Eigenvalues for Triangular Matrices ENGI 78: Linear Algebra Review Finding Eigenvalues and Diagonalization Adapted from Notes Developed by Martin Scharlemann The eigenvalues for a triangular matrix are the entries of the diagonal Reason: Suppose A is an upper triangular matrix and a kk is one of the diagonal entries Then B A a kk I is an upper triangular matrix with a 0 in the k th diagonal spot June 4, 06 So B is in echelon form with (at least) x k as a free variable The existence of a free variable means the null space is not empty The null space of B is exactly 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 matrix in general? Recall: Let B A λi be an n n matrix Then Example: Find all eigenvalues for the matrix A Answer: Find values of λ for which λ det(a λi ) det 0 λ 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 exactly the values λ for which det(a λi ) 0! Since det(a λi ) is a polynomial, this is a familiar problem /8 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
Example: Find all eigenvalues and corresponding eigenvectors for 6 the matrix A Solution: Step : Find the eigenvalues That is, find all the solutions to the characteristic equation for A: λ 6 det(a λi ) det 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 Hence eigenvalues are λ 7 and λ 4 /8 Step : Calculate eigenvectors for eigenvalue λ 7: This is the null space of 7 6 A 7 I 7 Solution: Eigenspace is all multiples of vector Check: 6 6 6 9 reduce 0 0 7 7 6/8 Step : Calculate eigenvectors for eigenvalue λ 4: This is the null space of + 4 6 A ( 4) I + 4 Solution: Eigenspace is all multiples of vector Check: 6 9 6 9 6 reduce 0 0 8 4 4 / Example: Find the eigenvalues for the matrix A /4 / det(a λi ) λ + There are two complex eigenvalues ±i We solve for the eigenspace in the usual way, although the matrices A ii and A + ii will now be complex 7/8 8/8
cos θ sin θ Final Example: Rotation: matrix R θ sin θ cos θ R (v) v 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 + Here is what we said before: In general, a rotated vector is moved to a new direction, so for most values of θ there will be no eigenvectors! (Hence no eigenvalues either) However, this is not exactly true! The question is whether the eigenvalues are complex Lets go ahead and determine them 9/8 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 complex whenever cos θ < So we only get real eigenvalues when θ is an integer multiple of π 0/8 Pushing further - can often get complete formula for A bazillion Why should we care about eigenvalues and eigenvectors? One reason: Suppose you plan to apply a matrix 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 Example (from above): Let A 6 What is A 0? Solution: Since A 7, 7 0 AAAAAAAAAA 7 7 7 7 0 /8 Example: Have shown A 7 and A 4 Combining these vector equations into a matrix 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 matrix 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 example, 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 But why would anyone ever need to calculate A 000? /8 Example Suppose w 0 is the number of wolves in the forest at time t 0, is the number of rabbits Consider the following simple biological model: 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 next year, the number of wolves w and rabbits r might be given by: w w 0 + r w 0 0 + The constants,, 0, and would come from biological experiments () () 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 In general, after k years we have w k wolves & r k rabbits: 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 matrix V 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!
Definition If A is a square matrix and there is an invertible matrix 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 non-example: Recall the shear matrix which yielded the following transformation: Question: When can you diagonalize A? Theorem Answer: Can diagonalize A if and only if A has n linearly independent eigenvectors 7/8 A has repeated eigenvalues λ 0 λ which both have the same eigenspace (multiples of ) 0 This matrix is not diagonalizable because diagonalizability requires n linearly independent eigenvectors, which is equivalent to having n distinct eigenvalues 8/8