# [3] (b) Find a reduced row-echelon matrix row-equivalent to ,1 2 2

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1 MATH Key for sample nal exam, August 998 []. (a) Dene the term \reduced row-echelon matrix". A matrix is reduced row-echelon if the following conditions are satised. every zero row lies below every nonzero row. all pivots are equal to. each pivot lies to the left of all pivots of rows below it. every other entry in the column of a pivot is equal to 0. [] (b) Find a reduced row-echelon matrix row-equivalent to,,,, 5 Applying the row operations R! R, R, R! R +R,we get 0, 0 Applying the row operations R! R, R, R! R, R,we get 0, 0, Finally, applying R \$ R, R! (,)R,we get

2 MATH Page of. Let A, 0, ; x x x x x x 5 5 ; and b 0 [] (a) Express the general solution of the homogeneous system Ax 0 as a linear combination of vectors in R 5. By performing on [Ajb] the row operations R! R + R, R! R +R, R! R,R, and R! ()R we get [Hjc] From this we see that the general solution of the homogeneous system is 5: x r [,;,; ; 0; 0] + r [0;,; 0;,; ] (r ;r R): [] (b) Write down the general solution of the nonhomogeneous system Ax b. From the reduced augmented matrix, we see that x [0;;0;;0] + r [,;,; ; 0; 0] + r [0;,; 0;,; ] (r ;r R):

3 MATH Page of. The matrix A in R nn is dened to be invertible if there exists B in R nn such that Let A; C R nn both be invertible. AB BA I: [] (a) Show that AC is invertible. Since A and C are invertible there exist B;D R nn such that AB BA I and CD DC I: Using the associativity of matrix multiplication, we have (AC)(DB) A(CD)B AIB AB I (DB)(AC) D(BA)C DIC DC I: Thus DB witnesses that AC is invertible. [] (b) Show that A T is invertible. Let B witness that A is invertible, i.e., AB BA I: Taking the transpose of each term we have B T A T A T B T I: Therefore B T witnesses that A T is invertible.

4 MATH Page of []. (a) Explain how to compute the rank of a matrix. By performing row operations on the given matrix A nd a reduced row-echelon matrix H row-equivalent to A. The rank of A is the number of nonzero rows of H. [] (b) Explain why it is true that rank(ab) min(rank(a); rank(b)) for all matrices A, B such that AB is dened. The crux of the matter is that the columns of AB are linear combinations of the columns of A indeed, if a ;::: ;a n are the columns of A and [b j ;::: ;b nj ] is the j-th column of B, then the j-th column of AB is b j a + :::+b nj a n. Therefore colspace(ab) colspace(a) and so rank(ab) dim(colspace(ab)) dim(colspace(a)) rank(a) : Applying this result to the product B T A T we obtain This is enough. rank(ab) rank, (AB) T rank(b T A T ) rank(b T )rank(b) :

5 MATH Page 5 of [] 5. Find a formula which denes a linear transformation F : R! R which satises F ([; ; ]) [0; ; ]; F ([;,;,])[;0;0] : The requirement above species F on a subspace W of dimension. To complete the specication of F on R we require that F ([0; ; 0]) [0; 0; 0]. Since F is linear, F ([; 0; 0]) F (()[[; ; ] + ()[[;,;,]) () ([0; ; ]+[;0;0]) F ([0; 0; ]) F ([; ; ], [; 0; 0], [0; ; 0])[0;;], [ ; ; ], [0; 0; 0] : Thus the standard matrix representation of F is Therefore a formula dening F is [F (e ) F (e ) F (e )] 0, 0 0 F ([x; y; z]) () [x, z; x + z; x + z] :

6 MATH Page of Denition of vector space A vector space over R is a set V of vectors together with a distinguished vector 0 in V and three functions which satisfy (u; v)! u + v; v!,v; (r; v)! rv (u; v V; r R) A (u + v) + w u +(v + w) associative law A u + v v + u commutative law A 0 + u u additive identity A u +(,u)0 additive inverse S r(u + v) ru + rv distributivity S (r + s)u ru + su distributivity S r(su) (rs)u associative law S u u scale preservation for all u; v; w V and r; sr. []. From the axioms for a vector space show that for all vectors u, v, w in V, u + v u + w ) v w : Suppose that u + v u + w. Then This is enough. v 0 + v by A (u+(,u)) + v by A (u+ v)+(,u) by A, A (u+ w)+(,u) hypothesis (u+(,u)) + w by A, A 0 + w by A w A :

7 MATH Page of. Let V R denote the vector space over R whose vectors are the matrices with entries from R. Let v 0 0 ; v 0 0 ; v 0 0 ; v 0 0 : [] (a) Show that fv ; v ; v ; v g is linearly dependent. By inspection, v + v v + v : [] (b) Find u V such that fv ; v ; v ; ug is a basis for V. 0 We can take u. Then u, v 0 0, u, v, u, v, v + u are the four matrices with one entry equal to and all the other entries equal to 0. Thus fv ; v ; v ; ug is a basis for V.

