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MATH 102
SOLUTIONS TO HW
Section 3.1, problem 6. First we find a basis for the space of vectors orthogonal to both a = (1, 1 , 1) and b = (1, − 1 , 0). this space is nothing other that the null-space of the matrix:
A =
[
]
To compute it, after the following Gaussian elimination steps:
(1) − 1 × {Row 1} + {Row 2} ⇒ {Row 2} (2) 12 × {Row 2} + {Row 1} ⇒ {Row 1} (3) − 12 × {Row 2} ⇒ {Row 2}
we are left with the reduces matrix:
R =
[
]
From this, a basis for the null-space of A is easily computed to be:
c =
To normalize these vectors, we just divide through by their lengths. Doing this yields the orthonormal set:
ˆa =
√^1 3 √^1 3 √^1 3
,^ ˆb^ =
√^1 2 − √^12 0
(^) , cˆ =
− √^16
− √^16
√^2 6
Section 3.1, problem 12. This problem again asks to compute the null-space of A. Recall that N (A) = [R(A)]⊥. After performing the reduction step:
(1) − 1 × {Row 1} + {Row 2} ⇒ {Row 2}
we have the reduced matrix:
R =
[
]
The basis for N (A) is:
R =
1
To decompose x = (3, 3 , 3) into these three vectors, we solve the linear system:
a b c
After performing the following Gaussian elimination steps:
(1) − 2 × {Row 1} + {Row 3} ⇒ {Row 3} (2) − 2 × {Row 2} + {Row 3} ⇒ {Row 3}
this system becomes equivalent to:
a b c
Thus, we have that c = −1, and therefore:
xn =
Subtracting this off gives xr = x − xn, that is:
xr =
Section 3.1, problem 14. This is a simple computation using FOIL. We compute that: (x − y)T^ (x + y) = xT^ x + xT^ y − yT^ x − yT^ y , = xT^ x − yT^ y , = ‖ x ‖^2 − ‖ y ‖^2.
This last line follows because of the symmetry xT^ y = yT^ x. Thus, (x−y) and (x+y) being orthogonal is equivalent to the condition ‖ x ‖^2 = ‖ y ‖^2. This is the same as saying ‖ x ‖ = ‖ y ‖ because these are both positive quantities.
Section 3.1, problem 38. If S only contains (0, 0 , 0) then S⊥^ = R^3 because every vector is orthogonal to zero.
If S is generated by (1, 1 , 1), then S⊥^ is the same as the null-space of the matrix: A =
[
]
This is easily computed to be the space spanned by the two special vectors:
e 1 =
(^) , e 2 =
the general equation P (V ⊥) = ~0, of P is the orthogonal projection onto V. Thus, P is never invertible unless it projects onto all of Rn. This is only if P = I, the identity matrix.
Section 3.2, problem 24. I will not draw this here. However, the two matrices for projection onto a 1 and a 2 are easily computed to be:
P 1 =
[
]
, P 2 =
[
]
This gives the projections:
P 1 b =
[
]
, P 2 b =
[
]
We have that P 1 b + P 2 b = (8/ 5 , 6 /5)T^6 = b.
