Linear Algebra/Eigenvalues and Eigenvectors/Solutions

Solutions Problem 1 For each, find the characteristic polynomial and the eigenvalues. <li> <li> <li> <li> <li> Answer <li> This simplifies to the characteristic equation . Because the equation factors into there is only one eigenvalue . <li> ; , <li> ; , <li> ; <li> ; Problem 2 For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors. <li> <li> Answer <li> The characteristic equation is . Its roots, the eigenvalues, are and . For the eigenvectors we consider this equation. For the eigenvector associated with , we consider the resulting linear system. The eigenspace is the set of vectors whose second component is twice the first component. (Here, the parameter is only because that is the variable that is free in the above system.) Hence, this is an eigenvector associated with the eigenvalue . Finding an eigenvector associated with is similar. This system leads to the set of vectors whose first component is zero. And so this is an eigenvector associated with . <li> The characteristic equation is and so the eigenvalues are and . To find eigenvectors, consider this system. For we get leading to this eigenspace and eigenvector. For the system is leading to this. Problem 3 Find the characteristic equation, and the eigenvalues and associated eigenvectors for this matrix. Hint. The eigenvalues are complex. Answer The characteristic equation has the complex roots and . This system For Gauss' method gives this reduction. (For the calculation in the lower right get a common denominator to see that it gives a equation.) These are the resulting eigenspace and eigenvector. For the system leads to this. Problem 4 Find the characteristic polynomial, the eigenvalues, and the associated eigenvectors of this matrix. Answer The characteristic equation is and so the eigenvalues are (this is a repeated root of the equation) and . For the rest, consider this system. When then the solution set is this eigenspace. When then the solution set is this eigenspace. So these are eigenvectors associated with and . Problem 5 For each matrix, find the characteristic equation, and the eigenvalues and associated eigenvectors. <li> <li> Answer <li> The characteristic equation is and so the eigenvalues are and also the repeated eigenvalue . To find eigenvectors, consider this system. For we get leading to this eigenspace and eigenvector. For the system is leading to this. <li> The characteristic equation is and the eigenvalues are and (by using the quadratic equation) and . To find eigenvectors, consider this system. Substituting gives the system leading to this eigenspace and eigenvector. Substituting gives the system (the middle coefficient in the third equation equals the number ; find a common denominator of and then rationalize the denominator by multiplying the top and bottom of the frsction by ) which leads to this eigenspace and eigenvector. Finally, substituting gives the system which gives this eigenspace and eigenvector. Problem 6 Let be Find its eigenvalues and the associated eigenvectors. Answer With respect to the natural basis the matrix representation is this. Thus the characteristic equation is . To find the associated eigenvectors, consider this system. Plugging in gives Problem 7 Find the eigenvalues and eigenvectors of this map . Answer , , Problem 8 Find the eigenvalues and associated eigenvectors of the differentiation operator . Answer Fix the natural basis . The map's action is , , , and and its representation is easy to compute. We find the eigenvalues with this computation. Thus the map has the single eigenvalue . To find the associated eigenvectors, we solve to get this eigenspace. Problem 9 Prove that the eigenvalues of a triangular matrix (upper or lower triangular) are the entries on the diagonal. Answer The determinant of the triangular matrix is the product down the diagonal, and so it factors into the product of the terms . Problem 10 Find the formula for the characteristic polynomial of a matrix. Answer Just expand the determinant of . Problem 11exer:CharPolyTransWellDefed Prove that the characteristic polynomial of a transformation is well-defined. Answer Any two representations of that transformation are similar, and similar matrices have the same characteristic polynomial. Problem 12 <li> Can any non- vector in any nontrivial vector space be a eigenvector? That is, given a from a nontrivial , is there a transformation and a scalar such that ? <li> Given a scalar , can any non- vector in any nontrivial vector space be an eigenvector associated with the eigenvalue ? Answer <li> Yes, use and the identity map. <li> Yes, use the transformation that multiplies by . Problem 13 Suppose that and . Prove that the eigenvectors of associated with are the non- vectors in the kernel of the map represented (with respect to the same bases) by . Answer If then under the map . Problem 14 Prove that if are all integers and then has integral eigenvalues, namely and . Answer The characteristic equation simplifies to . Checking that the values and satisfy the equation (under the condition) is routine. Problem 15 Prove that if is nonsingular and has eigenvalues then has eigenvalues . Is the converse true? Answer Consider an eigenspace . Any is the image of some (namely, ). Thus, on (which is a nontrivial subspace) the action of is , and so is an eigenvalue of . Problem 16 Suppose that is and are scalars. <li> Prove that if has the eigenvalue with an associated eigenvector then is an eigenvector of associated with eigenvalue . <li> Prove that if is diagonalizable then so is . Answer <li> We have . <li> Suppose that is diagonal. Then is also diagonal. Problem 17 Show that is an eigenvalue of if and only if the map represented by is not an isomorphism. Answer The scalar is an eigenvalue if and only if the transformation is singular. A transformation is singular if and only if it is not an isomorphism (that is, a transformation is an isomorphism if and only if it is nonsingular). Problem 18 <li> Show that if is an eigenvalue of then is an eigenvalue of . <li> What is wrong with this proof generalizing that? "If is an eigenvalue of and is an eigenvalue for , then is an eigenvalue for , for, if and then "?Strang 1980 Answer <li> Where the eigenvalue is associated with the eigenvector then . (The full details can be put in by doing induction on .) <li> The eigenvector associated with might not be an eigenvector associated with . Problem 19 Do matrix-equivalent matrices have the same eigenvalues? Answer No. These are two same-sized, equal rank, matrices with different eigenvalues. Problem 20 Show that a square matrix with real entries and an odd number of rows has at least one real eigenvalue. Answer The characteristic polynomial has an odd power and so has at least one real root. Problem 21 Diagonalize. Answer The characteristic polynomial has distinct roots , , and . Thus the matrix can be diagonalized into this form. Problem 22similarity transformation Suppose that is a nonsingular matrix. Show that the similarity transformation map sending is an isomorphism. Answer We must show that it is one-to-one and onto, and that it respects the operations of matrix addition and scalar multiplication. To show that it is one-to-one, suppose that , that is, suppose that , and note that multiplying both sides on the left by and on the right by gives that . To show that it is onto, consider and observe that . The map preserves matrix addition since follows from properties of matrix multiplication and addition that we have seen. Scalar multiplication is similar: . ? Problem 23 Show that if is an square matrix and each row (column) sums to then is a characteristic root of . Morrison 1967 Answer This is how the answer was given in the cited source. If the argument of the characteristic function of is set equal to , adding the first rows (columns) to the th row (column) yields a determinant whose th row (column) is zero. Thus is a characteristic root of . References Clarence C. (proposer) Morrison Quickie Mathematics Magazine 40 4 1967 232. Gilbert Strang Linear Algebra and its Applications Second Harcourt Brace Jovanovich 1980.