Math 20F Practice Final Solutions. Jor-el Briones

Math 2F Practice Final Solutions Jor-el Briones

Jor-el Briones / Math 2F Practice Problems for Final Page 2 of 6 NOTE: For the solutions to these problems, I skip all the row reduction calculations. Please DO NOT DO THAT ON THE EXAM. It helps to check things. Column Space, Null Space, Dimension Let A = 2 a) Give a basis for col(a). What is rank(a)? b) Give a basis for nul(a). What is the dimension of nul(a)? Solution: The key takeaway from this problem is to know how to find the column space and nullspace of a matrix. And like 8-9 2 a) Give a basis for col(a). What is rank(a)? You ll notice that I boxed the pivots. Why? BECAUSE PIVOT COLUMNS CORRE- SPOND TO LINEARLY INDEPENDENT COLUMNS IN THE ORIGINAL MATRIX (row operations preserve column depndence and independence)! Since ALL of the columns would span the column space, but eliminating any columns that would correspond to the non-pivot columns (which are inear combinations of pivot columns) would still allow the resulting set to span, we can form a basis. KEY PLAY: Pivot columns form a basis for the column space. IMPORTANT: Pivot columns OF THE ORIGINAL MATRIX are the ones that form a basis, NOT the pivot columns of the row reduced echelon form!! So we have our basis for col(a) is the pivot columns:, Since there are two vectors in this basis, rank(a)= dimcol(a) = 2

Jor-el Briones / Math 2F Practice Problems for Final Page of 6 b) Give a basis for nul(a). What is the dimension of nul(a)? Next, we solve the homogeneous system Ax =. It s pretty easy, since the row reduced augmented matrix is going to be similar to the row reduced coefficient matrix, but with a column of entries: 2 KEY PLAY: Solving the homogeneous system gives you a basis for the null space vectors, where each vector attached to a free variable is a vector in the basis. IMPORTANT: The basis of the nullspace is formed by ACTUALLY SOLVING THE HOMOGENEOUS SYSTEM. It is NOT made up of the non-pivot columns of the original or row reduced matrix!! So solving the homogeneous system, we get that x = t (afree veariable), x 2 = x = t, t and x = x = t. So the solution to Ax = is x = t = t t That means the basis of nul(a) is: (notice how the letter t is nowhere in this answer) And since there is only one vector in the nullspace, dimnul(a) =

Jor-el Briones / Math 2F Practice Problems for Final Page 4 of 6 2 Inverse Theorems Suppose A is an n n matrix and rref(a) = I. a) Show that Ax = has only the trivial solution. b) Show that Ax = b has a solution for all b in R n. Solution: a) Show that Ax = has only the trivial solution. Proof. rref(a) = I which means A has a pivot in every column of A, and every column of A is a pivot column. This means that, since pivot columns form a linearly independent set, the columns of A are linearly independent. But since the columns of A are linearly independent, Ax = has ONLY the trivial solution. Key parts of proof:. Know what I is, what rref is, and the implications of what that means 2. Know that rrefa is row equivalent to A and hence, the independence of their columns is the same.. Know that pivot columns form a linearly independent set. 4. Know the definition of linear independence/dependence, particularly regarding the solution to Ax = b) Show that Ax = b has a solution for all b in R n. Proof. rref(a) = I which means A has a pivot in every row of A. This means that the system corresponding to [A b is ALWAYS consistent for any choice of b in R n, and in particular, Ax = b has a solution for all b in R n. Key parts of proof:. Know what I is, what rref is, and the implications of what that means. 2. Know what it means for a COEFFICIENT MATRIX to have a pivot in every row.. Know chapter, theorem 4, and don t misuse it.

