Linear Algebra II (pdf, 500 kB)

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(3) For v ∈ V1, v ′ ∈ V2,
〈v ′ , f(v)〉 = 〈f(v), v ′ 〉 = 〈v, f ∗ (v ′ )〉 = 〈f ∗ (v ′ ), v〉 = 〈v ′ , (f ∗ ) ∗ (v)〉 ,
and the claim follows.
Now we characterize isometries.
9.15. Proposition. Let V and W be inner product spaces of the same finite dimension
over the same field. Let f : V → W be linear. Then the following are
equivalent.
(1) f is an isometry;
(2) f is an isomorphism and f −1 = f ∗ ;
(3) f ◦ f ∗ = idW ;
(4) f ∗ ◦ f = idV .
Proof. To show (1) ⇒ (2), we observe that for an isometry f and v ∈ V , w ∈ W ,
〈v, f ∗ (w)〉 = 〈f(v), w〉 = 〈f(v), f f −1 (w) 〉 = 〈v, f −1 (w)〉 ,
which implies f ∗ = f −1 . The implications (2) ⇒ (3) and (2) ⇒ (4) are clear. Now
assume (say) that (4) holds (the argument for (3) is similar). Then f is injective,
hence an isomorphism, and we get (2). Now assume (2), and let v, v ′ ∈ V . Then
〈f(v), f(v ′ )〉 = 〈v, f ∗ f(v ′ ) 〉 = 〈v, v ′ 〉 ,
so f is an isometry.
9.16. Theorem. Let V be a finite-dimensional inner product space and f : V → V
a linear map. Then we have
im(f ∗ ) = ker(f) ⊥
and ker(f ∗ ) = im(f) ⊥ .
Proof. We first show the inclusions im(f ∗ ) ⊂ ker(f) ⊥ and ker(f ∗ ) ⊂ im(f) ⊥.
So let z ∈ im(f ∗ ), say z = f ∗ (y). Let x ∈ ker(f), then
〈x, z〉 = 〈x, f ∗ (y)〉 = 〈f(x), y〉 = 〈0, y〉 = 0 ,
so z ∈ ker(f) ⊥ . If y ∈ ker(f ∗ ) and z = f(x) ∈ im(f), then
so y ∈ im(f) ⊥ . Now we have
〈z, y〉 = 〈f(x), y〉 = 〈x, f ∗ (y)〉 = 〈x, 0〉 = 0 ,
dim im(f) = dim V − dim im(f) ⊥
≤ dim V − dim ker(f ∗ )
= dim im(f ∗ )
≤ dim ker(f) ⊥
= dim V − dim ker(f)
= dim im(f) .
So we must have equality throughout, which implies the result.
Now we relate the notions of adjoint etc. to matrices representing the linear maps
with respect to orthonormal bases.