Name (Last name, First name): MTH 5 Linear Algebra Practice Final Exam April 6, 6 Exam Instructions: You have hours to complete the exam. There are a total of 9 problems. You must show your work and write in complete sentences. Partial credit may be given even for incomplete problems as long as you show your work.. Label the following statements as true or false. (a) If A is an n n matrix with rank n, then the reduced row echelon form of A is I n. Answer: True, it is invertible if it has rank n recall. (b) Any matrix can be put into reduced row echelon form by means of a nite sequence of elementary row operations. Answer: True, by Gaussian elimination, which all steps are done using elementary row operations, we reduced any m n matrix into its unique reduced row echelon form. (c) If (Ajb) is in reduced row echelon form, then the system Ax b is consistent. Answer: False. Consider the following example: A it is in reduced row echelon form. Consider the augmented matrix (Ajb) which is in reduced row echelon form, but the system with x x x Ax b is not consistent since it has no solution. ; b
. Use Gaussian elimination to solve the following systems of linear equations. x + x x 3 x + x + x 3 3x + 5x x 3 Solution We begin by forming the augmented matrix (Ajb) for this system: (Ajb) @ A : 3 5 Now we use Gaussian elimination to put this matrix into reduced row echelon form: (Ajb) @ A 3R+R3R3 @ A 3 5 R +R R R +R 3R 3 R +R R R 3+R R @ 3 3 3 3 A @ @ 3 3 4 @ 3 A : A R3R3 RR A 3 R3+RR @ 3 3 A @ 3 3 A @ 3 A Therefore the solution is x @ x x A @ 4 3 A : x 3
3. Prove: If A M nn (F ), where F R or F C, then Proof. We have A A T, that is, det (A ) det (A): (A ) ij A ji : We also have for any matrix B M nn (F ) that det (B) ( ) i+j B ij det ebij ; j i ( ) i+j B ij det ebij where e B ij is the (n )(n ) matrix obtained by deleting row i and column j from B. Thus, we will prove the statement by induction on n. For n, the statement is true since det (A ) A det (A): Suppose not the statement is true for some n, we will now prove its true for n +. We have det (A) ( ) i+j A ij det eaij and det (A ) j j j j j det (A); ( ) i+j A ij det eaij ( ) i+j A ij det ea ij ( ) i+j (A ) ij det fa ij ( ) i+j A ji det eaji where we used the induction hypothesis for det eaij det ea ij as well as the fact that fa ij eaji : Therefore, its true for n + and hence by induction is true for all n. completes the proof. This 3
4. (a) For the following matrix A M (R), test A for diagonalizability, and if A is diagonalizable, nd an invertible matrix Q and a diagonal matrix D such that Q AQ D. 4 A : 3 (b) Find an expression for A n, where n is an arbitrary positive integer. Solution (a) The eigenvalues of A are the solutions to the characteristic equation 4 det (A I) det 3 ( ) ( ) 3 + 3 + p p 49 3 49 ( 5) ( + ) Hence it has two distinct eigenvalues ; 5: Therefore, it is diagonalizable since dim R and as the eigenvalues of A are distinct then it has at least two linearly independent eigenvectors fv ; v g which are hence a basis for R implying A is diagonalizable. To any eigenvalue of A the set of eigenvectors of A corresponding to the eigenvalue is all the nonzero vectors in the nullspace N (A I). Thus, we compute this nullspace for each eigenvalue. First, N (A I) v R : (A I) v : Hence, we want to solve the equation @ A 4 a 3 b ( ) a + 4b : 3a + ( ) b Hence if then N (A I) t 3 : t R ; 4 if 5 then N (A I) t : t R 4
It follows that D Q AQ, where 5 Q : 3 4 (b) It now follows from this representation (and a proof by induction on n) that A QDQ A QDQ QDQ QD Q ; A 3 A A QD Q QDQ QD 3 Q ;. A n QD n Q : So that for all positive integers n. A n QD n Q A n ( Q n ) 5 n Q 5
5. (a) Find a counterexample to the statement: If A M nn (F ), where F R or F C, and ja ij j < then lim m A m. (b) Prove the following statement: If A M nn (F ), where F R or F C, and A is diagonalizable with each eigenvalue satisfying jj < then lim m A m. Solution 3 Let Then for all n. Thus, in this case, i.e., for all i, j. A A n : lim m Am A 6 lim m (Am ) ij (A) ij 6