Jor-el Briones / Math 2F Practice Problems for Final Page 5 of 6 That proof that s on the exam Let a, a 2,..., a k be k vectors that form a basis for a subspace H of R n. Let b, b 2,..., b r be r vectors that also form a basis for H. Let A be the n k matrix whose columns are a, a 2,..., a k. Let B be the n r matrix whose columns are b, b 2,..., b r. a) Show that there is a k r matrix C such that AC = B. Also, show that there is a r k matrix D such that BD = A b) Show that for any matrices X, Y, Z, where XY = Z, if the columns of Z are linearly independent, then the columns of Y are independent. c) Show that k = r, where k and r are the numbers of vectors in the bases previouly defined above (hint: Use parts a) and b)) Solution: a) Show that there is a k r matrix C such that AC = B. Also, show that there is a r k matrix D such that BD = A KEY PLAY: Know how matrix multiplication works and how to relate that to linear systems, and Ax = b Proof. Let C be some k r matrix. We want C to be a matrix such that AC = B. Now, we know that: AC = [ Ac Ac 2... Ac r where c, c 2,..., c r are the columns of C. We want: AC = [ Ac Ac 2... Ac r = [ b b 2... b r = B where b, b 2,..., b r are the columns of B. This can ONLY happen if Ax = b, Ax = b 2,..., Ax = b r all have solutions. But the columns of A form a basis for H, so that every vector in H is a linear combination of the columns of A, and b, b 2,..., b r are all vectors in H. So Ax = b, Ax = b 2,..., Ax = b r are all consistent. Let c be a solution to Ax = b, c 2 be a solution to Ax = b 2, and repeat all the way to c r, which we set as a solution to Ax = b r. In doing so, we achieve: And the desired matrix C exists. AC = [ Ac Ac 2... Ac r = [ b b 2... b r = B The proof for BD = A is similar:

Jor-el Briones / Math 2F Practice Problems for Final Page 6 of 6 Let D be some r k matrix. We want D to be a matrix such that BD = A. Now, we know that: BD = [ Bd Bd 2... Bd k where d, d 2,..., d k are the columns of D. We want: BD = [ Bd Bd 2... Bd k = [ a a 2... a k = A where a, a 2,..., a k are the columns of A. This can ONLY happen if Bx = a, Bx = a 2,..., Bx = a k all have solutions. But the columns of B form a basis for H, so that every vector in H is a linear combination of the columns of B, and a, a 2,..., a k are all vectors in H. So Bx = a, Bx = a 2,..., Bx = a k are all consistent. Let d be a solution to Bx = a, d 2 be a solution to Bx = a 2, and repeat all the way to d k, which we set as a solution to Bx = a k. In doing so, we achieve: BD = [ Bd Bd 2... Bd k = [ a a 2... a k = A And the desired matrix D exists. Key parts of proof:. KNOW HOW MATRIX MULTIPLICATION WORKS (particularly, the way Professor Harel went over in lecture) 2. KNOW THAT THE VECTORS THAT FORM A BASIS OF A SUBSPACE SPAN THAT SUBSPACE. Know how to tell if a system is consistent 4. Know that the corresponding columns of matrices that are equal must be equal b) Show that for any matrices X, Y, Z, where XY = Z, if the columns of Z are linearly independent, then the columns of Y are independent. KEY PLAY: Proving things by contradiction/ contrapositive Proof. Suppose that the columns of Y are linearly DEPENDENT (we still go with our assumption that the columns of Z are linearly independent). That means Y x = has some nontrivial solution, call it v. Next, we use matrix multiplication: Y v = = X(Y v) = X() = (XY )v = = Zv = But that means v is ALSO a nontrivial solution to the system Zx =, which means the coluns of Z are ALSO linearly dependent. Which is impossible. Contradiction.

