33A Linear Algebra and Applications: Practice Final Exam - Solutions

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1 33A Linear Algebra and Applications: Practice Final Eam - Solutions Question Consider a plane V in R 3 with a basis given by v = and v =. Suppose, y are both in V. (a) [3 points] If [ ] B =, find. (b) [3 points] If y =, find [ y] B. (c) [4 points] Now, consider the basis C for R 3 given by v =, v =, v 3 =. Give [ ] C and [ y] C, and eplain why this basis is useful if our task is to project vectors z R 3 into V. Answer 3 (a) If [ ] B =, this implies that = v + v = 3. (b) If y = = a v + b v, then this implies [by solving the set of linear equations], that a = and b =. Therefore, [ y] B =. (c) We already know that, y can be written solely in terms of the first two vectors, so [ ] C = and [ y] C =. In this basis, v 3 is orthogonal to V, and is therefore a basis for V (which a is -dimensional). This means that if we write any vector as [ z] C = b, this gives us eactly c the decomposition of z in terms of z = a v + b v and z = c v 3. Then, it is easy to find a [pro j V z] B = b. 33A/ Linear Algebra - Puck Rombach Final Eam

2 Question a b A positive transition matri is a matri A = such that a, b, c, d >, and a+c = b+d =. c d b (a) [4 points] Verify that both and are eigenvectors of A, and find the corresponding c eigenvalues. b + (b) [4 points] If we consider the discrete dynamical system (t) = A (t ), where () =, c and where A is a positive transition matri as given above, write a direct epression for (t), in which b and c are the only unknowns. v (c) [ points] Eplain why there is eactly one vector v = such that v + v = and A v = v. Answer (a) We check for v = and b and v c = that in both cases A v i = λ i v i. a b b A v = = c d c a b A v = = c d So, we have λ = and λ = a b. (b) We have a diagonalizable A, so we can write v b, c a b = (a b). a + b (t) = c λ t v + c λ t v, where () = c v + c v. It is easy to see that c = c =. So, (t) = b + (a b) t = c (Because b, c can be the only unknowns.) b + ( c b) t. c (c) This vector lies on the line v = b c v and the line v + v =. These straight lines intersect in eactly one point, which is our desired vector (the equilibrium distribution vector). 33A/ Linear Algebra - Puck Rombach Final Eam

3 3 Question 3 (a) [4 points] Without justification, cross out the items in the list which do not apply: For every square matri A, if we know the characteristic polynomial of a A, then we can infer the of A. eigenvalues eigenvectors eigenspaces trace determinant algebraic multiplicities geometric multiplicities invertibility rank (b) [ points] Find all the eigenvalues and eigenspaces of (a) See above. (b) [Eplicit calculation needed.] A = λ =, λ = 4, λ 3 = 5, E = span, E 4 = span, E 5 = span. 33A/ Linear Algebra - Puck Rombach Final Eam

4 4 Question 4 (a) [ points] Is it possible for any n n matri A to have im(a) = ker(a)? Hint: for eample, can you find a matri such that im(a) = ker(a) = span? (b) [3 points] Give an eample of a 3 3 matri A that has the plane V R 3 with equation = as its kernel. (c) [3 points] Give an eample of a 3 3 matri B that has the plane V R 3 from (b) as its image. (d) [ points] Without any calculation, using words, eplain what the transformation AB does (with A and B as defined in (b) and (c)). (a) That is possible. If we want ker(a) = span then we need a matri with rref equal to. And this already has the right image, too! Indeed, if A =, then ker(a) = span. (b) This works if we make all rows of A the coefficients of the equation, or scalar multiples thereof. Then, by definition V contains all vectors v such that A v =. Eample: A =. (c) The image is the span of the columns. V is -dimensional, and has as a basis, for eample and. So we can set B =. (d) For any R 3, we have B V = ker(a). So, for any R 3, we have AB =. Therefore, AB =. 33A/ Linear Algebra - Puck Rombach Final Eam

5 5 Question 5 (a) [3 points] Interpret the following transformations geometrically: T ( ) = A =, T ( ) = A = ( ) (b) [7 points] Consider the basis B =,. Find the B-matrices for both of the transformations, and interpret them geometrically. (a) T is a rotation over π/, and T is a horizontal shear with parameter. (b) Let S = = S. This change of basis mirrors R in the y-ais. Geometrically, this will result in a rotation over π/ and a horizontal shear with parameter -, respectively. We check that indeed B = S A S =, B = S A S =. 33A/ Linear Algebra - Puck Rombach Final Eam

