Preparation problems for the discussion sections on November 18th and 20th1. For each of the following matrices, determine the eigenvalues of the matrix and for eacheigenvalue, determine (a basis for) the eigenspace that is associated to that eigenvalue.a.4 0 −21 1 20 0 2,b.3 44 −3,c.1 1 11 1 11 1 1.Solution:a. For instance, by expanding along the second column, we find thatdet4 − λ 0 −21 1 − λ 20 0 2 − λ= (2 − λ)(1 − λ)(4 − λ).Hence, the eigenvalues of A are 2, 4, and 1. For λ = 2:2 0 −21 −1 20 0 0R2→R2−1/2R1,R1→1/2R1,R2→−R2−−−−−−−−−−−−−−−−−−−−−→1 0 −10 1 −30 0 0Hence, the corresponding eigenspace is span131.For λ = 4:0 0 −21 −3 20 0 −2R2→R2+R3,R1→R1−R3,R3→−1/2R2−−−−−−−−−−−−−−−−−−−−−−→0 0 01 −3 00 0 1Hence, the corresponding eigenspace is span310.For λ = 1:3 0 −21 0 20 0 1R2→R2−2R3,R1→R1+2R3,R1→R1−3R2−−−−−−−−−−−−−−−−−−−−−−−−→0 0 01 0 00 0 1Hence, the corresponding eigenspace is span010.b. We have:det3 − λ 44 −3 − λ= (3 − λ)(−3 − λ) − 16 = λ2− 25 = (λ − 5)(λ + 5)Hence, the eigenvalues of A are 5 and −5. For λ = 5:−2 44 −8R2→R2+2R1,R1→−1/2R1−−−−−−−−−−−−−−−→1 −20 01Hence, the corresponding eigenspace is span21.For λ = −5:8 44 2R2→R2−1/2R1,R1→1/8R1−−−−−−−−−−−−−−−→1120 0Hence, the corresponding eigenspace is span−12.c. We have:det1 − λ 1 11 1 − λ 11 1 1 − λ= (1 − λ)((1 − λ)(1 − λ) − 1) − (−λ) + (1 − (1 − λ))= (1 − λ)(−λ)(2 − λ) + 2λ = −λ((1 − λ)(2 − λ) − 2) = λ2(3 − λ)Hence, the eigenvalues of A are 0 and 3. For λ = 0:1 1 11 1 11 1 1R2→R2−R1,R3→R3−R1−−−−−−−−−−−−−−→1 1 10 0 00 0 0Hence, the corresponding eigenspace is span−101,−110.For λ = 3:−2 1 11 −2 11 1 −2R1↔R3,R2→R2−R1,R3→R3+2R1,R3→R3+R2−−−−−−−−−−−−−−−−−−−−−−−−−−−→1 1 −20 −3 30 0 0R2→−1/3R2,R1→R1−R2−−−−−−−−−−−−−−−→1 0 −10 1 −10 0 0Hence, the corresponding eigenspace is span111.2. LetA =2 1 0 00 2 1 00 0 2 10 0 0 3, B =2 0 0 00 2 1 00 0 2 10 0 0 3C =2 0 0 00 2 0 00 0 2 10 0 0 3Determine the eigenvalues of A, B, C and, for each eigenvalue, determine the eigenspace thatis associated to that eigenvalue.Solution: For A, we have:det2 − λ 1 0 00 2 − λ 1 00 0 2 − λ 10 0 0 3 − λ= (2 − λ)3(3 − λ)Hence, the eigenvalues of A are 2 and 3. For λ = 2:0 1 0 00 0 1 00 0 0 10 0 0 1R4→R4−R3−−−−−−−→0 1 0 00 0 1 00 0 0 10 0 0 02Hence, the corresponding eigenspace is span1000.For λ = 3:−1 1 0 00 −1 1 00 0 −1 10 0 0 0Hence, the corresponding eigenspace is span1111.For B, we have:det2 − λ 0 0 00 2 − λ 1 00 0 2 − λ 10 0 0 3 − λ= (2 − λ)3(3 − λ)Hence, the eigenvalues of B are 2 and 3. For λ = 2:0 0 0 00 0 1 00 0 0 10 0 0 1R4→R4−R3−−−−−−−→0 0 0 00 0 1 00 0 0 10 0 0 0Hence, the corresponding eigenspace is span1000,0100.For λ = 3:−1 0 0 00 −1 1 00 0 −1 10 0 0 0Hence, the corresponding eigenspace is span0111.For C, we have:det2 − λ 0 0 00 2 − λ 0 00 0 2 − λ 10 0 0 3 − λ= (2 − λ)3(3 − λ)3Hence, the eigenvalues of C are 2 and 3. For λ = 2:0 0 0 00 0 0 00 0 0 10 0 0 1R4→R4−R3−−−−−−−→0 0 0 00 0 0 00 0 0 10 0 0 0Hence, the corresponding eigenspace is span1000,0100,0010.For λ = 3:−1 0 0 00 −1 0 00 0 −1 10 0 0 0Hence, the corresponding eigenspace is span0011.3. (This question is not yet relevant for the third midterm exam.) LetA =1 11 1.Find a diagonal matrix D and an invertible matrix P such that A = P DP−1.Solution: We have to find the eigenvalues and corresponding eigenspaces. We have:det1 − λ 11 1 − λ= (1 − λ)(1 − λ) − 1 = −λ(2 − λ)Hence, the eigenvalues of A are 0 and 2. For λ = 0:1 11 1R2→R2−R1−−−−−−−→1 10 0Hence, the corresponding eigenspace is span−11.For λ = 2:−1 11 −1R2→R2+R1,R1→−R1−−−−−−−−−−−−−→1 −10 0Hence, the corresponding eigenspace is span11.The columns of P are (linearly independent) eigenvectors of A and D is the diagonal matrixwith eigenvalues of A on the main diagonal in the appropriate order (corresponding to columnsof P ). Therefore:P =−1 11 1D =0 00
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