Linear-Algebra-Done-Right · Unit 5 · Video 4 · Interactive Practice

The Fundamental Theorem of Linear Maps and Linear Systems

IKey Formulas

Formula	Name	Hypothesis
$\dim V = \dim \text{null}\, T + \dim \text{range}\, T$	Fundamental theorem of linear maps (rank-nullity)	$V$ finite-dimensional, $T \in \mathcal{L}(V, W)$
$\dim \text{null}\, T \geq \dim V - \dim W > 0$	Map to a smaller space is not injective	$\dim V > \dim W$
$\dim \text{range}\, T \leq \dim V < \dim W$	Map to a bigger space is not surjective	$\dim V < \dim W$
$T(x_1, \dots, x_n) = \left(\sum_{k=1}^{n} A_{1,k} x_k, \dots, \sum_{k=1}^{n} A_{m,k} x_k\right)$	System of $m$ equations in $n$ variables as $T\colon \mathbf{F}^n \to \mathbf{F}^m$	$n > m$ : nonzero homogeneous solution; $n < m$ : some constants unsolvable

Key Insight: Extend a basis $u_1, \dots, u_m$ of $\text{null}\, T$ to a basis $u_1, \dots, u_m, v_1, \dots, v_n$ of $V$ ; then $Tv_1, \dots, Tv_n$ is a basis of $\text{range}\, T$ . Only the counts matter, never the coefficients, and $\dim W$ never enters the formula.

IIBoth Sides of the Count

For every choice of coefficients $A_{j,k}$ , $\dim \text{null}\, T + \dim \text{range}\, T$ equals the number of variables $n = \dim \mathbf{F}^n$ .

Equations m Variables n

💡 Challenge: with $m = 3$ and $n = 4$ , try to make $T$ injective. The theorem says no entries can do it, because $\dim \text{null}\, T \geq 4 - 3 = 1$ .

IIIThe Proof in One Picture

A basis of $\text{null}\, T$ goes to $0$ ; the vectors added to reach a basis of $V$ go to a basis of $\text{range}\, T$ .

Proof step 1

Step 1 · Basis of null T for $T(x_1, \dots, x_5) = (x_1 + x_2,\ x_3,\ x_4 + x_5,\ 0)$

Tx = 0 \iff x_2 = -x_1,\ x_3 = 0,\ x_5 = -x_4

u_1 = (1, -1, 0, 0, 0)

u_2 = (0, 0, 0, 1, -1)

$\dim \text{null}\, T = 2$

IVCounting Equations Against Variables

Fewer equations than variables forces a nonzero homogeneous solution; more equations than variables leaves some constants unreachable.

System

💡 With two equations in one variable, $c = (0, 0)$ is always solvable by $x = 0$ : the theorem promises only that some choice of constants fails.

VQuiz Questions

Problem 1 · Apply the Fundamental Theorem

Given: $T \in \mathcal{L}(\mathbf{F}^7, \mathbf{F}^4)$ with $\dim \text{null}\, T = 3$ . Find $\dim \text{range}\, T$ .

✅ Correct!

7 = 3 + \dim \text{range}\, T

gives

4

; and a

4

-dimensional subspace of

\mathbf{F}^4

is all of

\mathbf{F}^4

, so

T

is surjective.

❌ Not quite. You computed

\dim W - \dim \text{null}\, T

. The theorem is about the domain:

\dim V = \dim \text{null}\, T + \dim \text{range}\, T

, and

\dim W

does not appear.

❌ Not quite. Adding

7 + 3

treats the null space as extra dimensions. The null space is part of

V

\dim \text{range}\, T = \dim V - \dim \text{null}\, T

. Also,

\text{range}\, T \subseteq \mathbf{F}^4

cannot have dimension

10

❌ Not quite. Solve

\dim V = \dim \text{null}\, T + \dim \text{range}\, T

with

\dim V = 7

for the range.

Show solution

Here $V = \mathbf{F}^7$ is finite-dimensional, so the fundamental theorem applies:

\dim \mathbf{F}^7 = \dim \text{null}\, T + \dim \text{range}\, T \quad\Longrightarrow\quad 7 = 3 + \dim \text{range}\, T

\dim \text{range}\, T = 4

Since $\text{range}\, T$ is a subspace of $\mathbf{F}^4$ with dimension $4 = \dim \mathbf{F}^4$ , it equals $\mathbf{F}^4$ : this $T$ is surjective.

Problem 2 · What the Dimensions Force

Given: an arbitrary $T \in \mathcal{L}(\mathbf{F}^3, \mathbf{F}^5)$ . Which statement must be true?

✅ Correct!

\dim \text{range}\, T \leq \dim \mathbf{F}^3 = 3 < 5

, so

\text{range}\, T \neq \mathbf{F}^5

❌ Not forced. A bigger target does not make a map injective: the zero map

\mathbf{F}^3 \to \mathbf{F}^5

has

\text{null}\, T = \mathbf{F}^3

and

\dim \text{range}\, T = 0

. Neither theorem has a converse.