6. Let A be an n n matrix with characteristic polynomial f (t) ( ) n t n + a n t n + + a t + a : (a) Prove that A is invertible if and only if a 6. (b) Prove that if A is invertible, then A a ( ) n A n + a n A n + + a I n : Proof. The characteristic polynomial of A is det (A ti n ) f (t) ( ) n t n + a n t n + + a t + a : Thus, det (A) f () a : Therefore, A is invertible if and only if det (A) 6 if and only if a 6. suppose that A is invertible. Then by the Cayley-Hamilton Theorem Now f (A) ( ) n A n + a n A n + + a A + a I and hence since a 6 and A exists we have A f (A) ( ) n A n + a n A n + + a I + a A so that we can solve for A to get A a ( ) n A n + a n A n + + a I n which is the desired result. 7
7. Let fv ; : : : ; v k g be an orthogonal set in V, and let a ; : : : ; a k be scalars. Prove that X k a iv i i X k ja ij kv i k : i Proof. Denote the inner product by hu; vi for u; v V. We proceed by induction on k, the number of orthogonal set of vectors from V and number of scalars. For k, X k a iv i i ka v k ja j kv k X k ja ij kv i k : i Let us also prove the statement for k. We have X k a iv i i ka v + a v k ha v + a v ; a v + a v i a a hv ; v i + a a hv ; v i + a a hv ; v i + a a hv ; v i ja j kv k + ja j kv k X k i ja ij kv i k : This proves the statement for k. Now suppose the statement is true for k let us prove it is true for k +, where k. Using the fact that its for k and k we have X k+ a Xk iv i i () a iv i + a k+ v k+ i Xk jj a iv i + ja k+ j kv k+ k i X k ja ij kv i k + ja k+ j kv k+ k i X k+ ja ij kv i k : i This proves the statement for k +. Therefore by induction the statement is true for all k. 8
8. Let V be a nite-dimensional inner product space V, and let T be an invertible linear operator on V. Prove T is invertible and T (T ). Proof. Let V be a nite-dimensional inner product space V, and let T be an invertible linear operator on V. Denote the inner product by hu; vi for u; v V. Then for any u, v V we have Du; T T E v DT u; T E v T T u; v hu; vi. Uniqueness of the adjoint implies T T IV. This implies (T ) T. This completes the proof. 9
9. Label the following statements as true or false. Assume that the underlying inner product spaces are nite-dimensional. (a) Every normal operator is diagonalizable by an orthonormal basis of eigenvectors. Answer: True, if F C but false if F R. (b) Every self-adjoint operator is normal and all its eigenvalues are real. Answer: True, see Lemma on p. 373 and if T is self-adjoint then T T so that T T T T T which means its also normal. (c) Every skew self-adjoint operator is normal and all its eigenvalues are purely imaginary. Answer: True, if T is skew self-adjoint then T T so that T T T T T which means its also normal. Also, if F C then it is self-adjoint so its eigenvalues are real implying the eigenvalues of T are purely imaginary. On other hand, if F R then it has no eigenvalues except for possibly, since it can be treated as an operator on F C whose restriction is the operator on F R so it cannot have any real eigenvalues since all eigenvalues of the operator on F R are eigenvalues of the operator on F C which has only purely imaginary eigenvalues. (d) The matrix 4 A 3 is normal. Answer: False, since 4 3 3 4 7 3 6 3 4 4 : 3 (e) The linear operator T : P (R) P (R) de ned by T (f (x)) f (x) is normal, where the inner product is hg (x) ; h (x)i Z g (t) h (t) dt, for g; h P (R). Answer: True, since for all g; h P (R) we have by integration by parts hg (x) ; T h (x)i Z g (t) h (t) dt h () g () g () h () Z g (t) h (t) dt
and hence This proves that ht g (x) ; h (x)i Z g (t) h (t) dt g (t) h (t) j t t Z g () h () g () h () g (t) h (t) dt Z g (t) h (t) dt g () h () g () h () hg (x) ; T h (x)i hg (x) ; T h (x)i : T T, i.e., T is skew self-adjoint, implying it is normal since T T T T T :