MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION Problem (a) For each matrix below (i) find a basis for its column space (ii) find a basis for its row space (iii) determine whether it is invertible and (iv) compute its inverse if possible A = B = C = Solution: We will row-reduce A to reduced row echelon form To compute the inverse of A (if it exists) we augment A by the identity matrix I on the right: 7 4 7 4 7 This shows that A is invertible (since it can be row-reduced to the identity matrix) and A = 4 7 The process for finding bases for the row and column spaces is explained in section of the textbook on page 8 In this case since A has a pivot position in each column and the pivot columns of A form a basis of Col A a basis for Col A is
MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION (In fact the column space of A is all of R so any basis of R is a possible answer to this part of the problem) The nonzero rows in the matrix we get by reducing A to row echelon form are a basis for Row A so a basis for Row A is (The row space is all of R so any basis of R would work here) For the matrix B we do the same thing (first augmenting B by the 4 4 identity matrix): where for the first equivalence row and row 4 were swapped the second equivalence is by a sequence of row-replacement operations the third equivalence is by rescaling rows and 4 and the final equivalence is by another sequence of row replacements So B is also invertible and B = As before B has pivot positions in every row and column so a basis for the column space is
MATH (LINEAR ALGEBRA ) FINAL EXAMFALL SOLUTIONS TO PRACTICE VERSION and a basis for the row space is (In fact the row space and column space of B both equal R 4 so any bases of R 4 would be acceptable answers to this part of the problem) The matrix C is not invertible because it is not square (it has rows and 5 columns) To find bases for the row and column spaces of C we row reduce to echelon form: C = As before a basis for Col C consists of the pivot columns of C (not of the row-reduced matrix) The pivots are in columns and so one basis for Col C is For a basis of Row C we can use the two nonzero rows of the row echelon matrix we found above: (b) Give an example of a matrix D which is not row-equivalent to the matrix A above Solution: We can pick any matrix D with a different reduced row echelon form as A recall from Chapter that the reduced row echelon form of a matrix is unique so this will work Any matrix with fewer than three pivot positions will work such as D = 7 (But there are infinitely many other possible correct answers to this problem!)
4MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION Problem (a) For each of the two matrices below (i) determine whether it is diagonalizable (ii) determine whether it is orthogonally diagonalizable and (iii) if it is diagonalizable find an invertible matrix P and a diagonal matrix D such that A = P DP HINT: The eigenvalues for the matrix A below are and A = B = 9 Solution: The matrix A is orthogonally diagonalizable since it is symmetric (by Theorem 55 in the textbook) To diagonalize it we first find eigenvectors of A (In the following solution I will find an orthonormal set of eigenvectors to show you how to orthogonally diagonalize A; however for the question as stated you could use any set of three linearly independent eigenvectors to get a correct answer) For the eigenvalue the associated eigenvectors are solutions to the linear system corresponding to whose solution is given by the equations x = x and x = x (with x as a free variable) So ( ) is an eigenvector We eventually want an orthonormal basis of R so we divide this eigenvector by its norm to get the unit eigenvector u = For the eigenvalue the associated eigenvectors are the solutions to the linear system corresponding to whose solution is given by the equation x = x x with x and x both free Plugging in ( ) and ( ) for the values of x and x in the above solution we get ( ) and ( ) two linearly independent eignevalues of A with
MATH (LINEAR ALGEBRA ) FINAL EXAMFALL SOLUTIONS TO PRACTICE VERSION5 eigenvalue Since we want an orthonormal set of eigenvectors we apply the Gram-Schmidt process to these two vectors to get the orthogonal set (see the solution to Problem 4 below for details on how to apply Gram- Schmidt) and then we make each of these vectors into unit vectors by dividing them by their norms to get u = u = Finally as explained in section 5 we can use the orthonormal set of eigenvectors { u u u } and the eigenvalues we have found above to construct matrices P and D that orthogonally diagonalize A: P = u u u = D = For B we again start by finding its eigenvalues B is upper-triangular so its eigenvalues are the entries on the main diagonal (as I explained in class) so the eigenvalues of B are and (Alternatively it is easy to find the eigenvalues of B by solving the characteristic equation which is = ( λ)( λ)) To find an eigenvector corresponding to the eigenvalue we solve: 9 9 The general solution is that x = and x is free so one possible eigenvector is ( ) To find an eigenvector corresponding to the eigenvalue we solve: 9 The general solution is that x = x and x is free so one possible eigenvector is ( ) So B is diagonalizable since the two eigenvectors we have found form a basis of R (by Corollary in section 5 of the textbook) Using the technique in section 5 we can let P = D =
MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION (The columns of P are the two eigenvectors we have found and the diagonal entries of D are the associated eigenvalues) However the matrix B is not orthogonally diagonalizable since the eigenspaces corresponding to the two different eigenvalues are not orthogonal (as we can see by taking a dot product of the eigenvectors we found: ( ) ( ) = ( ) + ( ) = which is not so these eigenvectors are not perpendicular) (b) Give an example of a matrix C which is not similar to the matrix A in part (a) Solution: Any two similar matrices have the same sets of eigenvalues (by Theorem 7 in the textbook) So we just have to find a matrix C whose set of eigenvalues is different from { }; there are many possible correct answers here such as C = (whose eigenvalues are and ) Problem For each of the following sets determine whether or not it is a subspace of R 4 If it is a subspace determine its dimension (a) The set of all x in R 4 such that B x = x where B is the matrix in Problem (a) above Solution: A vector x in R 4 is in this set if and only if it is in the eigenspace of B corresponding to the value (see the definition on page 9 of the textbook we re taking the number λ there to be ) which is the null space of the matrix B I 4 Since null spaces of 4 4 matrices are always subspaces of R 4 this set is a subspace of R 4 To find the dimension of this subspace note that since this space is the null space of B I 4 we can find its dimension by row reducing B I 4 (For a review of this see Example from section of the textbook; the technique described there generates a basis for Nul(A)) We row reduce B I 4 as follows: (by adding Row to Rows and 4 to create a pivot in the first column)
MATH (LINEAR ALGEBRA ) FINAL EXAMFALL SOLUTIONS TO PRACTICE VERSION7 (first swapping Rows and then rescaling Row by ) (first subtracting Row from Row and subtracting ( Row ) from Row 4 to create a pivot in the second column then rescaling Row by ) 4 By the Invertible Matrix Theorem (section Theorem 8) since B I has 4 pivots B I is invertible and (B I) x = has only the trivial solution x = In other words the null space of B is { x} (the set consisting of only the zero vector the zero subspace ) The dimension of this space is zero (See the definition of dimension on page 57 the zero subspace is defined to have dimension zero) (b) The row space of A where A is the matrix in Problem (a) above Solution: The matrix A is so its row space is a subspace of R but not a subspace of R 4 in fact it is not even a subset of R 4 and any subspace of R 4 must be a subset of R 4 (See the definition of subspace at the top of page plus Example 9 on the following page) So this set is not a subspace of R 4 (c) The set of all x in R 4 such that x Solution: Although this is a subset of R 4 and it contains the zero vector ( ) (which has length ) it is not a subspace of R 4 You could show this by showing that it violates either of the two conditions necessary to be a subspace (see Definition in the textbook) For example: the set is not closed under addition since the vectors e = ( ) and e = ( ) are both in the set (they each have length ) their sum e + e = ( ) is not because ( ) = + + + = > Alternatively you could show that this set is not a subspace of R 4 by showing that it is not closed under scalar multiplication for instance e = ( ) is in the set but is not in the set 7e = (7 ) = 7 + + + = 7 >
8MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION Problem 4 (a) Find an orthonormal basis for the subspace of R spanned by Solution: Call this subspace S First note that the set given is linearly independent (since neither vector is a multiple of the other) so it is a basis for S So we can use the Gram-Schmidt process (section page 4) to find an orthogonal basis of S: let v = v = = = Then v and v are orthogonal (since v v = but to get an orthonormal basis we must divide each vector by its norm The norm of v is and the norm of v is so an orthonormal basis of S is { v v } = (b) Extend the basis you found in part (a) to an orthonormal basis for R (by adding a new vector or vectors) Solution: This comes down to finding a vector in the orthogonal complement of S By the technique on page of the textbook (in section ) the orthogonal complement of S is equal to the nullspace of the matrix