Jor-el Briones / Math 2F Practice Problems for Final Page 7 of 6 Key parts of proof:. Proving things by contradiction: Assume that the conclusion (what you re trying to prove) is false, while your assumptions (what you are given) is still true. Show that this can t possibly happen. 2. Know how to manipulate matrices, and that you can MULTIPLY matrices on the left (or on the right) to both sides. of an equation.. KNOW WHAT IT MEANS TO BE LINEARLY DEPENDENT, IN PARTICULAR, IN RELATION TO THE HOMOGENEOUS SYSTEM, Ax = c) Show that k = r, where k and r are the numbers of vectors in the bases previouly defined above (hint: Use parts a) and b)) Proof. From part a), we know that there is some k r matrix C so that AC = B. But the columns of B form a basis for H, so its columns are linearly independent. But by part b), this means that the columns of C are linearly independent. But this means that C has no non-pivot columns, and every column is a pivot column. But this means there s a pivot in every column of C. Since there is AT MOST ONE pivot in each column and AT MOST ONE pivot in each row, there must be at least as many rows of C as there are columns of C. In other words, k r. Next, we do the same for BD = A: From part a), we know that there is some r k matrix D so that BD = A. But the columns of A form a basis for H, so its columns are linearly independent. But by part b), this means that the columns of D are linearly independent. But this means that D has no non-pivot columns, and every column is a pivot column. But this means there s a pivot in every column of D. Since there is AT MOST ONE pivot in each column and AT MOST ONE pivot in each row, there must be at least as many rows of D as there are columns of D. In other words, r k. Since k r AND r k, r = k Key parts of proof:. KNOW THAT THE VECTORS THAT FORM A BASIS OF A SUBSPACE ARE LINEARLY INDEPENDENT 2. Know that if the columns of a matrix are linearly independent, the columns must all be pivot columns, and there must be no non-pivot columns (because they are always linear combinations of pivot columns).. Know that there is AT MOST ONE pivot in each row and AT MOST ONE pivot in each column. 4. Know that you re allowed to use previously proved facts to prove new facts.

Jor-el Briones / Math 2F Practice Problems for Final Page 8 of 6 4 Inverses and Diagonalization Let A be an n n matrix with n different eigenvalues. a) Prove that A is diagonalizable (Hint: You may use the fact that each eigenvalue has at least one corresponding eigenvector) b) Prove that if none of the eigenvalues of A are, then A is invertible. Solution: a) Prove that A is diagonalizable (Hint: You may use the fact that each eigenvalue has at least one corresponding eigenvector) Proof. We have n different eigenvalues, and each of those eigenvalues has at least one corresponding eigenvector. So we have n different eigenvectors, each one corresponding to a different eigenvalue. But since all the eigenvectors correspond to different eigenvalues, they form a linearly indpendent set. So we have n linearly independent eigenvectors in R n, so they must span R n and form a basis for R n. We have a basis of eigenvectors, so the matrix A is diagonalizable. Key parts of proof:. Every distinct eigenvalue has at least one corresponding eigenvector. 2. Eigenvectors that correspond to different eigenvalues are linearly independent.. n linearly independent vectors in R n spans R n and therefore forms a basis for R n 4. An n n matrix with n different eigenvalues has a basis of eigenvectors. 5. A matrix is diagonalizable if and only if it has a basis of eigenvectors. If there isn t a basis of eigenvectors, then the matrix is not diagonalizable. b) Prove that if none of the eigenvalues of A are, then A is invertible. Proof. From part a), we know that A is diagonalizable. This means that there is some invertible matrix P and some diagonal matrix of eigenvalues Λ so that A = P ΛP. Since the matrices are equal, they must have the same determinant. Hence, by properties of the determinant: det(a) = det(p ΛP ) = det(p ) det(λ) det(p ) = det(λ) det(p ) det(p ) = det(λ) det(p P ) = det(λ) det(i) = det(λ) So A and Λ have the same determinant. But since Λ is a diagonal matrix, its determinant is just the product of entries on the diagonal. But since the entries on the diagonal of Λ are eigenvalues, none of which are, the determinant of Λ is non-zero. So det(a) = det(λ). And since A has a non-zero determinant, A is invertible.

Jor-el Briones / Math 2F Practice Problems for Final Page 9 of 6 Key parts of proof:. KNOW THAT THE DETERMINANT BEING NON-ZERO IMPLIES MATRIX IN- VERTIBILITY 2. Know how to manipulate determinants, and that det(ab) = det(a) det(b). Know that the determinant of a diagonal matrix (or triangular matrix) is the just the product of entries on the diagonal. 4. Know that you re allowed to use previously proved facts to prove new facts.

