Linear Algebra
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Other decompositions such as QR decomposition and Schur decomposition are available in Mathematica, but shall not be discussed in this book. Their respective command names are QRDecomposition and SchurDecomposition.
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12.41 Construct a 5 5 matrix whose eigenvalues are 2, 1, 0, 1, 2 with respective eigenvectors (1, 1, 0, 0, 0), (0, 1, 1, 0, 0), (0, 0, 1, 1, 0), (0, 0, 0, 1, 1), and (1, 0, 0, 0, 1).
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SOLUTION
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d = DiagonalMatrix[{ 2, 1, 0, 1, 2}]; p = Transpose[{{1, 1, 0, 0, 0}, {0, 1, 1, 0, 0}, {0, 0, 1, 1, 0}, {0, 0, 0, 1, 1}, {1, 0, 0, 0, 1}}]; a = p.d.Inverse[p] ; a // MatrixForm 0 2 2 2 2 1 3 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 3 2 2 2 2 2
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a b 2 2 12.42 Show that the matrix c d has real eigenvalues if and only if a + 4bc 2ad + d 0.
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m = {{a, b}, {c, d}}; Eigenvalues[m]
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{ 1 (a + d 2
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a2 + 4 b c 2 a d + d2 , 1 a + d + a2 + 4 b c 2 a d + d2 2
The eigenvalues will be real if and only if the expression inside the radical symbol is non-negative.
12.43 Construct a 7 7 matrix of random digits and show that the sum of its eigenvalues is equal to its trace and the product of its eigenvalues is equal to its determinant.
SOLUTION
a = Table[RandomInteger [9], {i, 1, 7}, {j, 1, 7}]; a // MatrixForm 2 2 3 6 4 4 1 9 3 8 3 2 1 5 8 5 5 9 9 7 8 0 1 0 3 9 6 3 7 9 8 6 6 7 9 6 5 3 5 9 3 9 2 1 1 9 4 7 0
eigenvalues = Eigenvalues[N[a]]
{34.6689, 1.97195 + 6.25152 , 1.97195 6.25152 , 6.08883,
eigenvalues[[i]]
2.35886, 0.138652 + 0.334612 , 0.138652 0.334612 }
or Total[eigenvalues]
22 + 0.
Linear Algebra
Tr[a] 22
eigenvalues[[i]]
or Product[eigenvalues[[i]], {i, 1, 7}]
2807. + 1.42585 10 13 Det[a] 2807
12.44 A matrix, P, is said to be orthogonal if PTP = I. If it is possible to find an orthogonal matrix P that diagonalizes A, then A is said to be orthogonally diagonalizable. However, only symmetric matrices are orthogonally diagonalizable. (A matrix is symmetric if AT = A. If A is symmetric, it can be shown that the eigenvectors corresponding to distinct eigenvalues are orthogonal.) Find a matrix P that orthogonally diagonalizes
3 1 A = 0 0 0
SOLUTION
1 3 0 0 0
0 0 2 1 1
0 0 1 2 1
0 0 1 1 2
3 1 a = 0 0 0
1 3 0 0 0
0 0 2 1 1
0 0 1 2 1
0 0 1 ; 1 2
{values,
vectors} = Eigensystem[a]
{{4, 4, 2, 1, 1}, {{0, 0, 1, 1, 1}, {1, 1, 0, 0, 0},{ 1, 1, 0, 0, 0}, {0, 0, 1, 0, 1}, {0, 0, 1, 1, 0}}} eigenspace1 = {vectors[[1]], vectors[[2]]}; eigenspace2 = {vectors[[3]]}; eigenspace3 = {vectors[[4]], vectors[[5]]}; v1 = Orthogonalize[eigenspace1]; v2 = Orthogonalize[eigenspace2]; v3 = Orthogonalize[eigenspace3]; p = Transpose[Join[v1, v2, v3]]; p // MatrixForm
There are five eigenvalues, two of which have multiplicity 2. We group their eigenvectors into three eigenspaces and apply the Gram-Schmidt process to each. The orthogonal matrix is the matrix whose columns are the vectors from Orthogonalize.
0 0 1 3 1 3 1 3
1 2 1 2 0 0 0
1 2 1 2 0 0 0
0 0 1 2 0 1 2
0 0 1 6 2 3 1 6
