Practice: Linear Independence, Rank, and Bases

Recognition · Classification

Four vectors are given in $R^{3}$ . What can be concluded before any computation?

1 hint available, least help first.

Hint 1: Retrieval cue

What is the largest possible rank of a matrix with three rows?

Classification

Consider the matrix whose columns are $a_{1} = (1, 2, 3)$ , $a_{2} = (2, 4, 6)$ and $a_{3} = (1, 0, 1)$ :

A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 4 & 0 \\ 3 & 6 & 1 \end{matrix}] .

Report the rank of $A$ .

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

Direct application · Construction

For each set below, decide whether the vectors are linearly independent, state the rank of the matrix formed from them, and where the set is dependent give an explicit combination expressing one vector in terms of the others. Verify any relation you give.

(a) $(\begin{matrix} 1 \\ 2 \end{matrix})$ , $(\begin{matrix} 3 \\ 1 \end{matrix})$

(b) $(\begin{matrix} 1 \\ 2 \end{matrix})$ , $(\begin{matrix} 2 \\ 4 \end{matrix})$

For the set that is independent, say whether the matrix it forms can serve as a basis and why.

Write your answer

Write your answer, then compare it with the worked solution.

2 hints available, least help first.

Hint 1: Retrieval cue

With exactly two vectors, what single question decides independence?

Hint 2: Concept cue

Form the two-by-two determinant in each case and compare with zero.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) Independent. The matrix $(\begin{matrix} 1 & 3 \\ 2 & 1 \end{matrix})$ has determinant $(1) (1) - (3) (2) = - 5 \neq 0$ , so rank 2. With two vectors, proportionality is also a valid check: scaling $(1, 2)^{T}$ by 3 gives $(3, 6)^{T}$ , not $(3, 1)^{T}$ . Because the columns are independent and the matrix is square with full rank, it is invertible and can serve as a basis: every right-hand side is reachable, in exactly one way. (b) Dependent. The second vector is exactly twice the first: $(2, 4)^{T} = 2 (1, 2)^{T}$ , checking $2 (1) = 2$ and $2 (2) = 4$ . The determinant is $(1) (4) - (2) (2) = 0$ , confirming it, and the rank is 1 because only one independent direction is present.

A complete answer does each of these:

Connects the result back to a basis
States the conclusion, not only the working
Exhibits the dependence relation itself
Uses a method suited to the problem
Reports the rank explicitly

Error diagnosis · Explanation

A student is given

a_{1} = (\begin{matrix} 1 \\ 2 \end{matrix}), a_{2} = (\begin{matrix} 3 \\ 1 \end{matrix}), a_{3} = (\begin{matrix} 2 \\ - 1 \end{matrix})

and writes:

" $a_{2}$ is not a multiple of $a_{1}$ , $a_{3}$ is not a multiple of $a_{1}$ , and $a_{3}$ is not a multiple of $a_{2}$ . No two are parallel, so the three vectors are linearly independent."

Each of the student's three observations is correct. Explain why the conclusion does not follow, give an explicit nontrivial combination of the three vectors equal to the zero vector, and state the argument that settles the question fastest.

Write your answer

Write your answer, then compare it with the worked solution.

2 hints available, least help first.

Hint 1: Retrieval cue

How many vectors are there, and how many components does each have?

Hint 2: Concept cue

Try to write $a_{3}$ as a combination of $a_{1}$ and $a_{2}$ , allowing both coefficients to vary.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

The three pairwise observations are all true, but independence is not a pairwise property. The definition asks whether any nontrivial choice of all three coefficients at once gives the zero vector, and a pairwise check never considers three vectors together. Here $a_{1} - a_{2} + a_{3} = 0$ : the first components give $1 - 3 + 2 = 0$ and the second give $2 - 1 - 1 = 0$ . Equivalently $a_{3} = - a_{1} + a_{2}$ , so the third vector adds no direction the first two do not already reach. The fastest settlement is the counting argument: three vectors in $R^{2}$ are always dependent, since rank cannot exceed the number of components, so no computation was needed at all.

A complete answer does each of these:

States the conclusion, not only the working
Exhibits the dependence relation itself
Uses a method suited to the problem

Transfer · Evaluation · Explanation

A colleague is implementing the simplex method. Their code selects $m$ columns to form the basis matrix $A_{B}$ , then computes the basic solution by solving $A_{B} x_{B} = b$ . They report:

"On one problem the solver crashed with a division-by-zero deep inside the linear solve. I have added a check that skips any column set where this happens and tries the next one, and now it runs."

Explain what property of the selected columns causes this, why the crash is the symptom rather than the disease, and what the code should test for before attempting the solve. Then say what would be wrong with treating the resulting point as a basic feasible solution if the check were omitted.

Write your answer

Write your answer, then compare it with the worked solution.

2 hints available, least help first.

Hint 1: Retrieval cue

What must be true of the columns of $A_{B}$ for $A_{B}^{- 1}$ to exist?

Hint 2: Strategy cue

Consider the equivalent conditions for a square matrix: independence, full rank, nonzero determinant, invertibility, unique solutions.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

The selected columns are linearly dependent, so $A_{B}$ is singular: its rank is below $m$ , its determinant is zero, and no inverse exists. The division by zero is where that shows up in the elimination arithmetic, but the cause is the column selection, not the solve. The code should test the candidate columns for independence before solving, by checking that the reduction produces $m$ pivots or equivalently that the determinant is nonzero, and reject the set as not a basis if it does not. Skipping failures does produce working behavior, but it discovers the defect by crashing rather than by testing, and it hides the reason. Treating the result as a basic feasible solution would be wrong because a dependent column set does not determine a point at all: $A_{B} x_{B} = b$ has either no solution or infinitely many, so there is nothing for the basis to designate, and the definition of a basic solution presupposes exactly the uniqueness that independence supplies.

A complete answer does each of these:

Exhibits the dependence relation itself
Connects the result back to a basis
States the conclusion, not only the working

Practice: Linear Independence, Rank, and Bases

Recognition · Classification

Classification

Direct application · Construction

Error diagnosis · Explanation

Transfer · Evaluation · Explanation

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