MATH (LINEAR ALGEBRA ) FINAL EXAMFALL SOLUTIONS TO PRACTICE VERSION9 This matrix row-reduces to and the solution to the associated homogeneous linear system is given by the equations x = x x and x = x (with x free) Plugging in for x we get the solution ( ) and we can normalize this to get the vector v = So { v v v } is an orthonormal basis of R (c) Is there a unique way to extend the basis you found in (a) to an orthonormal basis of R? Explain No the vector v found above is not the unique way to extend { v v } to an orthonormal basis of R : the vector v works too: it is also a unit vector orthogonal to v and v (But it turns out that v and v are the only two possibilities) Problem 5 (a) Write the quadratic form Q(x x ) = x + x + x x in the form Q( x) = x T A x for some symmetric matrix A Solution: Let A = 5 5 (This uses the method discussed in section 8 of the textbook) To check this answer note that x T A x = 5 x x x = x x 5 x x + 5x 5x + x = x (x + 5x ) + x (5x + x ) = x + x + x x (b) Find an orthogonal matrix P such that using the linear change of variables P y = x we can eliminate the cross-terms of Q; that is Q( x) = ay + by for some constants a and b in R Solution: As explained in section 8 the matrix P we want is a matrix which orthogonally diagonalizes the matrix A found above in part (a) We first compute the eigenvalues of A using the characteristic equation: = λ 5 5 λ = ( λ) 5 = λ 4λ = (λ 7)(λ + )
MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION so there are two eigenvalues 7 and Next we find an orthonormal set of eigenvectors of A For the eigenvalue 7 the corresponding eigenvectors are solutions to the system corresponding to the augmented matrix 5 5 5 5 so an eigenvector is any vector (x x ) where x = x for instance ( ) To get a unit eigenvector we can divide ( ) by its length (which is ) to get the vector u = Similarly for the eigenvalue the corresponding eigenvectors are solutions to the system 5 5 5 5 and as before one solution to this is the unit vector u = P = The set { u u } is an orthonormal basis of R (this is straightforward to check) So if (whose columns are the vectors u and u ) then just like in Example 4 from section 7 of the textbook if x y = P x y then Q(x x ) = 7y y Problem Let P be the vector space of all polynomials (with real coefficients) of degree at most Let T : P R be the linear transformation defined by the rule T (a + a t + a t a + a ) = a Let B = { t t } be the standard basis for P and let { } C = (a) Explain why C is a basis for R
MATH (LINEAR ALGEBRA ) FINAL EXAMFALL SOLUTIONS TO PRACTICE VERSION Solution: Notice that the first vector ( ) is nonzero and the second vector ( ) is not a scalar multiple of the first vector so the set B is linearly independent Since the vector space R has dimension and B is a linearly independent subset of R containing vectors it follows from Theorem in section of the textbook that B is a basis for R (In other words for this problem you do not need to check that Span(B) = R An alternate solution to this problem would be to row-reduce the matrix whose columns are the vectors ( ) and ( ) check that it has a pivot in each column and conclude that B does indeed span R ) (b) Find the matrix for T relative to the bases B and C Solution: For this part we apply formula () for the matrix on page 9 of the textbook in section 7; the matrix we are looking for is the one called R BB there (but here we call the second basis C instead of B ) So the matrix M we are looking for is M = T () C T (t) C T ( t ) C We first compute the values of T on the vectors in the basis B: T () = (since = + t + t ) T (t) = (since t = + t + t ) T (t ) = (since t = + t + t ) Next for each of the three vectors in R above we compute their coordinate vectors with respect to the basis C For T () this amounts to finding the (unique) solution to the linear system whose augmented matrix is where the two columns on the left are the vectors in C and the vector in the rightmost column is T () This can be row-reduced (by first subtracting two times Row from Row ) to: so we conclude that T () C = ( ) (If you re still having trouble computing coordinate vectors such as this one you should review section of the textbook especially the formula on page 7)
MATH (LINEAR ALGEBRA ) FINAL EXAM FALL SOLUTIONS TO PRACTICE VERSION Simlarly T (t) C = ( ) C = ( ) and T (t ) C is the solution to the linear system whose augmented matrix is 7 4 4 so therefore T (t ) C = (7 4) Putting this all together (using the equation for M above) we get that the matrix for T relative to B and C is 7 4 (c) Is T one-to-one? Does T map P onto R? Explain Solution: The linear transformation T is not one-to-one since T (t) = ( ) (as we found above) but also T () = ( ) so T maps two different polynomials to the same value ( ) The transformation T does map P onto R One way to see this is to row-reduce the matrix M for T relative to B and C (found in part (b)) to get 7 7 M = 7 4 4 so M has a pivot position in each row and therefore the matrix transformation x M x maps R onto R But this means that T also maps onto R since if v is any vector in R and v C = (x x ) then we can find some (y y y ) in R such that M y = x and then if p(t) = y + y t + y t it follows that and so T (p(t)) = v T (p(t)) C = M p(t) B = M y = x = v C