Linear-Algebra-Done-Right · Unit 4 · Video 2 · Interactive Practice

Dimension: Why Every Basis Has the Same Length

IKey Formulas

Formula	Name	Meaning
$\text{length}(B_1) = \text{length}(B_2)$	Bases have equal length	$B_1$ is independent and $B_2$ spans, so $\text{length}(B_1) \leq \text{length}(B_2)$ ; swapping the roles gives $\geq$
$\dim \mathbf{F}^n = n$ , $\ \dim \mathcal{P}_m(\mathbf{F}) = m + 1$	Dimension from a standard basis	Count $1, z, \dots, z^m$ : the constant polynomial $1$ is the extra vector
$\dim U \leq \dim V$ , and $\dim U = \dim V \implies U = V$	Subspace $U$ of finite-dimensional $V$	A basis of $U$ is an independent list in $V$ ; a subspace other than $V$ has $\dim U \lt \dim V$
$\text{length} = \dim V$ and (independent or spanning) $\implies$ basis	Right length	Extending an independent list adds no vectors; reducing a spanning list deletes none

Key Insight: Dimension depends on the scalars: $\dim \mathbf{R}^2 = 2$ over $\mathbf{R}$ , while $\dim \mathbf{C} = 1$ over $\mathbf{C}$ , even though $(x, y) \leftrightarrow x + iy$ matches the two sets and their additions.

IIThe Right Length: One Property Is Enough

In $\mathbf{R}^2$ , a list of length $2 = \dim \mathbf{R}^2$ is independent exactly when it spans; other lengths never give a basis.

IIISqueezing $\dim U$ When $p'(5) = 0$

For $U = \{p \in \mathcal{P}_3(\mathbf{R}) : p'(5) = 0\}$ , three facts squeeze $\dim U$ down to a single value.

Step 1

Step 1 — three polynomials in $U$

1' = 0

\big((x-5)^2\big)' = 2(x-5)

, which is

0

x = 5

\big((x-5)^3\big)' = 3(x-5)^2

, which is

0

x = 5

Step 2 — linearly independent

a + b(x-5)^2 + c(x-5)^3 = 0 \ \text{ for all } x

Only

c(x-5)^3

produces

x^3

, so

c = 0

; then only

b(x-5)^2

produces

x^2

, so

b = 0

; then

a = 0

An independent list in

U

is no longer than a basis of

U

3 \leq \dim U

Step 3 — the upper bound

3 \leq \dim U \leq \dim \mathcal{P}_3(\mathbf{R}) = 4

Step 4 — rule out $4$

x' = 1 \neq 0

x = 5

, so

x \notin U

and

U \neq \mathcal{P}_3(\mathbf{R})

\dim U = 4 = \dim \mathcal{P}_3(\mathbf{R})

would force

U = \mathcal{P}_3(\mathbf{R})

, so

\dim U = 3

Step 5 — a basis for free

1, (x-5)^2, (x-5)^3

is independent in

U

with length

3 = \dim U

, so it is a basis of

U

; spanning is never checked.

IVDimension Depends on the Scalars

With $(x, y)$ matched to $x + iy$ , the plane needs two vectors over $\mathbf{R}$ but only one over $\mathbf{C}$ .

VQuiz Questions

Problem 1 · Count a Standard Basis

Find $\dim \mathbf{F}^4$ and $\dim \mathcal{P}_4(\mathbf{F})$ .

✅ Correct! The standard basis of

\mathbf{F}^4

has four vectors, and

1, z, z^2, z^3, z^4

has five.

❌ Close, but… the list

1, z, z^2, z^3, z^4

has five vectors: the constant polynomial

1

is the one that is easy to forget.

❌ Not quite. The standard basis of

\mathbf{F}^4

(1, 0, 0, 0)

(0, 1, 0, 0)

(0, 0, 1, 0)

(0, 0, 0, 1)

: one vector per coordinate, four in all.

❌ Not quite. The two values are swapped:

\mathbf{F}^4

has four coordinates, while

\mathcal{P}_4(\mathbf{F})

has the four powers

z, \dots, z^4

plus the constant

1

Show solution

The dimension is the length of any basis, so it is enough to count one basis.