Jor-el Briones / Math 2F Practice Problems for Final Page of 6 5 Constructing matrices a) Construct a matrix that is not invertible and has no entries. Show that it is not invertible. b) Construct a 2 2 matrix where col(a) = nul(a) c) Construct a 2 2 matrix that is not diagonalizable. Solution: KEY PLAYS: Knowing the definitions of things and how to apply them a) Construct a matrix that is not invertible and has no entries. Show that it is not invertible. Key Play: Know the inverse theorems posted on TED One of the theorems posted on TED says that a mtrix is invertible if and only if rref(a) = I. So if the rref of a matrix isn t the identity, it is not invertible. So, one way to approach this problem: start with a non-identity rref matrix, and use row operations to work backwards into a matrix that has no zeros: rref(a) I, so it s not invertible = A b) Construct a 2 2 matrix where col(a) = nul(a) Key Play: Know what column space and nullspace are, and also how to manipulate matrices First, you should know that if two spaces are the same, they have the same dimensions, and the same basis (Note: The null space and the column space being the same can only happen with n n matrices, because then, the null space and column space live in the same larger space R n ). So in this case, since dim col(a) + dim nul(a) = n = 2, and dim col(a) = dim nul(a), we have that dim nul(a) = dim col(a) =. So our goal is to construct a matrix so that the column space and the nullspace are spanned bythe same [ vector. Let s just choose one, and go from there. So let s say col(a) is spanned by. [ [ [ [ Then every column of the matrix A is a multiple of. So we have: A = a b [ Where a and b are numbers we need to solve for. Next, we want to be in the NULLSPACE [ of A, so we multiply A by

Jor-el Briones / Math 2F Practice Problems for Final Page of 6 [ [ a b [ [ = a [ + b [ = a [ = [ () We have from here that a =, and that b can be anything SO THAT THE NULLSPACE IS STILL DIMENSION. Let b =. Then: A = [ Clearly the nullspace is still dimension, since there is a single free variable, and the column space is dimension, since there s a pivot. By construction, the nullspace and column space have the same basis, so the spaces are equal. c) Construct a 2 2 matrix that is not diagonalizable. Key Play: Know that a matrix is not diagonalizable if and only if it does NOT have a basis of eigenvectors. Also know that in order for a matrix to NOT be diagonalizable, it needs at least one repeated eigenvalue Tip/Trick: Diagonal and triangular matrices are easy to work with because the determinants of those matrices are the products of the entries on the diagonal, and the eigenvalues are the entries on the diagonal themselves. You need to know that a matrix is diagonalizable if and only if it has a basis of eigenvectors. So we need to choose a matrix that has not enough eigenvectors. Since every distinct eigenvalue has at least one eigenvector, we need to have a repeated eigenvalue (otherwise, we ll have two linearly independent eigenvectors from two separate eigenvalues). So let s say we have an eigenvalue of. To make things easier on ourselves, let s choose an upper triangular matrix. The reason why is that the eigenvalues are just the entries on the diagonal. Let s also choose an eigenvalue, like. A = [ x Next, we want to go with a matrix so that there isn t a basis of eigenvectors. Since there is only one eigenvalue (), we want it to that its corresponding eigenvectors are all multiples of the same vector (the dimension of the eigenspace is less than 2). So to figure that out, we solve the homogeneous system (A I)x =. (A I)x = = [ x x = = ( x )

Jor-el Briones / Math 2F Practice Problems for Final Page 2 of 6 Now we want at most free variable, so that the nullspace has dimension. In order to do that, x needs to be a pivot, and therefore, non-zero. So let x =. Then dim nul(a I) < 2, and since we have as the only eigenvalue, there is not a basis of eigenvectors (we only have dimension ) not enough eigenvectors. So we have the following non-diagonalizable matrix: A = [

