Linear Algebra

Transcription

1 . Linear Algebra Midterm Solutions. (pts) Consider a matrix A, andletb rref(a). (a) Is ker (A) necessarily equal to ker (B)? Explain. (b) Is im (A) necessarily equal to im (B)? Explain. (a) Yes. By construction of the reduced row-echlon form, the system A x and B x have the same solutions(the whole process of Gauss-Jordan elimination doesn t change the solutions of a system). (b) No. Choose A. Then, we have B rref (A). Hence, ½ im (A) span, ¾ ½ y ¾ where y R and im (B) span ½, ¾ ½ x ¾ where x R which tell us that im (A) 6 im(b).. (5pts) Consider the n n matrix M n which contains all integers,,,,n as its entries, written in sequence, column by column; for example, 5 M (a) Determine the rank of M. (b) Determine the rank of M n, for an arbitrary n. (c) For which integers n is M n invertible?

2 (a) By Gauss-Jordan elimination, we have 5 M Hence, the rank (M ). (b) By Gauss-Jordan elimination, we have n + n + (n ) n + n + n + (n ) n + M n n n n n n + n + (n ) n + n + (n +) n + (n +) (n ) n + ((n ) n +) n n (n +) n n (n +) n n ((n ) n +) (by using the fact: sn + t t (sn +) sn tsn (t ) sn for all s n and t n ) n + n + (n ) n + n n (n ) n (n ) n (n ) n (n ) (n ) n

3 (n ) (n ) Hence, the rank (M n ). (c) Since M n is invertible if and only if rank M n n. By part (b), we have M and M with are invertible. M n for n is not invertible.. (5pts) If A and B are two n n matrices such that BA I n.provethe following properties: (a) A and B are both invertible. (b) A B and B A. (c) AB I n.. (a) To prove A is invertible, we need to show that A x has exact one solution. Assume y be a solution of A x. Hence, we have A y. Thisimplies y I n y (BA) y B (A y). Therefore, is the only solution of A x. Thistellsusthat A is invertible. We will prove (b) first and use (b) to prove B is invertible. (b) Since A is invertible, there exists a A such that A A I n AA.And,wehave A I n A (BA) A B AA BI n B. By the definition of inverse linear transformation, we have if a linear transformation T is invertible, then so is T and (T ) T. Assume T ( x) A x. we have T ( y) A y B y. This implies B is invertible and B A A. This completes the proof of (a) and (b). (c) Since A B, wehaveab AA I n.

4 . (pts) Let T from R to R be the reflection in the plane given by the equation x +x +x. (a) Find the matrix B of this transformation with respect to the basis v, v, v (b) Use your answer in part (a) to find the standard matrix A of T. (a) Let P be the plane given by the equation x +x +x. We observe that v and v arebothontheplanep.sincet is areflection, it keeps v and v unchanged, that is, T ( v ) v and T ( v ) v. And, v is the normal vector of P. That means v is perpendicular to the plane P. Therefore, since T is a reflection in P,wehaveT ( v ) v. Withrespect. to the basis B { v, v, v },wehave[t ( v )] B [T ( v )] B and [T ( v )] B.So B B B [T ( v )] B [T ( v )] B [T ( v )] B (b) Write S v v v.. Bythetheorem in the textbook, we have A SBS. To find S,weuse Gauss-Jordan elimination on B,

5 Hence, 8 5 S and 6 A SBS (pts) Consider a linear transformation T from R n to R m. (a) Let v, v,, v q be vectors in R n.ift( v ),T ( v ),,T ( v q ) are linearly independent, are v, v,, v q linearly independent? How can you tell? (b) Let v, v,, v q be vectors in R n.if v, v,, v q are linearly independent, are T ( v ),T ( v ),,T ( v q ) linearly independent? How can you tell? (a) Yes! Assume a v + a v + + a q v q.. By applying T on both sides, we have ³ T (a v + a v + + a q v q )T. Since T is a linear transformation, a T ( v )+ + a q T ( v q )T (a v + + a q v q ). Since T ( v ),T ( v ),,T ( v q ) are linearly independent, we have a a a m. That tells us that v, v,, v q are linearly independent. 5

6 6 (b) No! Let T ( x) x be a linear transformation from R to R. e and are linearly independent in R.But, µ T ( e ) T are not linearly independent µ since we have T ( e ) T. 6. (pts) Given a matrix 5 A (a) Find a basis of kernel of A and dim (ker (A)). (b) Find a basis of image of A and dim (im (A)). By Gauss-Jordan elimination, we have A

7 rref (A). (a) Assume x x x x x x 5 ker A. Then we have A x. By Gauss-Jordan elimination(which implies ker (A) ker rref (A)), we have x x x x x 5. Assume x 5 t for all s, t R. We have solutions of the system, x x x x x 5 7t 5t t 7 t t t t

8 8 Assume v Moreover, for all k R, A (k v) k That means ker (A) span { v} This tells us that which span { v} ker (A) which implies ker (A) span{ v}. Since v isanonzerovector, v is linearly independent. Now, we can say { v} is a basis of ker (A). And, dim (ker (A)) equals the number of vectors in a basis. So, dim (ker (A)). (b) Set v, v, v, v, v 5 to be column vectors of A. Set w, w, w, w, w 5 to be column vectors of rref A. Weknowthat im (A) span{ v, v, v, v, v 5 }. From our reduced row-echlon form, we can read there is only one relation of w, w, w, w, w 5, 7 w + 5 w w 7 w w 5. This implies we have the same relation of v, v, v, v, v 5, 7 v + 5 v v 7 v v 5. And, that is the only relation of v, v, v, v, v 5.Thenwecan conclude that v, v, v, v are linearly independent and im (A) span{ v, v, v, v, v 5 } span{ v, v, v, v }. Now, we can say that { v, v, v, v } is a basis of im (A). And, dim (im (A)) equals the number of vectors in a basis. So, dim (im (A)).

9 7. (pts) Let L be a line in R that consists of all scalar multiples of the vector. (a) Find the orthogonal projection of the vector onto L. (b) Find a matrix A such that proj L ( x) A x for all x R. (a) Since + +, we have a unit vector of L, u. By the formula of projection, we have proj L ( x) ( u x) u. Hence the orthogonal projection of the vector is proj L µ + + µ 5 5. onto L (b) To calculate the matrix A of the orthogonal projection onto L, proj L ( ), wecalculateproj L ( e ), proj L ( e ) and proj L ( e ) first. We have proj L µ + + µ.

10 And, proj L And, proj L µ + + µ. µ + + µ So, the matrix A proj L ( e ) proj L ( e ) proj L ( e )..