Linear-Algebra-Done-Right · Unit 7 · Video 3 · Interactive Practice

Isomorphic Spaces and the Dimension of $\mathcal{L}(V, W)$

IKey Formulas

Formula	Name	Assumptions
$V \cong W \iff \dim V = \dim W$	Isomorphism test	$V$ , $W$ finite-dimensional over $\mathbf{F}$
$c_1 v_1 + \cdots + c_n v_n \mapsto (c_1, \dots, c_n)$	$V \cong \mathbf{F}^n$ ; e.g. $\mathcal{P}_m(\mathbf{F}) \cong \mathbf{F}^{m+1}$	$v_1, \dots, v_n$ a basis of $V$
$\mathcal{M}\colon \mathcal{L}(V, W) \to \mathbf{F}^{m,n}$ is an isomorphism	Matrix map	Bases of $V$ ( $n$ vectors) and $W$ ( $m$ vectors) fixed
$\dim \mathcal{L}(V, W) = (\dim V)(\dim W)$	Dimension of $\mathcal{L}(V, W)$	$V$ , $W$ finite-dimensional

Key Insight: An isomorphism is an invertible linear map; it relabels each $v$ as $Tv$ and carries sums and scalar multiples along. For finite-dimensional spaces the dimension alone decides whether such a relabelling exists — but which isomorphism you get depends on the bases you choose.

IIRelabelling $\mathcal{P}_2(\mathbf{F})$ as $\mathbf{F}^3$

The coordinate map sends $p + \lambda q$ to $T(p) + \lambda T(q)$ , whichever basis of $\mathcal{P}_2(\mathbf{F})$ defines it.

λ 1 Basis of P₂(F)

IIIIsomorphic $\iff$ Same Dimension

Sending basis vectors $v_k \mapsto w_k$ gives an isomorphism exactly when the two bases have equally many vectors.

V W

💡 The theorem promises that some isomorphism exists when the dimensions agree; a particular $T \in \mathcal{L}(V, W)$ , such as the zero map, can still fail to be invertible.

IVThe Matrix Map and $\dim \mathcal{L}(V, W)$

Each of the $mn$ entries of $\mathcal{M}(T)$ is a free scalar, and each choice is exactly one linear map.

dim V = n 2 dim W = m 3

VQuiz Questions

Problem 1 · Which Space Matches $\mathcal{P}_3(\mathbf{R})$ ?

Given: the space $\mathcal{P}_3(\mathbf{R})$ of real polynomials of degree at most $3$ — find the space isomorphic to it.

✅ Correct!

\dim \mathcal{P}_3(\mathbf{R}) = 4 = 2 \cdot 2 = \dim \mathbf{R}^{2,2}

, and equal dimension is all it takes.

❌ Off by one. The degree is

3

, but the basis

1, x, x^2, x^3

has four vectors, so

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

❌ Not quite.

\dim \mathbf{R}^{2,3} = 2 \cdot 3 = 6

, but

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

❌ Not quite.

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

has basis

1, x, \dots, x^4

, five vectors;

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

has dimension

4

Show solution

Finite-dimensional spaces are isomorphic exactly when their dimensions agree, so compute each dimension.

$\mathcal{P}_3(\mathbf{R})$ : basis $1, x, x^2, x^3$ , so dimension $3 + 1 = 4$ .
$\mathbf{R}^3$ : dimension $3$ .
$\mathbf{R}^{2,2}$ : dimension $2 \cdot 2 = 4$ .
$\mathbf{R}^{2,3}$ : dimension $2 \cdot 3 = 6$ .
$\mathcal{P}_4(\mathbf{R})$ : dimension $4 + 1 = 5$ .

Only $\mathbf{R}^{2,2}$ has dimension $4$ . An explicit isomorphism: $a_0 + a_1 x + a_2 x^2 + a_3 x^3 \mapsto \begin{pmatrix} a_0 & a_1 \\ a_2 & a_3 \end{pmatrix}$ .

Problem 2 · Equal Dimensions and a Given Map

Given: $\dim V = \dim W = 3$ and some $T \in \mathcal{L}(V, W)$ — decide which statement is correct.

✅ Correct! Equal dimension guarantees that some isomorphism exists; it says nothing about the particular map

T

❌ Not every map. The zero map

V \to W

is linear with

\text{null}\, 0 = V \neq \{0\}

, so it is not invertible, even though

\dim V = \dim W

❌ Not quite. Isomorphic means some isomorphism

V \to W

exists; one non-invertible

T

(say

T = 0

) does not rule out another map such as

v_k \mapsto w_k

❌ Not quite. Isomorphic spaces can be different sets:

\mathcal{P}_2(\mathbf{F}) \cong \mathbf{F}^3

, and their elements are polynomials and lists.

Show solution

Since $\dim V = \dim W = 3$ , the theorem gives $V \cong W$ : with bases $v_1, v_2, v_3$ of $V$ and $w_1, w_2, w_3$ of $W$ , the map $c_1 v_1 + c_2 v_2 + c_3 v_3 \mapsto c_1 w_1 + c_2 w_2 + c_3 w_3$ is an isomorphism.

That does not make every $T \in \mathcal{L}(V, W)$ invertible. The zero map has $\text{null}\, 0 = V$ , so it is not injective. Likewise the map with $v_1 \mapsto w_1$ , $v_2 \mapsto w_2$ , $v_3 \mapsto 0$ has $\dim \text{range}\, T = 2 \lt 3$ , so it is not surjective.

