c If A is produced from A by interchanging a pair of columns or a pair of rows

# C if a is produced from a by interchanging a pair of

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(c) If A is produced from A by interchanging a pair of columns or a pair of rows then det A = - det A . (d) If A is produced from A by adding a multiple of a row to another, then det A = det A . 3. If A is upper–triangular: det A = a 11 * . . . 0 a 22 * . . . 0 0 a 33 . . . . . . . . . 0 0 . . . a nn = a 11 · a 22 · a 33 · · · a nn Proof by induction on the size n : n = 1: For a 1 × 1 matrix A = [ a ], det A = a . Assume the statement is true for [ k - 1, k - 1] matrices. If A is a [ k , k ] matrix, then M kk is the determinant of an [ k - 1, k - 1] matrix, and by the inductive assumption, det A = 0 C k 1 + 0 · C k 1 + · · · + a kk · C kk = M kk a kk = ( a 11 · a 22 · · · a k - 1 k - 1 ) · a kk 4. det AB = det A det B . An important consequence is that determinants can be calculated by first row reducing using the properties (2), and then applying property (3) to the resultant upper triangular row echelon matrix. Just remember to keep the multiplicative factors which accumulate when constants are pulled out using properties (2). Another consequence is verifying one important aspect of Theorem 2, namely: A is an invertible matrix ⇐⇒ det A = 0. Proof. A is not invertible ⇐⇒ A has a zero row, where A is some row echelon matrix derived from A ⇐⇒ det A = 0 ⇐⇒ det A = 0 Subscribe to view the full document.

1.7. EIGENVALUES SND EIGENSPACES 19 1.7 Eigenvalues snd Eigenspaces The concept of eigenvalues and their associated eigenspaces (eigenvectors) leads to un- derstanding of the possible behaviour of the general linear operator A : V V . Definition 10. Given the linear operator A , a vector x V is called an eigenvector of A with eigenvalue λ , a scalar, if x is non-zero and the eigenvalue equation is satisfied: A x = λx (1.12) or equivalently ( λ I - A ) x = 0 For each eigenvalue λ , the subspace spanned by the λ -eigenvectors is called the λ -eigenspace . Example 3. If R is a rotation about an axis ω , then R ω = ω , so ω is an eigenvector with eigenvalue 1 . Any non-zero scalar multiple of ω is also a 1 -eigenvector. Theorem 4. If V is a complex vector space of dimension N , then any linear operator A : V V has at least one and at most N eigenvalues. Proof. (For matrices A : C N C N ) It is clear that x = 0 is always a solution of Ax = λ x . Parts 3. and 12. of our big theorem on invertibility, Theorem 2, implies a non-zero solution exists if and only if λ I - A is non-invertible if and only if det( λ I - A ) = 0. This equation is called the characteristic equation , and is satisfied by any eigenvalue λ . A bit of thought convinces you that det( λ I - A ) is a (complex) polynomial in λ of degree N , some of whose coefficients can be identified: det( λ I - A ) = λ N - Tr ( A ) λ N - 1 + · · · + ( - 1) N det( A ) where the trace is Tr ( A ) = N i =1 a ii Finally, the fundamental theorem of algebra says that it must have at least one and no more than n complex roots: these roots are the eigenvalues of A , the set of all such is called the spectrum of A .  • Winter '10
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