Jor-el Briones / Math 2F Practice Problems for Final Page of 6 6 Eigenvectors, Eigenvalues, Diagonalization Let A = 2 2 a) Find the characteristic equation of A b) Find the eigenvalues of A and their corresponding eigenvectors c) Is A diagonalizable? Justify your answer. d) Is the following system easily computible? If so, solve it: Solution: Y (T ) = AY (t) Y () = 5 a) Find the characteristic equation of A Key Play: Know what the characteristic equation is and how to find it. Tip/Trick: The determinant of a triangular matrix is just the product of entries on the diagonal. The characteristic equation is det(a λi) =. So first find A λi, then calculate the determinant. A λi = 2 2 λ = det(a λi) = = λ λ 2 2 λ λ λ 2 2 λ = ( λ)( λ)(2 λ) = The characteristic equation is boxed above. Since the matrix A λi is lower triangular, it was easy to calculate its determinant, since it is just the product of the entries on the diagonal. b) Find the eigenvalues of A and their corresponding eigenvectors IMPORTANT: Know how to find eigenvectors and eigenvalues.

Jor-el Briones / Math 2F Practice Problems for Final Page 4 of 6 First off, the eigenvalues are the solutions to the corresponding characteristic equation: ( λ)( λ)(2 λ) = So λ =,, 2. The corresponding eigenvectors are the basis for the nullspace for A λi for each distinct eigenvalue λ. So what finding eigenvectors amounts to is solving some number of homogeneous systems. Case: λ = : Solve (A I)x =. So Ax =, and: 2 2 2 So we have x =, x = t, x 2 = 2x = 2t, and x = t 2. We can choose our eigenvector to be v = 2 Case: λ = : Solve (A I)x =. So (A I)x =, and: 2 So we have x = t, x 2 = x = t, x = x 2 = t, and x = t our eigenvector to be v = Case: λ = 2:. We can choose t = and Solve (A 2I)x = : 2 2 So we have x = t, x 2 = x =, and x = t. We can choose our eigenvector to be

Jor-el Briones / Math 2F Practice Problems for Final Page 5 of 6 v = So our final answer is: λ =, with corresponding eigenvector(s) 2 λ =, with corresponding eigenvector(s) λ = 2, with corresponding eigenvector(s) c) Is A diagonalizable? Justify your answer. A is diagonalizable. We have different eigenvectors, each corresponding to a different eigenvalue. Eigenvalues that correspond to different eigenvalues are linearly independent, so these three eigenvectors are linearly independent. Since there are three linearly independent eigenvectors in R, they span R, and hence, form a basis of eigenvectors. In that case, A is diagonalizable, so it has a basis of eigenvectors. d) Is the following system easily computible? If so, solve it: Y (T ) = AY (t) Y () = 5 Key Play: You need to know how to solve a system of linear differential equations (the one that is defined above) The system is easily computible IF either the condition vector is an eigenvector, OR the condition vector is a linear combination of eigenvectors. But from part c), there is a basis of eigenvectors, so EVERY vector in R is a linear combination of eigenvectors. Hence, the problem is easily computible. FIRST! Find the exact linear combination of eigenvectors from part b) that result in our condition vector Y (). x 2 + x 2 + x = 5

Jor-el Briones / Math 2F Practice Problems for Final Page 6 of 6 So we solve the resulting augmented matrix system: 2 5 So x =, x 2 =, and x =, and 2 + + = Next, you MUST know that the solution to the system is Y (t) = e ta C = e ta. 5 From there, You should also know that for any eigenvector v of A, with corresponding eigenvalue λ, e ta v = e λt v. With these facts in mind, the solution to the system becomes: 5 Y (t) = e ta = e ta = e t = = 2 2 5 2 2 Y (t) = = e ta ( + e ta + e t + e t + e t e t e t 2 + + e ta + e 2t + e 2t e t 2 + e t + e t + e 2t + e 2t + )