Solved: Consider all 4 by 4 matrices A that are diagonalized by the same fixed

Questions 1-7 are about t~e eigenvalue and eigenvector matrices A and S.

Questions 1-7 are about t~e eigenvalue and eigenvector matrices A and S.

Questions 1-7 are about t~e eigenvalue and eigenvector matrices A and S.

Questions 1-7 are about t~e eigenvalue and eigenvector matrices A and S.

Questions 1-7 are about t~e eigenvalue and eigenvector matrices A and S.

Questions 1-7 are about t~e eigenvalue and eigenvector matrices A and S.

Questions 1-7 are about t~e eigenvalue and eigenvector matrices A and S.

Questions 8-10 are about Fibonacci and Gibonacci numbers.

Questions 8-10 are about Fibonacci and Gibonacci numbers.

Questions 8-10 are about Fibonacci and Gibonacci numbers.

True or false: If the eigenvalues of A are 2, 2, 5 then the matrix is certainly (a) invertible (b) diagonalizable (c) not diagonalizable.

True or false: If the only eigenvectors of A are mUltiples of (1, 4) then A has (a) no inverse (b) a repeated eigenvalue (c) no diagonalization SAS-1

Complete these matrices so that det A = 25. Then check that A = 5 is repeatedthe trace is 10 so the determinant of A - AI is (A - 5)2. Find an eigenvector with Ax = 5x. These matrices will not be diagonalizable because there is no second line of eigenvectors.

The matrix A = [~~] is not diagonalizable because the rank of A - 31 is __ Change one entry to make A diagonalizable. Which entries could you change?

Questions 15-19 are about powers of matrices.

Questions 15-19 are about powers of matrices.

Questions 15-19 are about powers of matrices.

Questions 15-19 are about powers of matrices.

Questions 15-19 are about powers of matrices.

Suppose A = SAS-l. Take determinants to prove detA = detA = AIA2 An This quick proof only works when A can be

Show that trace S T = trace T S, by adding the diagonal entries of STand T S : S = [~ ~] and Choose T as AS-I. Then SAS-1 has the same trace as AS-1 S = A. The trace of A equals the trace of A = sum of the eigenvalues.

AB - BA = I is impossible since the left side has trace = __ . But find an elimination matrix so that A = E and B = ET give A B - BA = [-b ~ ] which has trace zero.

If A = SAS- 1 , diagonalize the block matrix B = [~J]' Find its eigenvalue and eigenvector (block) matrices

Consider all 4 by 4 matrices A that are diagonalized by the same fixed eigenvector matrix S. Show that the A's form a subspace (cA and Al + A2 have this same S). What is this subspace when S = I? What is its dimension?

Suppose A2 = A. On the left side A multiplies each column of A. Which of our four subspaces contains eigenvectors with A = I? Which subspace contains eigenvectors with A = O? From the dimensions of those subspaces, A has a full set of independent eigenvectors. So a matrix with A 2 = A can be diagonalized.

(Recommended) Suppose Ax = AX. If A = 0 then X is in the nUllspace. If A i 0 then x is in the column space. Those spaces have dimensions (n - r) + r = n. So why doesn't every square matrix have n linearly independent eigenvectors?

The eigenvalues of A are 1 and 9, and the eigenvalues of Bare -1 and 9: A = [~ ~] and B = [~ ~]. Find a matrix square root of A from R = s..fA S-I. Why is there no real matrix square root of B?

(Heisenberg's Uncertainty Principle) AB - BA = I can happen for infinite matrices with A = AT and B = _BT. Then Explain that last step by using the Schwarz inequality. Then Heisenberg's inequality says that II Ax 11/ II x II times liB x II / II x II is at least!. It is impossible to get the position error and momentum error both very small.

If A and B have the same A'S with the same independent eigenvectors, their factorizations into are the same. So A = B.

Suppose the same S diagonalizes both A and B. They have the same eigenvectors in A = SA1S-1 and B = SA2S-I. Prove that AB = BA.

(a) If A = [g ~] then the determinant of A - AI is (A - a)(A - d). Check the "Cayley-Hamilton Theorem" that (A - aI)(A - dI) = zero matrix. (b) Test the Cayley-Hamilton Theorem on Fibonacci's A = U A], The theorem predicts that A2 - A - I = 0, since the polynomial det(A - AI) is A 2 - A-I.

Substitute A = SAS-1 into the product (A - )"1 I) (A - A2I) (A - AnI) and explain why this produces the zero matrix. We are substituting the matrix A for the number A in the polynomial peA) = det(A - AI). The Cayley-Hamilton Theorem says that this product is always peA) = zero matrix, even if A is not diagonalizable.

Find the eigenvalues and eigenvectors and the kth power of A. For this "adjacency matrix" the i, j entry of A k counts the k -step paths from i to j . 1's in A show edges between nodes

If A = [ij ~J and AB = BA, show that B = [~~J is also a diagonal matrix. B has the same eigen as A but different eigen . These diagonal matrices B form a two-dimensional subspace of matrix space. AB - BA = 0 gives four equations for the unknowns a, b, c, d-find the rank of the 4 by 4 matrix.

The powers Ak approach zero if all lAd < 1 and they blow up if any IAi I > 1. Peter Lax gives these striking examples in his book Linear Algebra: A = [i~] B = [_~ _;] C = [_~ _~] D = [_; 6.!] C 1024 =-C II D 1024 11 < 10-78 Find the eigenvalues A = eiB of Band C to show B4 = / and C 3 = -/.

The nth power of rotation through () is rotation through n(): An = [eos() -Sin()]n = [cosn() -Sinn()] sin () cos () sin n () cos n () . Prove that neat formula by diagonalizing A = SAS-1 . The eigenvectors (columns of S) are (I, i) and (i, 1). You need to know Euler's formula eiB = cos () + i sin ().

The transpose of A = SAS-1 is AT = (S-l)T AST. The eigenvectors in ATy = AY are the columns of that matrix (S-I)T. They are often called left eigenvectors. How do you multiply matrices to find this formula for A?

The inverse of A = eye(n) + ones(n) is A-I = eye(n) + C * ones(n). Multiply AA-I to find that number C (depending on n)