So statement (c) is the correct one: the dimensions decide whether an isomorphism exists, not whether a given map is one.

Problem 3 · A Subspace of $\mathcal{P}_3(\mathbf{R})$

Given: $U = \{p \in \mathcal{P}_3(\mathbf{R}) : p(1) = 0\}$ — find $\dim U$ and $\dim \mathcal{L}(U, \mathbf{R}^2)$ .

What is $\dim U$ ?

What is $\dim \mathcal{L}(U, \mathbf{R}^2)$ ?

✅ Correct!

\dim U = 4 - 1 = 3

, so

\dim \mathcal{L}(U, \mathbf{R}^2) = 3 \cdot 2 = 6

❌ Check $\dim U$ .

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

: the polynomial

1

has

1(1) = 1 \neq 0

, so

U

is a proper subspace and

\dim U \lt 4

❌ Check $\dim U$ .

U

is the null space of

p \mapsto p(1)

, a surjective linear map onto

\mathbf{R}

; apply the fundamental theorem of linear maps.

❌ The product, not the sum.

\dim \mathcal{L}(U, \mathbf{R}^2) = (\dim U)(\dim \mathbf{R}^2)

, not

\dim U + \dim \mathbf{R}^2

❌ Check $\dim U$ .

4 \cdot 2 = 8

uses

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

; the domain here is the smaller space

U

❌ Not quite. Use

\dim \mathcal{L}(U, \mathbf{R}^2) = (\dim U)(\dim \mathbf{R}^2)

with your value of

\dim U

Show solution

Step 1: $\dim U$ . Define $\varphi\colon \mathcal{P}_3(\mathbf{R}) \to \mathbf{R}$ by $\varphi(p) = p(1)$ . It is linear, and surjective because the constant polynomial $c$ maps to $c$ . Its null space is $U$ . By the fundamental theorem of linear maps,

\dim \mathcal{P}_3(\mathbf{R}) = \dim U + \dim \text{range}\, \varphi \implies 4 = \dim U + 1 \implies \dim U = 3.

(A basis: $x - 1,\ x^2 - 1,\ x^3 - 1$ .) So $U \cong \mathbf{R}^3$ .

Step 2: $\dim \mathcal{L}(U, \mathbf{R}^2)$ . With bases fixed, $\mathcal{M}$ is an isomorphism $\mathcal{L}(U, \mathbf{R}^2) \to \mathbf{R}^{2,3}$ , so

\dim \mathcal{L}(U, \mathbf{R}^2) = (\dim U)(\dim \mathbf{R}^2) = 3 \cdot 2 = 6.

Problem 4 · The Isomorphism Depends on the Basis

Given: the basis $1,\ x - 1,\ (x - 1)^2$ of $\mathcal{P}_2(\mathbf{R})$ and the isomorphism $T\colon \mathcal{P}_2(\mathbf{R}) \to \mathbf{R}^3$ , $T\big(c_1 \cdot 1 + c_2 (x - 1) + c_3 (x - 1)^2\big) = (c_1, c_2, c_3)$ — find $T(3 - x + 5x^2)$ .

✅ Correct!

7 + 9(x - 1) + 5(x - 1)^2 = 3 - x + 5x^2

: a different basis, a different isomorphism, a different list.

❌ That is the standard-basis list.

(3, -1, 5)

is the image under the isomorphism built from

1, x, x^2

; this

T

uses

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

❌ Shifted the wrong way.

(9, -11, 5)

gives

9 - 11(x + 1) + 5(x + 1)^2

, the expansion in powers of

x + 1

; substitute

x = y + 1

, not

y - 1

❌ Check the order.

c_1

is the coefficient of the first basis vector

1

, so the constant term of the expansion comes first.

Show solution

Write $p(x) = 3 - x + 5x^2$ in powers of $y = x - 1$ , i.e. substitute $x = y + 1$ :

p = 3 - (y + 1) + 5(y + 1)^2 = 3 - y - 1 + 5y^2 + 10y + 5 = 7 + 9y + 5y^2.

So $p = 7 \cdot 1 + 9(x - 1) + 5(x - 1)^2$ and $T(p) = (7, 9, 5)$ .

Check: $5(x - 1)^2 + 9(x - 1) + 7 = 5x^2 - 10x + 5 + 9x - 9 + 7 = 5x^2 - x + 3$ ✓

The standard basis $1, x, x^2$ sends the same polynomial to $(3, -1, 5)$ : both maps are isomorphisms $\mathcal{P}_2(\mathbf{R}) \to \mathbf{R}^3$ , and the theorem guarantees that one exists without singling one out.

Solved: 0 / 4

Isomorphic Spaces and the Dimension of L(V,W)\mathcal{L}(V, W)L(V,W)

IKey Formulas

IIRelabelling P2(F)\mathcal{P}_2(\mathbf{F})P2​(F) as F3\mathbf{F}^3F3

IIIIsomorphic ⟺ \iff⟺ Same Dimension

IVThe Matrix Map and dim⁡L(V,W)\dim \mathcal{L}(V, W)dimL(V,W)

VQuiz Questions

Isomorphic Spaces and the Dimension of $\mathcal{L}(V, W)$

IIRelabelling $\mathcal{P}_2(\mathbf{F})$ as $\mathbf{F}^3$

IIIIsomorphic $\iff$ Same Dimension

IVThe Matrix Map and $\dim \mathcal{L}(V, W)$