$\mathbf{F}^4$ has the standard basis $(1, 0, 0, 0)$ , $(0, 1, 0, 0)$ , $(0, 0, 1, 0)$ , $(0, 0, 0, 1)$ , of length $4$ , so $\dim \mathbf{F}^4 = 4$ .

$\mathcal{P}_4(\mathbf{F})$ has the standard basis $1, z, z^2, z^3, z^4$ : the constant $1$ plus one power of $z$ for each exponent from $1$ to $4$ . That is $4 + 1 = 5$ vectors, so $\dim \mathcal{P}_4(\mathbf{F}) = 5$ , matching $\dim \mathcal{P}_m(\mathbf{F}) = m + 1$ .

Problem 2 · Which Statement Is True?

Given: $V$ is a finite-dimensional vector space. Exactly one of these statements is always true. Which one?

✅ Correct!

\dim \mathcal{P}_3(\mathbf{R}) = 4

, so a spanning list of length

4

is a basis, and every basis is linearly independent.

❌ Not quite. The list

(1, 0), (2, 0)

\mathbf{R}^2

has length

2 = \dim \mathbf{R}^2

but is neither independent nor spanning. The right length plus one of the two properties gives a basis; length alone does not.

❌ Not quite. In

\mathbf{R}^2

, the subspaces

\{(x, 0) : x \in \mathbf{R}\}

and

\{(0, y) : y \in \mathbf{R}\}

both have dimension

1

but are different. The full-dimension theorem compares a subspace with the space that contains it.

❌ Not quite. Every basis of

V

has length exactly

\dim V

, so a list of length

\dim V + 1

is never a basis. A spanning list that long is linearly dependent.

Show solution

(a) is true. The standard basis $1, x, x^2, x^3$ shows $\dim \mathcal{P}_3(\mathbf{R}) = 4$ . A list that spans $\mathcal{P}_3(\mathbf{R})$ and has length $4 = \dim \mathcal{P}_3(\mathbf{R})$ is a basis, by the second right-length theorem, and a basis is linearly independent.

(b) is false: $(1, 0), (2, 0)$ is dependent, since $2(1, 0) - (2, 0) = 0$ , and its span is only the horizontal axis.

(c) is false: the two coordinate axes of $\mathbf{R}^2$ each have a basis of length $1$ but are unequal. The theorem " $\dim U = \dim V \implies U = V$ " needs $U$ to be a subspace of $V$ itself.

(d) is false: all bases of $V$ have the same length $\dim V$ , so no list of length $\dim V + 1$ is a basis.

Problem 3 · Polynomials Vanishing at 3

Given: $U = \{p \in \mathcal{P}_4(\mathbf{R}) : p(3) = 0\}$ , a subspace of $\mathcal{P}_4(\mathbf{R})$ . Find $\dim U$ and a basis of $U$ .

What is dim U?

Which list is a basis of U?

✅ Correct!

4 \leq \dim U \lt 5

gives

\dim U = 4

, and the independent list

(x-3), \dots, (x-3)^4

of length

4

is a basis without a spanning check.

❌ Check dim U. Bound it from both sides: find an independent list in

U

, then compare

U

with

\mathcal{P}_4(\mathbf{R})

❌ Not quite.

5 = \dim \mathcal{P}_4(\mathbf{R})

, and a subspace of full dimension equals the whole space. But

1 \notin U

since

1 \neq 0

x = 3

, so

U \neq \mathcal{P}_4(\mathbf{R})

❌ Not quite. One condition does not leave one dimension:

(x-3), (x-3)^2, (x-3)^3, (x-3)^4

are four independent vectors in

U

, so

\dim U \geq 4

❌ Close, but…

U

contains the four independent polynomials

(x-3), \dots, (x-3)^4

, and an independent list in

U

is no longer than a basis of

U

, so

\dim U \geq 4

❌ Check the basis. Each vector must lie in

U

, the list must be independent, and its length must be

\dim U

❌ Not quite. The constant polynomial

1

takes the value

1 \neq 0

x = 3

, so it is not in

U

❌ Close, but… this list is independent and lies in

U

, but its length

3

is less than

\dim U = 4

, and an independent list shorter than the dimension never spans.

