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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

1.1L1L4

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].

2.1step 1.1L2L3

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

3.1step 1.1L3∎

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

Depends on

Used by

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on 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