Linear Algebra 28 | Conservation of Dimension [dark version]

TL;DR

Dimension is invariant under bijective linear maps between subspaces of R^n.

Briefing Cornell Notes

Briefing

Dimension is preserved when two subspaces are connected by a bijective linear map: if a linear transformation gives a one-to-one correspondence between subspace U and subspace V, then U and V must have the same dimension. That matters because dimension isn’t just a count—it becomes an invariant under “structure-preserving” changes of coordinates. In practical terms, a bijective linear map can rename vectors and rearrange how subspaces sit inside R^n, but it cannot change how many independent directions the subspace contains.

The core equivalence is: dim(U) = dim(V) if and only if there exists a bijective linear map F: U → V. The forward direction constructs such an F when the dimensions match. Starting from a basis {u1, …, uk} of U and a basis {v1, …, vk} of V (with the same number k of vectors), the map is defined by sending each basis vector ui to vi. Linearity then forces the image of any vector x in U to be determined uniquely by its coordinates in the basis, so F is well-defined and linear. Bijectivity follows by building an inverse map that swaps the roles of the bases: it sends each vi back to the corresponding ui, and checking that composing the two maps returns the original vector.

The reverse direction starts with a bijective linear map F: U → V and deduces equality of dimensions. Because bijectivity splits into injectivity and surjectivity, it can be used to verify the two defining properties of a basis in V. Take a basis of U with k vectors. Applying F to each basis vector produces k vectors in V. Injectivity ensures these images remain linearly independent, while surjectivity ensures they span all of V: every y in V is hit by some x in U, and then y becomes a linear combination of the F-images of the U-basis vectors. With a linearly independent spanning family of size k, those images form a basis of V, so dim(V) = k = dim(U).

A final consequence sharpens the meaning of dimension in nested subspaces. If U is a subset of V and dim(U) = dim(V), then U must equal V. The reason is structural: a basis of U (with k vectors) lies inside V, and since the dimensions match, that same k-vector linearly independent set must already be a basis for V. Every vector in V is therefore a linear combination of vectors from U, so it must belong to U as well. Proper containment U ⊂ V forces a strict drop in dimension; you can’t have a genuinely smaller subspace sitting inside a larger one without losing independent directions.

Cornell Notes

A bijective linear map between subspaces preserves dimension. If dim(U)=dim(V), one can build a bijective linear transformation by matching basis vectors: send a basis of U to a basis of V and extend linearly; the inverse comes from reversing that basis correspondence. Conversely, if a bijective linear map F:U→V exists, applying F to a basis of U produces k vectors in V that stay linearly independent (injectivity) and still span V (surjectivity), so dim(U)=dim(V). When U⊆V and the dimensions coincide, U must equal V because a basis of U already functions as a basis for V. Dimension therefore acts as a complete invariant under bijective linear equivalence.

Why does matching basis vectors produce a well-defined linear map F:U→V?

Once U has a basis {u1,…,uk} and V has a basis {v1,…,vk}, defining F(ui)=vi for each i fixes F on the basis. Any x∈U has a unique coordinate expansion x=∑i ai ui. Linearity then forces F(x)=∑i ai F(ui)=∑i ai vi, so F(x) is determined uniquely by the coordinates of x. That guarantees the map is well-defined and linear.

How does bijectivity follow from the basis construction?

Bijectivity is shown by explicitly constructing an inverse. Define G:V→U by G(vi)=ui on the V-basis and extend linearly. Then for x∈U, (G∘F)(x)=x because x’s coordinates in the ui-basis are preserved through F to the vi-basis and then mapped back. Similarly, (F∘G)(y)=y for y∈V. Having two-sided inverses establishes that F is bijective.

Given a bijective linear map F:U→V, how can one prove the images of a basis of U form a basis of V?

Start with a basis {u1,…,uk} of U and consider {F(u1),…,F(uk)} in V. Injectivity implies linear independence is preserved: if ∑i ai F(ui)=0, then applying injectivity (via the fact that F(x)=0 only when x=0) forces ∑i ai ui=0, so all ai=0. Surjectivity implies spanning: for any y∈V, surjectivity gives y=F(x) for some x∈U, and x is a linear combination of the ui, so y becomes the corresponding linear combination of the F(ui). Thus the images span V and are linearly independent, hence form a basis.

Why does U⊆V together with dim(U)=dim(V) force U=V?

Take a basis of U with k vectors. Because U⊆V, those k vectors lie in V and remain linearly independent. Since dim(V)=k, any linearly independent set of size k in V must be a basis for V. Therefore every y∈V is a linear combination of vectors from U, meaning y∈U. So V⊆U, and with U⊆V already given, U=V.

What does the “conservation of dimension” mean in geometric terms?

A bijective linear map can change how a subspace sits inside R^n—rotating, shearing, or re-coordinatizing it—but it cannot change the number of independent directions available. Lines in R^2 (dimension 1) can be mapped bijectively to other lines, but a bijective linear map cannot turn a line into a plane (dimension 2) or vice versa.

Review Questions

If dim(U)=dim(V), what specific construction using bases guarantees a bijective linear map F:U→V?
How do injectivity and surjectivity each contribute to proving that F(basis of U) is a basis of V?
Why does proper containment U⊂V imply dim(U)<dim(V) when both are subspaces of the same R^n?

Key Points

1
Dimension is invariant under bijective linear maps between subspaces of R^n.
2
dim(U)=dim(V) holds exactly when there exists a bijective linear map F:U→V.
3
A bijective linear map can be constructed by sending a basis of U to a basis of V and extending linearly.
4
Injectivity preserves linear independence: images of a linearly independent set remain linearly independent.
5
Surjectivity ensures spanning: images of a basis of U span all of V.
6
If U⊆V and dim(U)=dim(V), then U=V; proper inclusion forces a strict dimension drop.
7
Dimension acts as a complete invariant for subspaces up to bijective linear equivalence.

Highlights

A bijective linear map can rename vectors and reshape subspaces, but it cannot change the number of independent directions—dimension stays fixed.

Constructing F becomes straightforward once bases are chosen: map each basis vector in U to a corresponding basis vector in V, then extend linearly.

Injectivity and surjectivity each play a distinct role: injectivity preserves independence, while surjectivity guarantees spanning.

Nested subspaces with equal dimension must coincide: if U⊆V and dim(U)=dim(V), then U=V.

Topics

Dimension Invariance
Bijective Linear Maps
Bases and Spanning
Linear Independence
Nested Subspaces