ssrs barcode font download The transpose makes the eigenvectors columns. in Software

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The transpose makes the eigenvectors columns.
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{{4, 0, 0, 0}, {0, 3, 0, 0}, {0, 0, 2, 0}, {0, 0, 0, 1}}
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p.d.Inverse[p] // MatrixForm 18 8 15 15 51 24 48 47 27 15 14 8 28 15 25 12
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p.d.Inverse[p]= a
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To summarize, MatrixForm[a] MatrixForm[p].MatrixForm[d] .MatrixForm[Inverse[p]] 18 8 15 15 51 24 48 47 27 15 3 0 14 8 1 0 = 28 15 2 0 25 12 3 1 1 1 3 2 3 2 3 2 4 0 0 0 0 3 0 0 0 0 2 0 0 0 0 1 3 9 5 3 5 15 9 6 1 2 1 1 1 4 2 1
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Unfortunately, not every matrix can be diagonalized. However, there is a standard form, called Jordan canonical form, that every matrix possesses. A Jordan block is a square matrix whose elements are zero except for the main diagonal, where all numbers are equal, and the superdiagonal, where all values are 1:
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0 0 . 0 0 1 0 0 .. 1 0 .. 0 1 .. . . .. 0 0 0 .. 0 0 0 .. 0 0 0 1 0 0 0 0 . 0 1
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If A is any n n matrix, there exists a matrix Q such that A = Q J Q 1 and
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J1 0 J= 0 . 0 0 J2 0 . 0 0 0 J3 . 0 .. 0 .. 0 .. 0 .. . . . Jk
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The Ji s are Jordan blocks. The same eigenvalue may occur in different blocks. The number of distinct blocks corresponding to a given eigenvalue is equal to the number of independent eigenvectors belonging to that eigenvalue.
EXAMPLE 31
5 4 3 a = 1 0 3 ; 1 2 1 Eigensystem[a] {{ 2, 4, 4}, {{ 1, 1, 1}, {1, 1, 1}, {0, 0, 0}}}
The eigenvalues are 2 and 4 with eigenvectors, respectively, ( 1, 1, 1) and (1, 1, 1). The vector {0, 0, 0} is not an eigenvector; its presence simply indicates that a third linearly independent eigenvector cannot be found. To construct Q, the standard procedure is to find a vector x such that (A 4 I) x = (1, 1, 1).
LinearSolve[a 4 IdentityMatrix[3], {1, 1, 1}] {1, 0, 0}
The matrices Q and J may now be constructed:
1 1 1 q = 1 1 0 ; 1 1 0 2 0 0 j= 0 4 1 ; 0 0 4
Linear Algebra
q.j.Inverse[q] // MatrixForm 5 4 3 1 0 3 1 2 1
JordanDecomposition[matrix] computes the Jordan canonical form of matrix. The output is a list {q, j} where q and j correspond to Q and J as described previously.
EXAMPLE 32 (Continuation of Example 31)
5 4 3 a = 1 0 3 1 2 1
{q, j} = JordanDecomposition[a]
q // MatrixForm 1 1 1 1 1 0 1 1 0 j // MatrixForm 2 0 0 0 4 1 0 0 4
EXAMPLE 33
Of course this agrees with the results of Example 31.
65 88 129 23 1 97 19 86 124 180 32 4 134 21 29 39 54 11 2 43 13 9 a = 36 50 77 11 3 56 5 ; 2 63 88 131 24 0 97 16 58 85 126 21 6 91 9 6 63 87 129 24 1 96 16
j} = JordanDecomposition[a];
MatrixForm[a] MatrixForm[q].MatrixForm[j].MatrixForm[Inverse[q]]
65 86 29 36 63 58 63
88 124 39 9 50 88 85 87
129 180 54 77 131 126 129
23 1 32 4 11 2 11 3 24 0 21 6 24 1
97 134 43 56 97 91 96
19 21 13 == 5 16 9 16 1
4 5 0 3 3 6 2
2 11 2 4 13 2 5 2 10 0
2 3 4 2 11 4 3 4 0 0
2 29 8 0 19 8 3 8 4 0
0 1
-1 1 2 2 4 1 1 1 2 0
1 1 1 0 2 0
2 0 0 0 0 0 0
1 2 0 0 0 0 0
0 1 2 0 0 0 0
0 0 1 2 0 0 0
0 0 0 0 3 0 0
0 0 0 0 1 3 0
0 0 0 0 0 1 3
17 2 19 2 9 2 5 17 27 19
25 2 25 2
18 19 9 10 36 55 38
7 2 2 7 2
0 0 0
27 2 14 13 2 7 27 41 29
3 3 3 2 1 5 7 7
6 6 25 37 26
2 0 7 0 10 1 7 1
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