# (a) If A is a 3 by 4 matrix, what does this tell us about its nullspace? Solution: dim N(A) 1, since rank(a) 3. Ax =

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1 . (5 points) (a) If A is a 3 by 4 matrix, what does this tell us about its nullspace? dim N(A), since rank(a) 3. (b) If we also know that Ax = has no solution, what do we know about the rank of A? C(A) does not span the entire R 3, so rank(a) 2. (c) If Ax = b and A T y = 0, find y T b by using those equations. This says that the space of A and the are. y T b = y T (Ax) = (A T y)x = 0. This says that the column space of A and the null space of A T are orthogonal.

2 2. (5 points) Suppose Ax = b reduces by the usual row operations to Ux = c : Ux = x x 2 x 3 x 4 = b b 2 b b 3 2b 2 + b = c. (a) Give a basis for the nullspace of A (that matrix is not shown) and a basis for the row space of A. N(A) = N(U), R(A) = R(U). Therefore we can read the bases directly from U: 3 2 N(A) = span 0, R(A) = span 6 4, (b) When does Ax = b have a solution? Give a basis for the column space of A. b C(A), equivalent to c C(U). 0 0 By looking at c as a function of b we can reconstruct A. Let E = 0. We have U = EA, c = Eb. Hence A = E U = = From here we see that C(A) = span, 2. 3 (c) Give a basis for the nullspace of A T. N(A T ) C(A). So N(A T ) = span 2 (we only need to find a vector orthogonal to both basis vectors we gave for C(A)). 2

3 3. (0 points) (a) Suppose q, q 2 are orthonormal in R 4, and v is NOT a combination of q and q 2. Find a vector q 3 by Gram-Schmidt, so that q, q 2, q 3 is an orthonormal basis for the space spanned by q, q 2, v. q 3 = v (vt q )q (v T q 2 )q 2 v (v T q )q (v T q 2 )q 2 (b) If p is the projection of b onto the subspace spanned by q and q 2 and v, find p as a combination of q, q 2, q 3. (You are solving the least squares problem Ax = b with A = [q, q 2, q 3 ].) p = A(A T A) A T b = AA T b = q (q T b) + q 2 (q T 2 b) + q 3 (q T 3 b) where we used the fact that the columns of A are orthonormal for A T A = I. (It is easy to see the final result just by thinking that we are actually projecting onto an orthonormal basis). 3

4 4. (0 points) (a) To solve a square system Ax = b when det A 0, Cramer s Rule says that the first component of x is x = det B det A with B = [b a 2... a n ]. So b goes into the first column of A, replacing a. If b = a, this formula gives the right answer x = det A det A =.. If b = a different column a j, show that this formula gives the right answer, x =. 2. If b is any combination x a + + x n a n, why does this formula give the right answer x?. x = 0, since det(b) = 0. Let us check that it gives the correct answer: the solution to Ax = a j is x = e j (it is unique, since det(a) 0, so A s columns are linearly independent). Hence x = For the same reason as above the first component of x is indeed x, the coefficient of a. Now let us see what Cramer s Rule gives. In this case det(b) = det([b a 2... a n ]) = det(x a a 2... a n ) = x det(a). So, indeed, x = det(b)/ det(a). (b) Find the determinant of A = Cofactor expansion by the first row: C = det = = So det(a) = ( 8) = C 2 = det = ( ) =

5 5. (5 points) (a) Suppose an n by n matrix A has n independent eigenvectors x,..., x n with eigenvalues λ,..., λ n. What matrix equation would you solve for c,..., c n to write the vector u 0 as a combination u 0 = c x + + c n x n? c c n [x x 2... x n ]. = u 0 (b) Suppose a sequence of vectors u 0, u, u 2,... starts from u 0 and satisfies u k+ = Au k. Find the vector u k as a combination of x,..., x n. Let u 0 = c x + + c n x n. Then u k = A k u 0 = n i= (λk i c i )x i. (c) State the exact requirement on the eigenvalues λ so that A k u 0 0 as k for every vector u 0. Prove that your condition must hold. λ i <, for all i. Clearly, if this holds, all the coefficients of x i in A k u 0 go to 0 as k. For the converse, we require that the coefficient λ k i c i of x i in A k u 0 to go to 0, for any choice of u 0. Equivalently, we need λ k i c i 0 for any c i. Hence λ i <. 5

6 6. (0 points) (a) Find the eigenvalues of this matrix A (the numbers in each column add to zero). A = The number in each column add to zero, hence N(A T ), so dim N(A T ) > 0, and thus rank(a) = rank(a T ) < 3. So λ = 0. We can easily spot λ 2 = as another eigenvalue, since subtracting A + I has two equal columns, and hence det(a + I) = 0. Looking at the trace we get that λ 3 = tr(a) λ λ 2 = 2. (b) If you solve du dt () u(t) goes to zero?. = Au, is () or (2) or (3) true as t? (2) u(t) approaches a multiple of (what vector?) (3) u(t) blows up? Approaches a multiple of 2. Observe that 2 is an eigenvector for the 0 eigenvalue, and that the only nonzero eigenvalue of e At as t is e 0 t =, with the same eigenvector. Also, u(t) = e At u(0), and that the only nonzero eigenvalue of e At (as t ) is, with the same eigenvector. So lim t u(t) is a projection of u(0) on the line spanned by 2, hence a multiple of it. 6

7 7. (0 points) Every invertible matrix A equals an orthogonal matrix Q times a positive definite matrix S. This famous fact comes directly from the SVD for the square matrix A = UΣV T, by choosing Q = UV T. (a) How can you prove that Q = UV T is orthogonal? Q is orthogonal if and only if Q T Q = I. Notice that since A is invertible, U and V are both square matrices of full rank. Q T Q = (UV T ) T (UV T ) = V U T UV T = V (U T U)V T = V V T = I We used the fact that U is orthogonal, hence U T U = I, and that V T is orthogonal because V s columns are eigenvectors of a symmetric matrix A T A, so V T = V. (b) Substitute Q and A to write S = Q A in terms of U, V and Σ. How can you tell that this matrix S is symmetric positive definite? S = Q A = (UV T ) A = (V T ) U UΣV T = V ΣV T = (Σ /2 V T ) T (Σ /2 V T ) 7

8 8. (5 points) A 4-node graph has all six possible edges. Its incidence matrix A and its Laplacian matrix A T A are A = A T A = (a) Describe the nullspace of A. N(A) = span (Recall that applying A to a vector of potentials gives the potential drops along edges, so in order for a vector of potentials to be in the null space, all the potentials within one connected component must be the same.) (b) The all-ones matrix B = ones (4) has what eigenvalues? Then what are the eigenvalues of A T A = 4I B? B = T = 4( /2)( /2) T, where is the all-ones vector in R 4. So B has eigenvalues 4, 0, 0, 0. I and B diagonalize in the same eigenbasis, so λ i (4I B) = λ i (4I) λ i (B) = 4λ i (I) λ i (B) for all i. So the eigenvalues of A T A are 0, 4, 4, 4. (c) For the Singular Value Decomposition A = UΣV T, can you find the nonzero entries in the diagonal matrix Σ and one column of the orthogonal matrix V? σ i = λ i (A T A), so the nonzero singular values are 2, 2, 2. We only need to find one eigenvector of A T A. An obvious one is /2, since all the rows sum up to 0. 8

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