First I'll show that (im T)⊥ = ker T*. Since dim (im T)⊥ = dim V - dim (im T)
= dim ker T, this will prove that dim ker T = dim ker T*.
Let v be an element of (im T)⊥. Then <v, Tu> = 0 for all u in V. So <T*(v), u> =
<v, Tu> = 0 for all u in V. Let {e1,...,em} be an orthonormal basis for V. Then
T*(v) = <T*(v), e1> e1 + ... + <T*(v), em> em
------- = <v, Te1> e1 + ... + <v, Tem> em
------- = 0e1 + ... + 0em = 0.
Thus v is in ker T*. So (im T)⊥ is contained in ker T*. Conversely, suppose v is an element of ker T*. Then T*(v) = 0 and <v, Tu> = <T*(v), u> = <0, u> = 0 for all u in V. Thus v is in (im T)⊥. Hence ker T* is contained in (im T)⊥. By the two containments, we have (im T)⊥ = ker T*.
Comments
First I'll show that (im T)⊥ = ker T*. Since dim (im T)⊥ = dim V - dim (im T)
= dim ker T, this will prove that dim ker T = dim ker T*.
Let v be an element of (im T)⊥. Then <v, Tu> = 0 for all u in V. So <T*(v), u> =
<v, Tu> = 0 for all u in V. Let {e1,...,em} be an orthonormal basis for V. Then
T*(v) = <T*(v), e1> e1 + ... + <T*(v), em> em
------- = <v, Te1> e1 + ... + <v, Tem> em
------- = 0e1 + ... + 0em = 0.
Thus v is in ker T*. So (im T)⊥ is contained in ker T*. Conversely, suppose v is an element of ker T*. Then T*(v) = 0 and <v, Tu> = <T*(v), u> = <0, u> = 0 for all u in V. Thus v is in (im T)⊥. Hence ker T* is contained in (im T)⊥. By the two containments, we have (im T)⊥ = ker T*.
What is the explicit meaning of T* in this case?