Solutions to Midterm 2 Practice Problems Written by Victoria Kala Last updated 11/10/2015

Transcription

1 Solutions to Midterm 2 Practice Problems Written by Victoria Kala vtkala@math.ucsb.edu Last updated //25 Answers This page contains answers only. Detailed solutions are on the following pages (a) A (g) 7 5 cos θ sin θ 4. (b) T (2 ) ( 2 ) sin θ cos θ (c) ker(t ) {} (d) range(t ) span 2. (a) Linear (b) Not linear (c) Linear (d) Not linear 9 4. (a) 8 7 (b) 5 2 (c) Undefined (d) (e) Undefined 8 28 (f) (a) det(g) (b) G does not exist 6. (a) det(j) 5 (b) J / / /5 /5 /7 /7 7. U is not a subspace 8. See detailed solution (a) col(k) span (b) null(k) span 4 5

2 Detailed Solutions. Let T be the linear transformation defined by the formula T (x x 2 ) (x 2 x x + x 2 x x 2 ). (a) Find the standard matrix A for the linear transformation such that T (x) Ax. Solution. Recall that the standard matrix A is given by A ( T (e ) T (e 2 ) ) where e ( ) and e 2 ( ). We need to find T ( ) and T ( ). Using the formula above we see that T ( ) ( ) and T ( ) ( ). Therefore A. Let s check: x 2 x x x 2 x + x 2 x x 2 This is the same as T (x x 2 )! (b) Find the image of (x x 2 ) (2 ). Solution. Use the formula above to find T (2 ) (or you can use matrix multiplication using the matrix in (a)): T (2 ) ( 2 ). (c) Find the kernel of T (Hint: This is the null space of A). Solution. The kernel of T is the set of all vectors x such that T (x). But since T (x) Ax then we solve Ax (this is the null space of A): R2+R R R2+R4 R4 R+R R R+R4 R4 2

3 R2 R R R This shows x x 2 hence x. This is the only solution to Ax hence ker(t ) null(a) {}. (d) Find the range of T (Hint: This is the column space of A). Solution. The range of T is the column space of A. The column space of A is the span of the columns: col(a) span. This is the solution. However here is another question: is this a basis of the column space? The answer is yes! We found the reduced row echelon form of A to be. The first and second column each have a pivot hence the first and second column of A are linearly independent and span the space. 2. Determine whether T : R 2 R 2 is a linear operator where (a) T (x (2x + y x Solution. Linear operator means linear transformation. For (a) (d) we need to show that: (( ) ( x w x w T + T + T and T ( ( ( x x c ct. ) Let s look at addition: x w 2x + y 2w + 2x + y + 2w + T + T + y x y w x y + w (( ) x w x + w 2(x + w) + (y + ) T + T y + (x + w) (y + ) We see that T (( ) ( x w x w + T + T and so the first property holds.

4 Let s look at scalar multiplication: ( ) x cx 2cx + cy T c T y cy cx cy (( x 2x + y c(2x + ct c ) x y c(x ( ) x x We see that T c ct and so the second property holds. y y Thus T is a linear transformation. (b) T (x (x + Solution. Let s look at addition: ( x w x + w + x + w + 2 T + T + y y + ( ) x w x + w x + w + T + T y y + y + Therefore T (( ) ( x w x w + T + T and so the first property doesn t hold. This is enough to show that the transformation is not linear. It also doesn t hold for scalar multiplication: ( ( x cx cx + T c T ) cy cy ( x x + cx + c ct c y cy Therefore T ( ( ( x x c ct and so the second property doesn t hold. ) Since neither property holds then T is not a linear transformation. (Note: you only need to show that one property fails so choose whichever one seems easiest to you.) (c) T (x (y Solution. Let s look at addition: ( ( x w y T + T (( ) x w T + T We see that T ( x + y ) w T x + T y ( + ) ) ( x + w y + y + y + y + y + w and so the first property holds. 4

5 Let s look at scalar multiplication: ( ) x cx cy T c T y cy cy ( ( x y cy ct c cy ( ) x x We see that T c ct and so the second property holds. y y Thus T is a linear transformation. (d) T (x ( x Solution. Let s look at addition: ( ( x w ) ( x ) ( w T + T + ) x + w y y + ( ) ( x w x + w ) x + w T + T y y + y + Therefore T (( ) ( x w x w + T + T and so the first property doesn t hold. This is enough to show that the transformation is not linear. It also doesn t hold for scalar multiplication: ( ( ( x cx ) cx T c T ) cy cy ( x ) ( ) x c x ct c y y c y Therefore T ( ( ( x x c ct and so the second property doesn t hold. ) Since neither property holds then T is not a linear transformation. (Note: you only need to show that one property fails so choose whichever one seems easiest to you.). Consider the matrices A 2 4 B C D Compute the following (where possible). If the operation is not defined explain why. (a) B 2 2B + I. 5

6 Solution. B is a 2 2 matrix. B 2 will also be a 2 2 matrix 2B will be a 2 2 matrix and I will be a 2 2 matrix therefore B 2 2B + I is defined and B B + I (b) A T C Solution. A is a 2 matrix and so A T will be a 2 matrix. C is also 2 matrix hence A T C is defined and T A T C ( ) (c) BD Solution. B is a 2 2 matrix D is a matrix. The number columns of B are not the same as the number rows of D hence BD is not defined. (d) (AC)D Solution. A is a 2 matrix C is a 2 matrix. AC is defined and will be a 6

7 matrix. D is a matrix and so (AC)D is defined. 5 2 (AC)D (e) CB 2A Solution. C is a 2 matrix B is a 2 2 matrix. The number of columns of C is not the same as the number of rows of B so CB is undefined. Therefore CB 2A is not defined. (f) B Solution. B (B ). B will be a 2 2 matrix and will be defined since det(b) and B (B ) will be a 2 2 matrix and ( ) (B )