6 Question Consider the following data points: (, 5), (, 3), (3, 4). We would like to fit a function f (t) = + t to this data using least squares. (a) [4 points] Write down the inconsistent system A = b that we wish to solve, by giving the eact values for A and b. (b) [3 points] Write the least-squares solution in terms of A and b. How do you know that your epression is valid? (i.e., will not give an error message if you input it into a computer) 4.3 (c) [3 points] In general, define the error of a least-squares solution. Given that A = 4., find 3. the error of this solution. (Your answer may look like 4.) (a) The inconsistent system comes from a series of equations that look like 5 = +, 3 = +,... and looks like 5 = (b) We have = (A T A) A T b. This is only a valid epression if A T A is indeed invertible. We know that ker(a T A) = ker(a), so, we only need to check that ker(a) = [which is easy from the rref of A...] (c) The error of the least-square solution is the magnitude of the difference between A and b. This describes how close our solution is to the vector the inconsistent system was looking for. So, we have error = A b = = = (.7) +. + (.4)..4 33A/ Linear Algebra - Puck Rombach Final Eam

7 7 Question 7 Let V R 3 be the span of and. (a) [ points] Give an orthonormal basis for V. (b) [3 points] Give the matri A of the projection onto V in R 3, and find the projection of 9 e 3. (c) [ points] Find a (not necessarily orthonormal) basis for V. (d) [3 points] The matrices of both the projection and reflection in V can be written as a product S BS, where rank(s ) = 3 and B is a diagonal matri. Give a matri S and a matri B in both cases. (a) [Use G-S to find:] û = 3, û =. (b) We use the orthonormal basis to define / 3 / Q = / 3 / / 3 /. Since the columns of Q are an orthonormal basis for V, we have / 3 / A = QQ T = / 3 / ( ) / 3 / 3 / 3 / 3 / / / / = / /, / / where A is the matri of the projection onto V in R 3. Then is the projection of 9 e 3 into V. A9 e 3 = 9/ 9/ (c) We have V = (im(q)) = ker(q T ) = ker = ker = span 33A/ Linear Algebra - Puck Rombach Final Eam

8 8 (d) In order to diagonalize the projection/reflection, we need S to be the matri of eigenvectors, and B a diagonal matri with eigenvalues on the diagonal (in the correct order with respect to S ). We have that for the projection, the basis for V are the eigenvectors with eigenvalue, and the basis for V are the eigenvectors with eigenvalue. For the reflection, the basis for V are the eigenvectors with eigenvalue, and the basis for V are the eigenvectors with eigenvalue -. So, we have that in both cases, we can let S =. We then have and B pro j = B re f l =. 33A/ Linear Algebra - Puck Rombach Final Eam

9 9 Question 8 (a) [4 points] Suppose we have nonzero deviation vectors, y of two characteristics, such that = c y. Show carefully what this implies about the correlation coefficient r between the two characteristics. (b) [ points] Find the correlation coefficient between the daily profit and number of paintings Shop Profit (in s) Paintings A - inside three CA coffee shops. B - C 3 (a) We have that r = y y = c( y y) c y y = c c =, if c >,, if c <. (b) [ points] We find deviation vectors = and y =. This gives r = y y = 3 = 3. 33A/ Linear Algebra - Puck Rombach Final Eam

10 Question 9 State whether the following statements are TRUE or FALSE, by putting a cross under T or F in the table below. You do not need to justify your answer. (a) [ points] It is possible for a system of linear equations A = y to have eactly solutions for. (b) [ points] If the matri AB is invertible, then ker(b) = { }. (c) [ points] The rank of a matri A is the largest number of linearly independent rows it has. (d) [ points] If the vectors,, 3 form an independent set, then the vectors, 3 form an independent set. (e) [ points] Vectors of the form, where a R form a subspace of R3. a (f) [ points] Every nonzero subspace of R n has an orthonormal basis. (g) [ points] If A is similar to B, then there eists an invertible matri S such that S AS = B. (h) [ points] There eists a 5 5 matri A such that ker(a) = im(a). (i) [ points] If an n n matri A has n distinct eigenvalues, then A is diagonalizable. (j) [ points] If A is diagonalizable and A B, then B must be diagonalizable. a b c d e f g h i j T F 33A/ Linear Algebra - Puck Rombach Final Eam