DETERMINANTS AND EIGENVALUES 1. Introduction Gauss ...

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94 II. DETERMINANTS AND EIGENVALUESHence, the eigenvalues are λ =1,λ = 6, and λ = −4. We proceed to find theeigenspaces for each of these eigenvalues, starting with the largest.First, take λ = 6, and put it in (6) to obtain the system⎡⎣ 1 − 6 4 3⎤⎡4 1−6 0 ⎦⎣ v ⎤⎡1v 2⎦ =0 or ⎣ −5 4 3⎤⎡4 −5 0⎦⎣ v ⎤1v 2⎦ =0.3 0 1−6 v 3 3 0 −5 v 3To solve, use Gauss-Jordan reduction⎡⎣ −5 4 3⎤ ⎡⎤ ⎡⎤−1 −1 3 −1 −1 34 −5 0⎦ → ⎣ 4 −5 0⎦ → ⎣ 0 −9 12⎦3 0 −5 3 0 −5 0 −3 4⎡⎤ ⎡⎤−1 −1 3 −1 −1 3→ ⎣ 0 0 0⎦ → ⎣ 0 3 −4⎦0 −3 4 0 0 0⎡→ ⎣ 1 1 −3⎤ ⎡0 1 −4/3⎦ → ⎣ 1 0 −5/3⎤0 1 −4/3⎦.0 0 0 0 0 0Note that the matrix is singular, and the rank is smaller than 3. This must be thecase because the condition det(A − λI) = 0 guarantees it. If the coefficient matrixwere non-singular, you would know that there was a mistake: either the roots ofthe characteristic equation are wrong or the row reduction was not done correctly.The general solution isv 1 =(5/3)v 3v 2 =(4/3)v 3with v 3 free. The general solution vector is⎡v = ⎣ (5/3)v ⎤ ⎡3(4/3)v 3⎦ = v 3⎣ 5/3⎤4/3 ⎦ .v 31Hence, the eigenspace is 1-dimensional. A basis may be obtained by setting v 3 =1as usual, but it is a bit neater to put v 3 = 3 so as to avoid fractions. Thus,⎡v 1 = ⎣ 5 ⎤4 ⎦3constitutes a basis for the eigenspace corresponding to the eigenvalue λ = 6 . Notethat we have now found all eigenvectors for this eigenvalue. They are all the nonzerovectors in this 1-dimensional eigenspace, i.e., all non-zero multiples of v 1 .Next take λ = 1 and put it in (6) to obtain the system⎡⎣ 0 4 3⎤⎡4 0 0⎦⎣ v 1⎦ =0.3 0 0⎤v 2v 3
4. EIGENVALUES AND EIGENVECTORS FOR n × n MATRICES 95Use Gauss-Jordan reduction⎡ ⎤ ⎡ ⎤⎣ 0 4 34 0 0⎦ →···→⎣ 0 1 3/4⎦.3 0 00 0 0The general solution isv 1 =0v 2 =−(3/4)v 3with v 2 free. Thus the general solution vector is⎡ ⎤ ⎡0v = ⎣ −(3/4)v 3⎦ = v 3⎣ 0⎤−3/4 ⎦ .v 31Put v 3 = 4 to obtain a single basis vector⎡v 2 = ⎣ 0⎤−3 ⎦4for the eigenspace corresponding to the eigenvalue λ = 1. The set of eigenvectorsfor this eigenvalue is the set of non-zero multiples of v 2 .Finally, take λ = −4, and put this in (6) to obtain the system⎡⎣ 5 4 3⎤⎡4 5 0⎦⎣ v ⎤1v 2⎦ =0.3 0 5 v 3Solve this by Gauss-Jordan reduction.⎡⎣ 5 4 3⎤ ⎡4 5 0⎦ → ⎣ 1 −1 3⎤ ⎡4 5 0⎦ → ⎣ 1 −1 3⎤0 9 −12 ⎦3 0 5 3 0 5 0 3 −4⎡→ ⎣ 1 −1 3⎤ ⎡0 3 −4⎦ → ⎣ 1 0 5/3⎤0 1 −4/3⎦.0 0 0 0 0 0The general solution isv 1 = −(5/3)v 3v 2 =(4/3)v 3with v 3 free. The general solution vector is⎡v = ⎣ −(5/3)v ⎤ ⎡3(4/3)v 3⎦ = v 3⎣ −5/3⎤4/3 ⎦ .v 3 1
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Page 11 and 12: 2. DEFINITION OF THE DETERMINANT 81
Page 13 and 14: 3. SOME IMPORTANT PROPERTIES OF DET
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Page 31 and 32: 5. DIAGONALIZATION 101are eigenvect
Page 33 and 34: 5. DIAGONALIZATION 103Example 3. Co
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Page 37 and 38: 6. THE EXPONENTIAL OF A MATRIX 107(
Page 39 and 40: 7. REVIEW 109⎡3. A = ⎣ 0 1 1⎤

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