8 MATH Page 8 of [] 8. (a) Evaluate the determinant Performing the row operations R 5! R 5, R and R! R, R yields a matrix in which rows and 5 are the same. These row operations do not change the value of the determinant. So the given determinant is 0. [] (b) Let A be a square matrix. State the relationship between det(a) and rank(a). Let A be in R nn. Then det(a) 0 () rank(a) n:

9 MATH Page 9 of 9. Let A denote the matrix [] (a) Find the eigenvalues of A. The characteristic polynomial is, 0 0, 0, 0,, 0,,,,, 0, 0 0, 0, 0 0,, 0,,, (, ),,, 0,, 0, 0, 0 0, 0,, 0,,,,, (, ), 0, 0,,,,,, 0, 0,, 0, 0, 0 0, 0,, 0,,,,,,,(, ), 0,,, 0, (, ) (, ), : Therefore the eigenvalues are 0;; p. [] (b) Find a matrix C such that C, AC is a diagonal. By inspection eigenvectors belonging to 0, are [; ;,;,], [;,; ; 0] respectively. We compute an eigenvector belonging to + p by nullspace p ( + )I, A 0 p 0 0,,,, p p nullspace B 0 p p C, 0 +, 5A sp([; ; ; ]) : Hence [; ; ; p ] is an eigenvector belonging to + p, and similarly [; ; ;, p ] is an eigenvector belonging to, p. Therefore we may take C,,, 0 p, p 5

10 MATH Page 0 of [] 0. (a) Find the projection of [;,; ] on sp([; ;,]). p [;,; ] [; ;,] [; ;,](,)[; ;,] [; ;,] : [; ;,] [; ;,] [] (b) Find a formula for the projection of b [b ;b ;b ] on the subspace sp([; ;,]; [,; ; ])., Let A denote the matrix, W sp([; ;,]; [,; ; ]) is T. The projection matrix for the subspace P A(A T A), A T A A 8,, So the projection of b on W is P b,, b, b, A T A 8 0, 0 8 0, 0 ;b ; b,b. 8 A T

11 MATH Page of []. (a) State three conditions on a matrix A R nn which are equivalent to Abeing an orthogonal matrix. Any three of the following ve conditions are acceptable:. The columns of A form an orthonormal basis of R n.. The rows of A form an orthonormal basis of R n.. A T A I.. kaxk kxk for all x R n. 5. (Ax) (Ay) x yfor all x; y R n. [] (b) Let T : R! R be an orthogonal linear transformation such that T ([; 0]) [; p ]: Explain why there are only two possibilities for T and describe them. The standard matrix representation of T is an orthogonal matrix and has columns T ([; 0]) and T ([0; ]). From condition we see that T ([; 0]) and T ([0; ]) form an orthonormal basis of R. Since T ([; 0]) [; p ], we see that, if T ([0; ])[a ;a ], then a + a and ()a +( p )a 0. Therefore T ([0; ]) is either [ p ;,] or [, p ; ].

12 MATH Page of [8]. The following data points are given: (,;,8); (,;,8); (; 0); (; 0); (; ); (; 8) By using a method from linear algebra nd the least-squares linear t for these data points. Your answer should make it clear what method you are using. Let A,, (A T A), A T 0 T and b,8, T. Then, 0,, 9 8,,,8 0 According to the theory, the coecients of the least-squares linear t are given by r0 r (A T A), A T b So the least-squares linear t is y5+x A T :

13 MATH Page of. Let R R denote the vector space over R consisting of all functions f : R! R. Let V denote the subspace of R R. spanned by f; sin x; cos xg. Let B h; sin x; cos xi; B 0 hsin x; cos x; sin x cos xi: Let F : V! V be the unique linear transformation which maps B to B 0 in the sense that F () sin x, F (sin x) cos x, and F (cos x) sin x cos x. [] (a) Find a matrix C R such that, for all v in V, Cv B v B 0 Note that sin x + cos x sin x sin x cos x cos x, sin x + cos x: Thus the required matrix C is C B;B 0 0, [] (b) Find the matrix [F ] B;B which represents F with respect to B; B. Notice that the matrix of F with respect to B; B 0 is I the identity matrix. Therefore the matrix of F with respect to B; B is [F ] B;B C B 0 ;B [F ] B;B 0 (C B;B 0),,,, 0 0 0,, 0

14 MATH Page of [5]. Consider the curve inr whose equation is x + p xy +y : Show that this curve is an ellipse and nd the length of its major and minor axes. The symmetric matrix of coecients of the quadratic form x + p xy +y is A p p : The characteristic polynomial is (,9)(,). Letting C be the orthogonal diagonalizing x x matrix, the change of variables C converts the equation to y y 9x +y : This is clearly the equation of an ellipse. The major axis has length, while the minor axis has length. [] 5. Consider the surface S in R whose equation is x + y + z +xy +yz +zx, x + z : Show that S is cylindrical in the sense that there is a unit vector u such that S is invariant under translation by any scalar multiple of u. Here the symmetric matrix of coecients has eigenvalues, 0, where 0 has multiplicity. So the associated orthogonal transformation converts the equation to the form x, p x + ay + bz : The exact values of a and b do not matter except that one of them is nonzero. Making a rotation about the x-axis we can convert the equation to the form x, p x + cy : This is enough. The unit vector u is the one in the direction of the z-axis.

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