LINEAR-ALGEBRA · Unit 9 · Video 2 · Interactive Practice

The Goldilocks Principle: Basis and Dimension

IKey Formulas

Formula	Name	Description
$c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots = \mathbf{0} \Rightarrow$ all $c_i = 0$	Linear Independence	Only the trivial solution makes the zero vector
$\text{Span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} = V$	Spanning Set	Every vector in $V$ is a linear combination
$n$ vectors form a basis for $\mathbb{R}^n \iff \det \neq 0$	Basis Test	Matrix invertibility checks both properties at once
$\dim(V) =$ number of vectors in any basis	Dimension Theorem	This count is the same for every basis of $V$

IIVisualization 1 — The Parallelogram Test

Two vectors form a basis for $\mathbb{R}^2$ exactly when their parallelogram has nonzero area — and that area is $|\det|$ .

v₁ x 2 v₁ y 1 v₂ x −1 v₂ y 2

💡 Challenge: make $\mathbf{v}_2$ a multiple of $\mathbf{v}_1$ — say $\mathbf{v}_1 = (2, 1)$ , $\mathbf{v}_2 = (4, 2)$ . The area collapses to $0$ : two dependent vectors span only a line.

IIIVisualization 2 — The 3×3 Determinant Test

Fix $\mathbf{v}_1, \mathbf{v}_2$ and move the third vector: the determinant is nonzero exactly when $\mathbf{v}_3$ escapes their plane.

v₃ x 3 v₃ y 3 v₃ z 8

A = [ v₁ | v₂ | v₃ ]

💡 $(3, 4, 7) = \mathbf{v}_1 + \mathbf{v}_2$ lies in the plane, so $\det = 0$ ; $(3, 3, 8)$ escapes it — three independent directions and $\det = 2$ .

IVVisualization 3 — Too Few, Just Right, Too Many

A basis for $\mathbb{R}^2$ needs exactly two vectors — one too few can't span, one too many must be redundant.

💡 That just-right count — $2$ for $\mathbb{R}^2$ , $3$ for $\mathbb{R}^3$ — is the dimension: every basis of a space has the same size.

VQuiz Questions

Question 1

The matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 3 \\ 2 & 5 & 8 \end{bmatrix}$ has $\det(A) = 2$ .

Do the columns of $A$ form a basis for $\mathbb{R}^3$ ?

✅ Correct! For an $n \times n$ matrix, $\det \neq 0$ confirms both independence and spanning in one computation.

❌ Not quite. Recall the key theorem: for $n$ vectors in $\mathbb{R}^n$ , matrix invertibility ( $\det \neq 0$ ) is the single test for a basis.

Show solution

Solution:

For $n$ vectors in $\mathbb{R}^n$ , the single basis test is:

$n \text{ vectors form a basis for } \mathbb{R}^n \iff n \times n \text{ matrix is invertible} \iff \det \neq 0$

Since $\det(A) = 2 \neq 0$ , the matrix is invertible. This simultaneously guarantees:

Linear independence: the null space contains only $\mathbf{0}$
Spanning: every vector in $\mathbb{R}^3$ is reachable

One check replaces two. The answer is Yes.

Note: the value of the determinant (2) tells us the volume scaling factor, not how many vectors are independent!

Question 2

Vectors $\mathbf{v}_1 = (1, 1, 2)$ and $\mathbf{v}_2 = (2, 3, 5)$ span a plane in $\mathbb{R}^3$ . The vector $(3, 4, 7)$ equals $\mathbf{v}_1 + \mathbf{v}_2$ .

What is the dimension of $\text{Span}\{\mathbf{v}_1, \, \mathbf{v}_2, \, (3, 4, 7)\}$ ?

✅ Correct! A dependent vector doesn't add a new dimension — the span remains a 2D plane.

❌ Not quite. Since $(3,4,7)$ already lies in the span of the first two vectors, adding it doesn't change the span at all.

Show solution

Solution:

Since $(3, 4, 7) = \mathbf{v}_1 + \mathbf{v}_2$ , the third vector is a linear combination of the first two. Adding a dependent vector does not expand the span:

$\text{Span}\{\mathbf{v}_1, \mathbf{v}_2, (3,4,7)\} = \text{Span}\{\mathbf{v}_1, \mathbf{v}_2\}$

The span is still the same 2D plane. The set $\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for that plane (2 independent vectors), so:

$\boxed{\dim = 2}$

This is the "Too Many" scenario from the Goldilocks principle — the third vector is dead weight.

Question 3

True or False: A vector space $V$ can have one basis with 4 vectors and another basis with 5 vectors.

✅ Correct! The Dimension Theorem guarantees that all bases for the same space have the same number of vectors.

❌ Not quite. Recall the Dimension Theorem — the number of vectors in a basis is an invariant of the space, not a choice!

Show solution

Solution:

False. This directly contradicts the Dimension Theorem:

Every basis for a given vector space has the same number of vectors.

This invariant count is the dimension of the space. If one basis has 4 vectors, then $\dim(V) = 4$ , and every other basis for $V$ must also have exactly 4 vectors.

A set of 5 vectors in a 4-dimensional space must contain at least one dependent (redundant) vector, so it cannot be a basis.

Question 4

You know that $\dim(\mathbb{R}^4) = 4$ . You have a set of 4 vectors in $\mathbb{R}^4$ . According to the basis verification algorithm from the video, what is the minimum you need to check to confirm this is a basis?

✅ Correct! When the vector count matches the known dimension, checking independence alone is sufficient — spanning follows automatically.

❌ Not quite. When you know the dimension and have the right count of vectors, the basis verification algorithm provides a powerful shortcut.

Show solution

Solution:

Since we know $\dim(\mathbb{R}^4) = 4$ and we have exactly $k = 4$ vectors, the algorithm goes:

✅ Know dimension? Yes, $d = 4$ .
✅ Count matches? Yes, $k = d = 4$ .
Check independence only: form the $4 \times 4$ matrix and verify $\det \neq 0$ .

The powerful shortcut from the Dimension Theorem: when you have exactly $d$ vectors in a $d$ -dimensional space, independence automatically guarantees spanning. So checking independence alone is sufficient!

For $\mathbb{R}^n$ specifically, the matrix invertibility test ( $\det \neq 0$ ) is the single computation that confirms everything.

Note: orthogonality is a stronger condition than independence — it's sufficient but not necessary for a basis.

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