❌ Not forced. The bound

\dim \text{null}\, T \geq \dim V - \dim W

here reads

\dim \text{null}\, T \geq 3 - 5 = -2

, which says nothing. The inclusion

(x_1, x_2, x_3) \mapsto (x_1, x_2, x_3, 0, 0)

has

\text{null}\, T = \{0\}

❌ Not quite. Compare

\dim V = 3

with

\dim W = 5

: a map to a bigger space can never reach all of it.

Show solution

By the fundamental theorem and $\dim \text{null}\, T \geq 0$ :

\dim \text{range}\, T = 3 - \dim \text{null}\, T \leq 3 < 5 = \dim \mathbf{F}^5

Equal spaces have equal dimensions, so $\text{range}\, T \neq \mathbf{F}^5$ and $T$ is not surjective, whatever $T$ is.

The other statements hold for some maps and fail for others: the zero map is not injective and has $\dim \text{range}\, T = 0 \neq 3$ ; the inclusion $(x_1, x_2, x_3) \mapsto (x_1, x_2, x_3, 0, 0)$ has $\dim \text{null}\, T = 0 < 2$ .

Problem 3 · Size of a Solution Space

Given: a homogeneous system of $4$ linear equations in $6$ variables, written as $T\colon \mathbf{F}^6 \to \mathbf{F}^4$ so that its solution set is $\text{null}\, T$ .

Smallest possible dimension of the solution set?

If $\dim \text{range}\, T = 3$ , what is the dimension of the solution set?

✅ Correct!

\dim \text{null}\, T = 6 - \dim \text{range}\, T \geq 6 - 4 = 2

, and with

\dim \text{range}\, T = 3

it is exactly

6 - 3 = 3

❌ Not quite. The solution set cannot be just

\{0\}

: with more variables than equations,

\dim \text{null}\, T \geq 6 - 4 = 2 > 0

❌ Check the lower bound.

\dim \text{range}\, T \leq \dim \mathbf{F}^4 = 4

, so

\dim \text{null}\, T = 6 - \dim \text{range}\, T \geq 6 - 4

❌ Not quite.

4 - 3

subtracts from the number of equations. The theorem subtracts from the dimension of the domain:

\dim \text{null}\, T = 6 - 3

❌ Check the exact count. Use

\dim \mathbf{F}^6 = \dim \text{null}\, T + \dim \text{range}\, T

with

\dim \text{range}\, T = 3

Show solution

Lower bound. $\text{range}\, T$ is a subspace of $\mathbf{F}^4$ , so $\dim \text{range}\, T \leq 4$ . The fundamental theorem gives

\dim \text{null}\, T = 6 - \dim \text{range}\, T \geq 6 - 4 = 2.

The bound is attained, for example by $T(x_1, \dots, x_6) = (x_1, x_2, x_3, x_4)$ , whose null space is spanned by $e_5, e_6$ .

Exact count. If $\dim \text{range}\, T = 3$ , then

\dim \text{null}\, T = 6 - 3 = 3.

The solution set is a $3$ -dimensional subspace of $\mathbf{F}^6$ ; and since $\text{range}\, T \neq \mathbf{F}^4$ , some choice of constants $c_1, \dots, c_4$ makes the non-homogeneous system unsolvable as well.

Problem 4 · Beyond Fⁿ: Polynomials

Given: $T \in \mathcal{L}(\mathcal{P}_4(\mathbf{R}), \mathbf{R}^3)$ defined by $Tp = \big(p(1), p(2), p(3)\big)$ , and $T$ is surjective. Find $\dim \text{null}\, T$ .

✅ Correct!

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

and

\dim \text{range}\, T = 3

, so

\dim \text{null}\, T = 5 - 3 = 2

❌ Close, but…

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

has basis

1, x, x^2, x^3, x^4

, so its dimension is

5

, not

4

❌ Not quite. A map from a

5

-dimensional space to

\mathbf{R}^3

is never injective:

\dim \text{null}\, T \geq 5 - 3 = 2

❌ Not quite. Surjective means

\text{range}\, T = \mathbf{R}^3

; then subtract its dimension from

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

Show solution

$\mathcal{P}_4(\mathbf{R})$ has basis $1, x, x^2, x^3, x^4$ , so it is finite-dimensional with dimension $5$ . Surjectivity gives $\text{range}\, T = \mathbf{R}^3$ , so $\dim \text{range}\, T = 3$ . The fundamental theorem:

5 = \dim \text{null}\, T + 3 \quad\Longrightarrow\quad \dim \text{null}\, T = 2.

Check directly. $p \in \text{null}\, T$ means $p(1) = p(2) = p(3) = 0$ , so

p(x) = (x - 1)(x - 2)(x - 3)(a + bx), \qquad a, b \in \mathbf{R},

and the two polynomials $(x-1)(x-2)(x-3)$ and $x(x-1)(x-2)(x-3)$ form a basis of $\text{null}\, T$ : dimension $2$ .

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