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February 17th, 2010, 10:27 PM  #1 
Newbie Joined: May 2008 Posts: 29 Thanks: 0  Symmetric matrix and eugenvalues..
I just dont get the following problem: Find a real 2 x 2 symmetric matrix A with eugenvalues: l = 1 and l = 4 and eigenvector u = (1,1) belonging to l = 1; well i looked to definition which says that if A is real symmetric matrix with eugenvectors u and v belonging to l1 and l2, then u and v are orthogonal... from my problem i know u... but i do not know eigenvector v... and even if id knew that...how to derive A then? kinda messed up a little... any ideas? thank u! 
February 17th, 2010, 11:42 PM  #2 
Member Joined: Dec 2009 Posts: 31 Thanks: 0  Re: Symmetric matrix and eugenvalues..
Ans: A = [5/2 3/2] [3/2 5/2]. Proof: Let a be a 2x2 real symmetric matrix. Then A = [a b]. Since A is symmetric, then A = ATranspose = [a c]. This implies that b = c, so we can now [c d] [b d] now write A = [a b]. Now, since we know that [1,1]transpose is the eigenvector for l = 1, then we also know that [a1 b] *u[1 1]transpose = [0 0]transpose. [b d] [b d1] So, the above system tells us that (a1)*u + bu = 0 and bu + (d1)*u = 0. Multiplying the second equation by 1, we obtain bu + (1d)*u = 0. Adding this equation to the first equation in our series gives us (a1)*u + (1d)*u = 0, dividing by u, we obtain a1+1d = ad = 0, which gives us a = d, which means we can now rewrite A = [a b]. [b a] Now, we are also given that l1 = 1 and l2 = 2, which means that the characteristic polynomial of the matrix must be (l1)(l4) = l^2  5l + 4. So the det(A  lI) = l^2  5l +4, and since we know the form of A, we know that the det(AlI) is also equal to (al)^2  b^2 = l^2  2al +a^2  b^2. So l^2  2al +a^2  b^2 = l^2  5l +4. Subtracting l^2 off of each side, we now have 2al + a^2  b^2 = 5l + 4. Then from here, we see that 2al = 5l and a^2  b^2 = 4. The first equation gives us a = 5/2. Then, this gives us in turn that (5/2)^2  b^2 = 4, which implies that b^2 = +/(3/2). With further validation (which is not shown here), you can actually rule out 3/2, because if you check the eigenvector of l=1 with b = c = 3/2, you actually get u = [1, 1], which contradicts our given assumption. Therefore, A = [5/2 3/2] [3/2 5/2]. QED 
February 18th, 2010, 08:01 AM  #3 
Newbie Joined: May 2008 Posts: 29 Thanks: 0  Re: Symmetric matrix and eugenvalues..
based on ur proof i have figured it all out and understood concepts. thank u very much!


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eugenvalues, matrix, symmetric 
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