Review• The determinant is characterized by det I = 1 and the effect of row op’s:◦ replacement: does not change the det er minant◦ intercha n g e: reverses the sign of the determinant◦ scaling row by s: multiplies the determina nt by s•1 2 3 40 2 1 50 0 2 10 0 0 3= 1 · 2 · 2 · 3 = 12• det (A) = 0A is not invertible• det (A B) = det (A)det(B)• det (A−1) =1det (A)• det (AT) = det (A)• What’s wrong?!det (A−1) = det1ad − bcd −b−c a=1ad − bc(da − (−b)(−c)) = 1The corrected calculation is:det1ad − bcd −b−c a=1(ad − bc)2(d a − (−b)(−c)) =1ad − bcThis is compatible with det (A−1) =1det (A).Example 1. Suppose A is a 3 × 3 ma tr ix with d e t (A) = 5. What i s det (2A)?Solution. A has three rows.Multiplying all3 of them by 2 produces 2A.Hence,det (2A) = 23det (A) = 40.Armin [email protected] “ba d” way to compute determinantsExample 2. Compute1 2 03 −1 22 0 1by cofactor expansion.Solution. We expand by the second column:12 03−1 220 1= −2 ·−3 221+ (−1) ·10+2 1− 0 ·1032−= − 2 · (−1) + (−1) · 1 − 0 = 1Solution. We expand by the third column:1 203 −122 01= 0 ·+3 −12 0− 2 ·1 2−2 0+ 1 ·1 23 −1+= 0 − 2 · (−4) + 1 · (−7) = 1Why i s the method of cofactor expansion not practical?Be cause to compute a largen × n determinant,• one reduces to n determinants of size (n − 1) × (n − 1),• then n (n − 1) determinants of size (n − 2) × (n − 2),• and so on.In the end, we haven! = n(n − 1)3 · 2 · 1 many nu mbers to add.WAY TOO MUCH WORK! Already25! = 15511210043330985984000000 ≈ 1.55 · 102 5.Context: today’s fastest computer, Tianhe-2, runs at34 pflops (3.4 · 101 6op’s per second).By the way: “fastest” is mea sured b y co mputed LU decompositions!Example 3.First off , say hello to a new friend: i, the imaginary unitIt is infamous f ori2= −1.|1| = 11 ii 1= 1 − i2= 21 ii 1 ii 1= 11 ii 1− ii 0i 1= 2 − i2= 31 ii 1 ii 1 ii 1= 11 ii 1 ii 1− ii 0i 1 ii 1= 3 − i21 ii 1= 5Armin [email protected]1 ii 1 ii 1 ii 1 ii 1= 11 ii 1 ii 1 ii 1− i21 ii 1 ii 1= 5 + 3 = 8The Fibonacci numbers!Do you know about the connection of Fibonacci numbers and rabbits?Eigenvectors and eigenvaluesThroughout,A wil l be an n × n ma t rix.Definition 4. An eigenvector of A is a nonzero x such thatAx = λx for some scalar λ.The scalar λ is the corresponding eigenvalue.In words: eigenvectors are thosex, f or which Ax is parallel to x.Example 5. Verify that1−2is an eig envector of A =0 −2−4 2.Solution.Ax =0 −2−4 21−2=4−8= 4xHence, x is an eigenvector of A with eigenv alue 4.Armin [email protected] 6. Use your geometric understanding to findthe eige nvectors and eigenvalues ofA =0 11 0.Solution. Axy=yxi.e. multiplication with A is reflection through the line y = x.• A11= 1 ·11So: x =11is an eigenvector with eigenvalue λ = 1.• A−11=1−1= −1 ·−11So: x =−11is an eigenvector with eigenvalue λ = −1.Practi ce problemsProblem 1. Let A be an n × n matrix.Express the foll owing in terms ofdet (A):• det (A2) =• det (2A) =Hint: (unless n = 1) this is not just 2 det (A)Armin
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