8 (g) CC T Solution. C is a 2 matrix and C T is a 2 matrix. The number of columns of C match the number of rows of C T so CC T is defined and T cos θ sin θ 4. Find the inverse of. sin θ cos θ Solution. The inverse is well defined since cos θ sin θ sin θ cos θ cos2 θ + sin 2 θ and cos θ sin θ cos θ sin θ sin θ cos θ sin θ cos θ sin θ Let s check our solution: ( cos θ sin θ cos θ sin θ cos 2 θ + sin 2 θ sin θ cos θ sin θ cos θ sin θ cos θ + sin θ cos θ The other direction works as well Let G 4. 6 (a) Find det(g). ( cos θ sin θ cos θ Solution. Let s use cofactor expansion by crossing out the first column: ( ) ( 24) ( 2 + 2) ). ) sin θ cos θ + sin θ cos θ sin 2 θ + cos 2 θ 8

9 Thus det(g). (b) Does G exist? If so find it. Solution. No G does not exist since det(g). 6. Let J (a) Find det(j). Solution. J is a lower triangular matrix hence the determinant of J is the product of the diagonal entries: det(j) (b) Does J exist? If so find it. Proof. We set up the augmented matrix ( J I ) and reduce to get ( I J ) : R R R R R2 R R4 R R R2 R R R R R4 R R R4 R2 R4 R2 R / / /5 / / / 5 7 9

10 5 R R / / /5 /5 /7 /7 Therefore J / / /5 /5. /7 /7 Let s check our solution: / / /5 /5 5 / + / / /5 + /5 /5 + /5 5/5 /7 /7 5 7 /7 + /7 /7 + /7 5/7 + 5/7 The other direction works as well. 7. Let U {(x : x 2 y } be a subset of R 2. Is U a subspace of R 2? Why or why not? Solution. You should always sketch the subset whenever possible. U can easily be sketched out (see the figure below U is the dark shaded set). We need to verify look at the three properties of a subspace to see if U is a subspace: ( ( (i) The ero vector of R 2 is. U since 2 and. ) ) If you are looking at this graphically you can clearly see that the ero vector is in the set U. ( x w (ii) We need to pick two vectors in U and add them together. Let U. Then x( 2 ( w) 2 ( y ) and. When we add the two vectors together we have x w x + w +. The inequalities will also add together and we will have y + x + w 4 and( y + ) 2. Since these inequalities are not preserved (they aren t 2 x + w and ) then U hence U is not closed under addition. y + If you are looking at this graphically you need to find two vectors that when added ( together they are no longer in the set U. One such example would be the vectors ( ) ) 2 2 and which are both in U. But their sum is which is not in U. (These 2 vectors are plotted in the figure below.)

11 x (iii) We need to pick a vector in U and multiply it by a scalar. Let U and c R. ( y ) cx Then x 2 and y. If we multiply c by our vector we have. If c then our cy inequalities will be cx 2c and cy c; if c < then our inequalities will be cx 2c and cy c. Since these inequalities are not preserved then U is not closed under scalar multiplication. If you are looking at this graphically you need to find a vector and a scalar such that a scalar multiplied by this vector will no ( longer ) be ( in) the set U. One such example would be the vector and the scalar 5: 5 U. (These vectors are plotted in 5 the figure below.) Above we showed that U fails under addition and scalar multiplication therefore U is not a subspace. Showing just one of these fails is enough to show that U is not a subspace. 8. Let v ( 2 ) v 2 (2 9 ) v ( 4). Show that the set S {v v 2 v } is a basis for R. Solution. To show that S is a basis we need to show that (i) all the vectors in S are linearly independent and (ii) the vectors in S span the space which in this case is R. (i) Linearly Independent: We need to take an arbitrary linear combination and set it equal to the ero vector: c v + c 2 v 2 + c v.

12 We set up an augmented matrix with our vectors as the columns and reduce to solve for c c 2 c : R+R2 R2 2 5 R+R R R+R2 R2 2 2 R R 2R2+R R 2 2 This is enough for us to solve the system. This shows that c c 2 c. This proves that the set of vectors in S are linearly independent. (ii) Span: In the previous part we showed that the matrix formed by v v 2 v has a pivot in every row. This implies that these vectors span R. You can also use the fact that a set of n linearly independent vectors spans R n. 9. Let K (a) Find a basis for the column space of K. Solution. The column space of K is the spanning set of all the columns of K: col(k) span However we are asked to find the basis of col(k) which means we only want to find the linearly independent vectors in this set. We will do this by reducing K: R+R2 R R+R R 2 6 R+R R

13 R2+R R R+R2 R R+R R 4 5 4R2+R R 4 5 The columns with pivots are the st rd and 5th columns. This means that the st rd and 5th columns of K will form the basis of the column space of K: 4 5 col(k) span (b) Find a basis for the null space of K. Solution. We need to find all the vectors x such that Kx. We do this by reducing the augmented matrix ( K ). We already reduced the matrix in part (a): The columns with pivots are associated with x x x 5. The free variables are x 2 x 4 x 6. We set x 2 s x 2 t x u for s t u R. We now solve for x x x 5 in terms of the free variables. From the third row we have: From the second row we have: From the first row we have: x 5 + 5x 6 x 5 5u x + x 4 + 4x 6 x t 4u x x 2 4x 4 7x 6 x s + 4t + 7u We write our solution out in parametric form: x s + 4t + 7u 4 7 x 2 s x x 4 t 4u t s + t + 4 u x 5 5u 5 x 6 u

14 This is the spanning set for null space of K: null(k) span