Foundations of Data Science

The first integral is just the standard integral of the beta function and its value is s!(k−2−s)!
(k−1)!
.
To bound the second integral, note that for z ∈ [1, µ], |z − 1| ≤ 1
k−2 and
z k−s−2 ≤ ( 1 + ( 1/(k − 2) )) k−s−2
≤ e (k−s−2)/(k−2) ≤ e.
(
So, |E((x i − µ) s (k − 1)s!(k − 2 − s)! e(k − 1)
1
)| ≤ +
(k − 1)! (k − 2) ≤ s!Var(y) s+1 k − 4 + e )
≤ s!Var(x).
3!
Now, apply the first inequality of Theorem 12.5 with s of that theorem set to k − 2 or
k − 3 whichever is even. Note that a = εE(x) ≤ √ 2nσ 2 (since ε ≤ 1/k 2 ). The present
theorem follows by a calculation.
12.7 Eigenvalues and Eigenvectors
Let A be an n×n real matrix. The scalar λ is called an eigenvalue of A if there exists a
nonzero vector x satisfying the equation Ax = λx. The vector x is called the eigenvector
of A associated with λ. The set of all eigenvectors associated with a given eigenvalue form
a subspace as seen from the fact that if Ax = λx and Ay = λy, then for any scalers c
and d, A(cx + dy) = λ(cx + dy). The equation Ax = λx has a nontrivial solution only if
det(A − λI) = 0. The equation det(A − λI) = 0 is called the characteristic equation and
has n not necessarily distinct roots.
Matrices A and B are similar if there is an invertible matrix P such that A = P −1 BP .
Theorem 12.8 If A and B are similar, then they have the same eigenvalues.
Proof: Let A and B be similar matrices. Then there exists an invertible matrix P
such that A = P −1 BP . For an eigenvector x of A with eigenvalue λ, Ax = λx, which
implies P −1 BP x = λx or B(P x) = λ(P x). So, P x is an eigenvector of B with the same
eigenvalue λ. Since the reverse also holds, the theorem follows.
Even though two similar matrices, A and B, have the same eigenvalues, their eigenvectors
are in general different.
The matrix A is diagonalizable if A is similar to a diagonal matrix.
Theorem 12.9 A is diagonalizable if and only if A has n linearly independent eigenvectors.
Proof:
(only if) Assume A is diagonalizable. Then there exists an invertible matrix P
and a diagonal matrix D such that D = P −1 AP . Thus, P D = AP . Let the diagonal
elements of D be λ 1 , λ 2 , . . . , λ n and let p 1 , p 2 , . . . , p n be the columns of P . Then
AP = [Ap 1 , Ap 2 , . . . , Ap n ] and P D = [λ 1 p 1 , λ 2 p 2 , . . . , λ n p n ] . Hence Ap i = λ i p i . That
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