❌ Not quite. These polynomials take the values

3, 9, 27, 81

x = 3

, so none of them is in

U

Show solution

In $U$ : each of $(x-3), (x-3)^2, (x-3)^3, (x-3)^4$ has value $0$ at $x = 3$ .

Independent: suppose $a(x-3) + b(x-3)^2 + c(x-3)^3 + d(x-3)^4 = 0$ for all $x$ . Only the last term produces $x^4$ , so $d = 0$ ; then only $c(x-3)^3$ produces $x^3$ , so $c = 0$ ; similarly $b = 0$ and then $a = 0$ .

Lower bound: an independent list in $U$ is no longer than a basis of $U$ , so $4 \leq \dim U$ .

Upper bound: $U$ is a subspace of $\mathcal{P}_4(\mathbf{R})$ , so $\dim U \leq 5$ . The constant $1$ lies in $\mathcal{P}_4(\mathbf{R})$ but not in $U$ , so $U \neq \mathcal{P}_4(\mathbf{R})$ , and $\dim U = 5$ would force $U = \mathcal{P}_4(\mathbf{R})$ . Hence $\dim U = 4$ .

Basis: $(x-3), (x-3)^2, (x-3)^3, (x-3)^4$ is an independent list in $U$ of length $4 = \dim U$ , so it is a basis of $U$ .

Problem 4 · The Same Set, Two Fields

Given: $\mathbf{C}^2$ is a vector space over $\mathbf{C}$ . Keeping the same addition but allowing only real scalars makes $\mathbf{C}^2$ a vector space over $\mathbf{R}$ . Find its dimension in each case.

✅ Correct! Over

\mathbf{C}

the basis is

(1, 0), (0, 1)

; over

\mathbf{R}

it is

(1, 0), (i, 0), (0, 1), (0, i)

❌ Not quite. With only real scalars,

(i, 0)

is not a multiple of

(1, 0)

, so

(1, 0), (0, 1)

no longer spans: the dimension depends on the field.

❌ Not quite. The values are swapped. Complex scalars give more multiples of each vector, so fewer vectors are needed over

\mathbf{C}

, not more.

❌ Not quite.

\dim \mathbf{C} = 1

over

\mathbf{C}

, but

\mathbf{C}^2

\mathbf{F}^n

with

\mathbf{F} = \mathbf{C}

and

n = 2

, so its dimension over

\mathbf{C}

2

; over

\mathbf{R}

each coordinate needs two real directions.

Show solution

Over $\mathbf{C}$ : $\mathbf{C}^2 = \mathbf{F}^2$ with $\mathbf{F} = \mathbf{C}$ , whose standard basis $(1, 0), (0, 1)$ has length $2$ .

Over $\mathbf{R}$ : write a vector as $(a + bi, c + di)$ with $a, b, c, d \in \mathbf{R}$ . Then

(a + bi, c + di) = a(1, 0) + b(i, 0) + c(0, 1) + d(0, i),

so the list $(1, 0), (i, 0), (0, 1), (0, i)$ spans with real scalars. If this combination is $0$ , then $a + bi = 0$ and $c + di = 0$ , so $a = b = c = d = 0$ : the list is independent. It is a basis of length $4$ , so the dimension over $\mathbf{R}$ is $4$ , twice the dimension over $\mathbf{C}$ , just as $\dim \mathbf{C} = 1$ over $\mathbf{C}$ while $\mathbf{R}^2$ has dimension $2$ .

Solved: 0 / 4

Dimension: Why Every Basis Has the Same Length

IKey Formulas

IIThe Right Length: One Property Is Enough

IIISqueezing dim⁡U\dim UdimU When p′(5)=0p'(5) = 0p′(5)=0

IVDimension Depends on the Scalars

VQuiz Questions

IIISqueezing $\dim U$ When $p'(5) = 0$