Eigenvalues for Triangular Matrices ENGI 78: Linear Algebra Review Finding Eigenvalues and Diagonalization Adapted from Notes Developed by Martin Scharlemann June 7, 04 The eigenvalues for a triangular matri are the entries of the diagonal Reason: Suppose A is an upper triangular matri and a kk is one of the diagonal entries Then B A a kk I is an upper triangular matri with a 0 in the k th diagonal spot So B is in echelon form with (at least) k as a free variable The eistence of a free variable means the null space is not empty The null space of B is eactly the eigenspace for the eigenvalue a kk /8 /8 Defining Equation for Eigenvectors and Eigenvalues A v λ v How do you find the eigenvalues for a matri in general? Recall: Let B A λi be an n n matri Then Nul B 0 B invertible det B 0 Translation: let λ R and B A λi λ is eigenvalue for A Nul B 0 det B 0 In other words, the eigenvalues of A are eactly the values λ for which det(a λi ) 0! Since det(a λi ) is a polynomial, this is a familiar problem /8 Eample: Find all eigenvalues for the matri A Answer: Find values of λ for which λ det(a λi ) det 0 λ In other words, find roots of the quadratic equation λ λ + 4 0 We ve known since high school how to do this! λ λ + 4 (λ 4)(λ ) 0 λ or 4 Check: Both A 4I non-empty null spaces and A I clearly have 4/8
Eample: Find all eigenvalues and corresponding eigenvectors for 6 the matri A Solution: Step : Find the eigenvalues That is, find all the solutions to the characteristic equation for A: det(a λi ) det λ 6 0 λ Calculate: λ 6 det ( λ)( λ) 8 λ λ 8 λ Note that λ λ 8 is called the characteristic polynomial of A λ λ 8 (λ 7)(λ + 4) 0 λ 7 or λ 4 Step : Calculate eigenvectors for eigenvalue λ 7: This is the null space of 7 6 6 6 A 7 I 7 9 reduce 0 0 Solution: Eigenspace is all multiples of vector Check: 6 7 7 Hence eigenvalues are λ 7 and λ 4 /8 6/8 Step : Calculate eigenvectors for eigenvalue λ 4: This is the null space of + 4 6 9 6 9 6 A ( 4) I + 4 reduce 0 0 Solution: Eigenspace is all multiples of vector Check: 6 8 4 4 / Eample: Find the eigenvalues for the matri A /4 / det(a λi ) λ + There are two comple eigenvalues ±i We solve for the eigenspace in the usual way, although the matrices A ii and A + ii will now be comple 7/8 8/8
cos θ sin θ Final Eample: Rotation: matri R θ sin θ cos θ R (v) Here is what we said before: Every vector is moved to a new direction, so no eigenvectors! (Hence no eigenvalues either) However, this is not eactly true! Lets determine the eigenvalues v 9/8 R θ λi cos θ λ sin θ sin θ cos θ λ det(r θ λi ) (cos θ λ) + sin θ cos θ cos θλ + λ + sin θ λ cos θλ + Setting this to 0 and solving for λ we get, λ cos θ ± 4 cos θ 4 λ cos θ ± cos θ Try some particular values of θ In general, the eigenvalues will be comple whenever cos θ < So we only get real eigenvalues when θ is an integer multiple of π 0/8 Why should we care about eigenvalues and eigenvectors? One reason: Suppose you plan to apply a matri linear transformation A to a vector v lots of times: AAAAAA v A bazillion v If v is an eigenvector and λ is its eigenvalue, then A bazillion v λ bazillion v So it s easy to calculate Eample (from above): Let A Solution: Since A 7, 6 What is A 0? 7 0 AAAAAAAAAA 7 7 7 7 0 /8 Pushing further - can often get complete formula for A bazillion Eample: Have shown A 7 and A 4 Combining these vector equations into a matri equation: A 7 4 7 0 7 4 0 4 Diagonalization We can generalize the left and right sides of the above equation: AV VD where V is composed of the eigenvectors of A arranged as columns and D is a diagonal matri with the corresponding eigenvalues along the diagonal Its often useful to rewrite this equation as the following diagonalization of A: A VDV
Returning to our eample, we can compute A 000 using the relation A VDV If we denote V we have AV V 7 0 0 4 A 000 V A V 7 0 V 0 4 7 000 0 0 ( 4) 000 V Why would anyone ever need to calculate A 000? Eample Suppose w 0 is the number of wolves in the forest, is the number of rabbits Consider what happens: Wolves make more wolves If there are rabbits to eat, they make even more wolves! Similarly rabbits make (lots) more rabbits But rabbits also get eaten by wolves So, if in year zero there are w 0 wolves & rabbits, then net year, the number of wolves w and rabbits r might be given by: w w 0 + r w 0 0 + () () /8 4/8 In other words, w r 0 w0 If the same model applies the following year, then after that second year there are w wolves & r rabbits where ( ) w w w0 r In other words, 0 r w r 0 A w0 0 /8 More generally, after k years have w k wolves & r k rabbits where : wk A k w0 Characteristic equation of Roots of λ λ + 79 00 λ ± r k 0 is λ λ + 79 00 0 can be found via the quadratic formula: 4 79 ± 00 Now find eigenvectors for each, and assemble into the matri V Then A 000 V ( + 00 )000 0 V 0 ( 00 )000 Multiply by the initial condition vector to get the number of wolves and rabbits afte00 years! 6/8
Definition If A is a square matri and there is an invertible matri P so that A PBP then A and B are similar Two matrices, A and B, which are similar will share the same eigenvalues In the diagonalization A VDV the matrices A and D are similar by design Question: When can you diagonalize A? Theorem Answer: Can diagonalize A if and only if A has n linearly independent eigenvectors 7/8 non-eample: Recall the shear matri which yielded the following transformation: rectangle e T(e ) Ae A(rectangle) T(rectangle) A has repeated eigenvalues λ 0 λ which both have the same eigenspace (multiples of ) 0 This matri is not diagonalizable because diagonalizability requires n linearly independent eigenvectors, which is equivalent to having n distinct eigenvalues 8/8