Exercise. Given the system of equations 8 < : x + y + z x + y + z k x + k y + z k where k R. a) Study the system in terms of the values of k. b) Find the solutions when k, using the reduced row echelon form of the matrix. c) For k, nd the inverse matrix of the principal minor, using Gauss-Jordan s method. a) Let us consider the matrix form of the system, focusing on the augmented matrix: @ k @ k F+(k! k k k k + )F!F From the last row, take! F F F!F F!F k k @ k k k + k + We obtain cases: - If k the system has only one solution (Rank of the matrix of the system, rank of the augmented matrix ). - If k the system @ has no solution. (Rank of the matrix of the system, rank of the augmented matrix ). - If k the system @ has in nite solutions. (Rank of the matrix of the system, rank of the augmented matrix ). b) When k the system we have obtained is @ but we are asked to obtain the reduced row echelon form, so we continue the process @! @! @ F+F!F Finally, from the equivalent system we can write F!F F!F z y + x
Exercise. Given the matrices and B being a matrix with all the elements di erent from zero and verifying B identity matrix), obtain: a) Values a and b such that a + bi. b) Values p and q such that B pb + qi, checking that B has inverse matrix. c) Values x and y for which B xb + yi. B + I (I is the a) a + bi a + b a + b a a a + b therefore a a + b and a b b) B veri es the relation B B + I B + B I B (B + I) I Therefore B B + I and the values we were asked for are p q c) B B B ( B + I) B B + B ( B + I) + B 9B I + B B I and the values are x and y.
Exercise. Given the basis B f(; ) ; (; )g B f(; ) ; (; )g a) Calculate the change matrices B C and C B. b) Use them to calculate B B. c) Find the coordinates of v R respect to base B if v B a) b) We know that B C B C B B C B B C! C B c) Finally v B B B v B Exercise. Let S and T be two subspaces de ned as a) Calculate dim S, dim T and dim (T + S). b) The sum S + T is direct? a) S (x; y; z) R : x + y T hf(; ; ) ; (; ; ) ; (; ; )gi S (x; y; z) R : x + y (x; y; z) R : y x f(x; x; z) : x; z Rg hf(; ; ) ; (; ; )gi dim S For T we need to check if the vectors are L.I. and since
vectors are L.D., but the two rst vectors (for example) are L.I., Finally T hf(; ; ) ; (; ; )gi dim T T + S hf(; ; ) ; (; ; ) ; (; ; ) ; (; ; )gi and we need to check how many vectors L.I. are there. s before, since the three rst vectors verify that b) Since S + T hf(; ; ) ; (; ; ) ; (; ; )gi dim (S + T ) S + T R dim (S + T ) dim S + dim T dim (S \ T ) dim (S \ T ) + dim (S \ T ) and the sum is not direct (dimension should be for the sum to be direct). Exercise. Let H Im (f) image subspace of f : R! R given by f (x; y; z) (x + y; x + y + z; x + y + mz) a) Calculate the dimension of H in terms of m. b) If m is such that dim H is the lowest, calculate H? and ker (f). Im f hff (; ; ) ; f (; ; ) ; f (; ; )gi hf(; ; ) ; (; ; ) ; (; ; m)gi Since m m! m if m, the vectors are L.D. and dim Im f. If m vectors are L.I. and dim Im f. b) Take m (so dim Im f ) and Im f hf(; ; ) ; (; ; )gi. Therefore and H? hf(; ; )gi. (; ; ) (; ; ) (; ; ) Finally in order to calculate ker f let us equal function to : (x + y; x + y + z; x + y z) (; ; )
x + y x + y + z x + y z and solving the system we get x y z and ker f hf( ; ; )gi dim ker f Exercise. a) Given S hf( ; ; ; ) ; (; ; ; )gi. Find the projection of v ( ; ; ; ) onto S. b) Using least-squares method nd the straight line that t best the points ( ; ) ; (; ) ; (; ) ; (; ) a) pply Gram-Schmidt ( ; ; ; ) u k( ; ; ; )k p p p ; ; ; w v (v u ) u (; ; ; ) (; ; ; ) (; ; ; ) ; ; ; ; ; ; u ; ; ; p p p p ; ; ; ; ; Projection: proj S ( ; ; ; ) ( ; ; ; ) + ( ; ; ; ) ; ; ; + p p ; ; ; p p p p p ; p ; ; p p ; p p p p p ; ; ; ; ; ; + p p p p p p p p p!! p p p p p p p p! ; ; ; ; ; ; ; ; ; ; ; ; b) We are searching for a straight line y mx + n such that m + n m + n m + n m + n
In matrix form B @ C m n B @ Multiplying by the transpose of the matrix of the system B C m @ n m n C B @ C Solving the system m + n m + n the solution is so the straight line we were looking for is m n y x Exercise. Consider a;b @ b a Find the values a and b for which the matrix a;b is diagonalizable. If possible diagonalize ; and calculate ( ; ) n. if j ab Ij b a a ( ) ( ) (a ) Then: - If a ; all the eigenvalues are di erent and the matrix is diagonalizable. - If a is double and is a simple root. Consider and calculate the eigenvectors ( ) x! @ b @ x y z @
and from this system we only get one eigenvector for any value of b (we would need two parameters to obtain two eigenvectors and match the number of times is solution of the characteristic equation). Therefore the matrix is not diagonalizable. - If a we get two eigenvalues: (double) and (simple). Consider and calculate the eigenvectors (we need two associated to this eigenvalue) ( ( ) ) x! @ b In this case, we can see that the rst and the third row are the same (x ) and if b we get two parameters and therefore two eigenvectors, and if b we only get one parameter and only one eigenvector. Therefore the matrix is diagonalizable if b and a. Finally, if a and b the matrix is ; @ @ x y z @ and is easy to check (homework) that the eigenvalues and eigenvectors are 8 9 8 < < @ : ; $ ; @ 9 8 9 < : ; $ ; @ : ; $ and Therefore Finally D ; P DP @ @ ( ; ) n P D n P @ P @ @ @ @ n ( ) n n @