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Math 232 Final Examination Solution April 10 , 2007
- Consider the matrix
(a) (3 points) Compute det(A).
Answer
We use row reduction to compute the determinant:
det
= 2 det
= 2 det
2 det
(b) (3 points) Compute A
− 1 .
Answer
We use row reduction on an augmented matrix. We join the computations after the first
two steps made above:
1
2
3
2
1
2
1
2
3
2
9
2
5
2
11
2
5
2
, hence A
− 1
11
2
5
2
(c) (3 points) Solve the system
2 x 1 +8x 2 − 2 x 3 = − 2
3 x 1 +1 3 x 2 − 4 x 3
− 1 x 1 − 7 x 2 +3x 3
Answer
With A
− 1 this is really easy:
x 1
x 2
x 2
= A
− 1
11
2
5
2
- Consider the matrix
A =
(a) (3 points) Compute the characteristic polynomial of A.
Answer
χA(λ) = det
10 − λ 12
6 − 7 − λ
= (10 − λ)(− 7 − λ) + 72 = λ
2 − 3 λ + 2 = (λ − 1)(λ − 2)
(b) (4 points) Compute the eigenvalues of A and the corresponding eigenspaces.
Answer
The eigenvalues are the roots of the characteristic polynomial, i.e., 1 , 2.
The eigenspace for λ = 1 is:
Nul
= Nul
= Span{
The eigenspace for λ = 2 is:
Nul
= Nul
= Span{
(c) (2 points) Give an invertible matrix P and a diagonal matrix D such that A = P
− 1 DP.
Answer
Since the columns of P =
are eigenvectors of A for the eigenvalues 1 , 2 respectively,
we have that for D =
we have A = P
− 1 DP.
- Consider the following basis for R
3 : {
(a) (2 points) Give the definition of an orthogonal basis.
Answer
An orthogonal basis of a vector space W ⊂ R
n is a set of vectors B = {b 1 ,... , b r } such
that
- B spans W , i.e., W = Span = {b 1 ,... , b r
- B is linearly independent
- The elements of B are mutually orthogonal, i.e., b i · b j = 0 for all i %= j.
In fact, if you know that none of the vectors in B is the zero vector and that they are
mutually orthogonal then the linear independence of B follows.
(b) (2 points) Show that the given basis for R
3 is not an orthogonal basis.
Answer
The vectors are not mutually orthogonal:
(c) (4 points) Perform the Gram-Schmidt orthogonalisation process on the given basis to
obtain an orthogonal basis for R
3
. You may rescale vectors if it makes the arithmetic
easier for you.
Answer
Given the vectors x 1
, x 2
, x 3
we perform the Gram-Schmidt
process to produce an orthogonal basis v 1 , v 2 , v 3
v 1 = x 1 =
; v 2 = x 2 −
x 2 · v 1
v 1 · v 1
v 1 =
v 3 = x 3 −
x 3 · v 1
v 1 · v 1
v 1 −
x 3 · v 2
v 2 · v 2
v 2 =
v 3 =
(d) (2 points) Give the orthogonal projection of
onto the orthogonal complement of
Span{
}. Explain your answer.
Answer
From part (c) we know that the projection of x 3 onto Span{x 1 , x 2 } = Spanv 1 , v 2 is x 3 −v 3 ,
so the projection onto the orthogonal complement is
v 3
- (3 points) Let A be an n × m matrix. Prove that if v ∈ Nul(A) then v is orthogonal to all
vectors in the row space of A.
Answer
Let {a 1 ,... , a n } be the vectors that make up the rows of A, i.e., A =
− a
T
1
− a
T
n
.^ If
v ∈ Nul(A) then
0 = Av =
− a
T
1
− a
T
n
v^ =
a
T
1
v
a
T
n
v
a 1 · v
a n · v
so, we see that v is orthogonal to each a i
. But then, if w is in the row space of A, then
w = c 1 a 1
- · · · + c n a n and therefore,
v · w = v · (c 1 a 1 + · · · + cnan) = c 1 v · a 1 + · · · + cnv · an = 0
so indeed, v is orthogonal to any vector in Row(A)
- We consider the vector space R
3 × 3 of 3 × 3 matrices. We consider the subset of symmetric
matrices:
V = {A ∈ R
3 × 3 : A
T = A}
(a) (2 points) What properties should V satisfy to be a subspace?
Answer
(a) 0 ∈ V
(b) If A, B ∈ V then A + B ∈ V
(c) If A ∈ V and c ∈ R then cA ∈ V.
(b) (3 points) Prove that V is a subspace.
Answer
(a) The zero vector in R
3 × 3 is the 0-matrix. The transpose of that is again the zero matrix,
so indeed, it is an element of V.
(b) We need to show that if A
T = A and B
T = B then (A + B)
T = A + B. Indeed,
(A + B)
T = A
T
T = A + B, assuming A, B ∈ V.
(c) We need to show that if A
T = A and c ∈ R then (cA)
T = cA. Indeed,
(cA)
T = c(A
T ) = cA, assuming A ∈ V.
(c) (2 points) What dimension does V have? Give a basis for V.
Answer
A general symmetric matrix is of the form
A =
a b c
b d e
c e f
= a
d
The 6 given matrices are clearly linearly independent (each has a 1 in a position where all
others have a 0).
Bonus question: Let W ⊂ R
n be a subspace and let proj W
: R
n → R
n be the orthogonal
projection onto W. Prove that proj W
can only have the eigenvalues 0 and 1 and describe the
corresponding eigenspaces in terms of W.
Answer
In the lecture we have proved that applying an orthogonal projection twice does not acoomplish
anything new:
proj W
(proj W
(y)) = proj W
(y)
If we have a non-zero vector v such that proj W
(v) = λv then
λv = proj W
(v) = proj W
(proj W
(v)) = proj W
(λv) = λproj W
(v) = λ
2 v
This means that λ = λ
2 , i.e., λ = 0 or λ = 1.
Remember that we can inquely write y = yˆ + z where yˆ ∈ W and z ∈ W
⊥
. With this notation,
proj W
(y) = ˆy.
Hence, if y is an eigenvector corresponding to the eigenvalue λ = 1 , then yˆ = proj W (y) = y,
so y ∈ W. Hence, the eigenspace corresponding to λ = 1 is W.
If y is an eigenvector for λ = 0 then
by = 0 , so z = y − 0 = y. In that case, y ∈ W
⊥ , so W
⊥
is the eigenspace corresponding to λ = 0.
