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$\ker T^{} = (im T)^{⊥}$ and $im T^{} = (\ker T)^{⊥}$ in finite dimension

Statement

For a linear map $T : V \to W$ between finite-dimensional inner product spaces,

$\ker T^{*} = (im T)^{⊥}, im T^{*} = (\ker T)^{⊥} .$

Equivalently, $\ker T = (im T^{*})^{⊥}$ and $im T = (\ker T^{*})^{⊥}$ .

Facts & Assumptions

Given: A finite-dimensional linear map $T : V \to W$ .

[L1]

The adjoint identity is $⟨ T v, w ⟩ = ⟨ v, T^{*} w ⟩$ for all $v, w$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ ).

[L2]

Double adjoints satisfy $T^{* *} = T$ (Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ ).

[L3]

In finite dimension, $U^{⊥ ⊥} = U$ for every subspace $U$ (In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ ).

[L4]

A vector lies in $U^{⊥}$ exactly when it pairs to zero with every vector of $U$ (The orthogonal complement $W^{⊥} = {v : ⟨ v, w ⟩ = 0 for all w \in W}$ ).

Proof

technique · direct

1.1

A vector $w \in W$ lies in $\ker T^{*}$ exactly when $⟨ v, T^{*} w ⟩ = 0$ for every $v \in V$ . By [L1], this is exactly $⟨ T v, w ⟩ = 0$ for every $v$ , hence exactly $w \in (im T)^{⊥}$ by [L4].

L1L4

2.1

Apply step 1.1 to $T^{*}$ and use [L2]: $\ker T = (im T^{*})^{⊥}$ . Taking orthogonal complements and applying [L3] gives $(\ker T)^{⊥} = im T^{*}$ .

step 1.1L2L3

3.1

Taking orthogonal complements in step 1.1 and using [L3] also gives $im T = (\ker T^{*})^{⊥}$ .

step 1.1L3∎

Depends on

Used by

For a linear map T:V→ W between finite-dimensional inner-product spaces, x minimises ‖ Tx-b‖ if and only if T^*(Tx-b)=0, equivalently T^*Tx=T^*b; minimisers exist and any two differ by an element of ker T Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., result 7.6 (standard reference, not scraped)
Sergei Treil, Linear Algebra Done Wrong, Theorem 5.5.1 (standard reference, not scraped)

ker⁡T∗=(im⁡T)⊥kerT∗=(imT)⊥ and im⁡T∗=(ker⁡T)⊥imT∗=(kerT)⊥ in finite dimension