2012 Sol
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Linear algebra II 2012 final exam solutions 1a. The eigenvalues satisfy λ 2 - 4λ + 3 = 0, so λ =1, 3. The eigenvectors are v 1 = [ 1 -1 ] , v 2 = [ 1 1 ] . 1b. Let B be the matrix whose columns are v 1 and v 2 . Then B -1 AB is a diagonal matrix with diagonal entries 5 and 2, so B -1 AB = [ 5 0 0 2 ] = ⇒ A = [ 1 -2 1 1 ][ 5 0 0 2 ][ 1/3 2/3 -1/3 1/3 ] = [ 3 2 1 4 ] . 2a. The characteristic polynomial is det(A - λI )= -λ 3 +5λ 2 - 7λ + 3 = (3 - λ)(λ - 1) 2 and we need only worry about the double eigenvalue. Using row reduction, we get A - I = -1 2 3 -1 0 1 -1 2 3 -→ 1 0 -1 0 1 1 0 0 0 so there is only one free variable and one Jordan block. Thus, the Jordan form is J = 3 1 1 1 . 2b. Based on the characteristic polynomial, the eigenvalues are λ =0, 0, 0, 0, 1, 1. Based on the minimal polynomial, the largest Jordan block is 2 × 2. This already determines the Jordan blocks for the double eigenvalue. The null space of B has dimension 2, so there are exactly two Jordan blocks with λ = 0. Thus, the Jordan form is J = 1 1 1 0 1 0 0 1 0 .
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sol
Transcript of 2012 Sol
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Linear algebra II2012 nal exam solutions
1a. The eigenvalues satisfy 2 4+ 3 = 0, so = 1; 3. The eigenvectors are
v1 =
1
1; v2 =
11
:
1b. Let B be the matrix whose columns are v1 and v2. Then B1AB is a diagonal matrix
with diagonal entries 5 and 2, so
B1AB =5 00 2
=) A =
1 21 1
5 00 2
1=3 2=3
1=3 1=3=
3 21 4
:
2a. The characteristic polynomial is
det(A I) = 3 + 52 7+ 3 = (3 )( 1)2
and we need only worry about the double eigenvalue. Using row reduction, we get
A I =241 2 31 0 11 2 3
35 !241 0 10 1 10 0 0
35so there is only one free variable and one Jordan block. Thus, the Jordan form is
J =
243 1 11
35 :2b. Based on the characteristic polynomial, the eigenvalues are = 0; 0; 0; 0; 1; 1. Based
on the minimal polynomial, the largest Jordan block is 2 2. This already determinesthe Jordan blocks for the double eigenvalue. The null space of B has dimension 2, sothere are exactly two Jordan blocks with = 0. Thus, the Jordan form is
J =
266666641 1
10 1
00 1
0
37777775 :
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3a. We need to nd the values of a for which the matrix
Q =
242 a 0a a+ 4 a 40 a 4 a+ 1
35is positive denite. According to Sylvester's criterion, this is equivalent to
det
2 aa a+ 4
= (a+ 2)(a 4); detQ = (a 1)(a 4)(a+ 6)
being both positive. In particular, Q is positive denite if and only if 1 < a < 4.
3b. When a = 0, the eigenvalues are = 2; 5p73
2so the signature is (2; 1).
4a. Suppose A is real symmetric and v an eigenvector with eigenvalue . Then
jvj2 = (v)tv = (Av)tv = vtAv = vtAv = jvj2:
Writing = a+ bi with a; b real, we conclude that
a+ bi = = = a bi =) b = 0 =) = a 2 R:
4b. If A is such a matrix, then I = AAt = A2 and so (A + I)(A I) = 0. Since all theeigenvalues of A are positive, 1 is not an eigenvalue, so A+ I is invertible and
(A+ I)(A I) = 0 =) A I = 0 =) A = I:
