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As we saw in class, eigenvalues and eigenvectors of an arbitrary square matrix A are not necessarily well-behaved. In particular, it is possible for A
As we saw in class, eigenvalues and eigenvectors of an arbitrary square matrix A are not necessarily well-behaved. In particular, it is possible for A to have complex eigenvalues/eigenvectors or even to be defective. Fortunately, it turns out that if A is symmetric then its eigenvalues and eigenvectors are especially well-behaved. In this problem, we will show that a symmetric matrix has only real eigenvalues and eigenvectors. It is also true that a symmetric matrix A cannot be defective, but we will not prove that here For this problem, we will need to use some basic facts and notation about comples numbers: A complex number can always be written as z-a +bi, where a and b are real numbers The conjugate of a complex number z is -a - bi. That is, the conjugate switches the sign of the imaginary part. A real number is a complex number with b-0. A complex number z = a+bi is real if and only if z-2 . If z = a + bi, then zz = zz = a2 + b2, which is real. A complex vector is a vector z with complex entries. Likewise, a complex matrix is a matrix B with complex entries. The notation z or B means to take the conjugate of each entry of the vector z or matrix B The conjugate of a product is the product of conjugates. That is, if A and IB are complex matrices then AB = AB. Likewise, if z is a complex vector then . z . z = z"z is a real number In the rest of this problem, A is a symmetric matrix with real entries (a) Prove that (Ax) y x (Ay) for any complex vectors x and y (b) Prove that 7 Az is real for any complex vector z (c) Let be an eigenvalue of A and let z be one of its eigenvectors. That is. AZ-AZ Prove that is real and that there is a real eigenvector z. (Once you have a real eigenvector, it is easy to show that there are many non-real eigenvectors as well. We are only interested in real solutions.) As we saw in class, eigenvalues and eigenvectors of an arbitrary square matrix A are not necessarily well-behaved. In particular, it is possible for A to have complex eigenvalues/eigenvectors or even to be defective. Fortunately, it turns out that if A is symmetric then its eigenvalues and eigenvectors are especially well-behaved. In this problem, we will show that a symmetric matrix has only real eigenvalues and eigenvectors. It is also true that a symmetric matrix A cannot be defective, but we will not prove that here For this problem, we will need to use some basic facts and notation about comples numbers: A complex number can always be written as z-a +bi, where a and b are real numbers The conjugate of a complex number z is -a - bi. That is, the conjugate switches the sign of the imaginary part. A real number is a complex number with b-0. A complex number z = a+bi is real if and only if z-2 . If z = a + bi, then zz = zz = a2 + b2, which is real. A complex vector is a vector z with complex entries. Likewise, a complex matrix is a matrix B with complex entries. The notation z or B means to take the conjugate of each entry of the vector z or matrix B The conjugate of a product is the product of conjugates. That is, if A and IB are complex matrices then AB = AB. Likewise, if z is a complex vector then . z . z = z"z is a real number In the rest of this problem, A is a symmetric matrix with real entries (a) Prove that (Ax) y x (Ay) for any complex vectors x and y (b) Prove that 7 Az is real for any complex vector z (c) Let be an eigenvalue of A and let z be one of its eigenvectors. That is. AZ-AZ Prove that is real and that there is a real eigenvector z. (Once you have a real eigenvector, it is easy to show that there are many non-real eigenvectors as well. We are only interested in real solutions.)
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