54 Quiz 8 Solutions GSI: Morgan Weiler
√ 7
5
3
diagonalizable? If so, diagonalize it.
Problem 1 (6 pts). Is A = √
5 3 −3
Solution: It is, because it is symmetric. In fact it is orthogonally diagonalizable.
To find the eigenvalues, factor the characteristic polynomial:
(7 − λ)(−3 − λ) − 25(3) = −21 − 4λ + λ2 − 75 = λ2 − 4λ − 96 = (λ − 12)(λ + 8)
so the eigenvalues are 12 and −8.
For λ = 12, the eigenvector satisfies
√ √ 7
5
x
12x
7x
+
5
12x
3
3y
√
=
⇔ √
=
y
12y
12y
5 3 −3
5 3x − 3y
solving the second equation gives
√
1
5 3x = 15y ⇔ √ x = y
3
which also satisfies the first equation:
√ 1
7x + 5 3 √ x = 7x + 5x = 12x
3
√ 3
so an eigenvector for λ = 12 is
.
1
For λ = −8, the eigenvector satisfies
√ √ 7
5
3
x
−8x
7x
+
5
3y
−8x
√
=
⇔ √
=
y
−8y
−8y
5 3 −3
5 3x − 3y
solving the second equation gives
√
√
5 3x = −5y ⇔ − 3x = y
which also satisfies the first equation:
√
√
7x + 5 3(− 3)x = 7x − 15x = −8x
1
√ .
so an eigenvector for λ = −8 is
− 3
Since A is symmetric it should be orthogonally diagonalizable. To save myself having to invert the
matrix of eigenvectors (though this isn’t that hard in this situation since it’s only 2 by 2 – in general it
is harder) I know that if I have orthonormal eigenvectors I can invert the matrix of eigenvectors simply
by taking its transpose. The eigenvectors are already orthogonal. Normalizing them gives the
diagonalization
#
#
√ " √3
" √3
1
1
7
5
3
12
0
2√
2
2√
√
= 21
3
3
1
0 −8
5 3 −3
−
−
2
2
2
2
though you could come up with slightly different ones.
1
Problem 2 (7 pts). Find the least-squares solution x for the problem minx ||Ax − b||, for

 

1
1 −1 0



2
0 −1
1
and
b
=
A=
−2
−1 0 −2
1
1
1
0
There are many ways to do this. One hint: check ai · aj where ai and aj are columns of A.
Solution: The columns of A are orthogonal, so if A = a1 a2 a3 then
 b·a1   1+2+2+1   
6/7
a1 ·a1
7
b·a2 




0
= 0 
x = [projColA b]{a1 ,a2 ,a3 } = a2 ·a2 =
−1+4
b·a3
3/5
5
a ·a
3
3
However, solving using AT Ax = AT b is not hard either, since AT A is diagonal.
2
Problem 3 (7 pts). Find an orthogonal basis for P2 with the inner product
hp, qi = p(−2)q(−2) + p(0)q(0) + p(1)q(1).
Solution: Use Gram-Schmidt on the basis {1, t, t2 } for P2 :
v1 = 1
ht, 1i
1
h1, 1i
(−2)(1) + (0)(1) + (1)(1)
1
=t−
(1)(1) + (1)(1) + (1)(1)
1
=t+
3
ht2 , t + 13 i
ht2 , 1i
1
v3 = t2 −
1−
1
1 (t + )
h1, 1i
3
ht + 3 , t + 3 i
v2 = t −
= t2 −
(4)(− 53 ) + (0)( 13 ) + (1)( 34 )
1
(4)(1) + (0)(1) + (1)(1)
1−
5
1 1
4 4 (t + )
5
3
3
(− 3 )(− 3 ) + ( 3 )( 3 ) + ( 3 )( 3 )
5 − 16
1
= t − − 423 (t + )
3
3
9
5
1
8
= t2 − + 21 (t + )
3
3
3
9 8
= t2 − + t
7 7
2
3