrix}\, +7\, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$$
You may have taken a transpose where not appropriate. In both cases, four columns, the vector created by summing $$w c_1 + x c_2 + y c_3 + z c_4$$ is exactly the result of multiplying the matrix on the left times the column vector $$V = \left( \begin{array}{c} w \\ x \\ y \\ z \end{array} \right)$$ Finding the reduced row echelon form of the rectangular matrix leads to a way to find a null vector. For the first matrix, 3 by 4, it is guaranteed there is at least one nonzero vector $$V$$ that is a null vector. Write out $$V,$$ any nonzero entry is a column in the rectangular matrix that can be written in terms of the others.
In the 4 by 4 case, it is possible that the only $$V$$ that works is the zero vector. Not sure. If there is a nonzero $$V,$$ same as before
$$V = \left( \begin{array}{c} 4 \\ -4 \\ 7 \\ -1 \end{array} \right)$$ or any nonzero multiple of that, so that any one of the four columns can be written in terms of the other three. Put the other way, with the four columns in the 3 by 4 matrix, we have $$4 c_1 - 4 c_2 + 7 c_3 - c_4 = 0$$