MATH 51: A SYMMETRIC MATRIX PROBLEM[Here is an example of how one might use the spectral theorem of symmetricmatrices. It is adapted from q5 on midterm II of Winter 2010 - that question wasbadly written and the published solution is not quite complete. Throughout, textin usual font forms a model answer, text in italics is my commentary. Thanks toFrederick (a fellow TA) and Elizabeth for exploring this question with me.]Let B be a symmetric matrix such thatB001=001, B570= 3570and det B = 7. Find the eigenvalues of B and a basis of R3consisting of eigen-vectors of B.We are given that•001is an eigenvector of B with eigenvalue 1•570is an eigenvector of B with eigenvalue 3We also know that B is symmetric, so it can be diagonalised. Hence its determi-nant is the product of its eigenvalues. So 7 = det B = 1 ∗ 3 ∗ λ, where λ is aneigenvalue of B. This gives λ =73. So the eigenvalues of B are 1, 3,73.[Here is a proof that “the determinant of a diagonalisable matrix is the productof its eigenvalues”: if A is diagonalisable, then A is similar to a diagonal matrixD. The diagonal entries of D are the eigenvalues of A, possibly with repetition.Similar matrices have the same determinant, so det A = det D, and the deter-minant of a diagonal matrix is the product of its diagonal entries. Hence det Ais the product of the eigenvalues of A, possibly with repetition.1MATH 51: A SYMMETRIC MATRIX PROBLEM 2“Product of its eigenvalues” is a shorthand phrase; you should think of thedeterminant of a diagonalisable n-by-n matrix as “the product of n numbers, allof which are eigenvalues, and each eigenvalue appears at least once amongst thesen numbers”. It is important here that the matrix be diagonalisable, although itdoesn’t need to be symmetric.]Let v =v1v2v3be an eigenvector of B with eigenvalue73. B is symmetric,so by spectral theorem, eigenvectors corresponding to distinct eigenvalues areorthogonal. Hencev.001= 0 and v .570= 0which means v3= 0 and 5v1+ 7v2= 0. This forcesv = c−750for some c, and c = 0 since v = 0. So−750=vc. As any nonzero multiple of an eigenvector is also an eigenvector,−750is a73-eigenvector of B.So001,570,−750are eigenvectors of B corresponding to distincteigenvalues, so they are linearly independent. Since any linearly independentset of three vectors in R3is a basis,001,570,−750is an eigenbasiswith respect to B.MATH 51: A SYMMETRIC MATRIX PROBLEM 3[It turns out that, if we did not know det B = 7 (as in the original question),the same three vectors give an eigenbasis, but the possibility of repeated eigen-values creates some subtle difficulties with the logic. It has taken me 10 daysto understand this snag, so don’t panic if it’s confusing. I can’t imagine such adifficult question coming up on the exams, at least not intentionally (I believe thepeople who wrote this question did not see these additional complications).Recall that Levandosky states the spectral theorem as “a symmetric matrix hasan orthonormal eigenbasis”, but hidden in the proof is the stronger statement that“eigenvectors of a symmetric matrix corresponding to different eigenvalues areorthogonal”. If you look closely, the induction proves even more than this, it showsthat “any orthogonal set of eigenvectors of a symmetric matrix can be extendedto an orthogonal eigenbasis” and using this version of the spectral theorem allowsthe above computation to go through.When I first attempted this question, I didn’t see this stronger version of thespectral theorem, so I analysed the cases where 1 and 3 are repeated eigenvaluesseparately. It’s a bit messy, but not too difficult, I can show you if you want.]Amy Pang,
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