Lecture Notes for Math 414: Linear Algebra II
Fall 2015, Michigan State University
Matthew Hirn
September 3, 2015
Beginning of Lecture 1
1
Vector Spaces
What is this course about?
1. Understanding the structural properties of a wide class of spaces which

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 24
Now we use the fact that V = U
V onto U .
U ? to dene the orthogonal projection of
Denition 46. Suppose U is a nite dimensional subspace of V .
The orthogonal projection of V onto U is the oper

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 22
Denition 44. A function ' is a linear functional on V if ' 2 L(V, F).
Examples:
Fix an arbitrary u 2 V . Then:
':V !F
v 7! '(v) = hv, ui
is a linear functional on V .
Fix an arbitrary continu

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 23
6.C Orthogonal Complements and Minimization Problems
Denition 45. If U V , then the orthogonal complement of U is:
U ? = cfw_v 2 V : hv, ui = 0, 8 u 2 U
Geometrical Examples:
If U is a line i

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 19
Theorem 19 (Triangle Inequality). Suppose u, v 2 V . Then:
ku + vk kuk + kvk,
with equality if and only if u = cv for c
0.
Proof. For the rst part:
ku + vk2 = hu + v, u + vi
= hu, ui + hv, vi +

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 27
Proposition 55. If T is self-adjoint and hT v, vi = 0 for all v 2 V , then
T = 0 (even if F = R).
Proof. If F = C then we already proved this. So assume that F = R, and
suppose that T is self-a

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 26
The null space and range of T are related to the null space and range of
T through the orthogonal complement, as we now prove.
Proposition 50. If T 2 L(V, W ), then:
1. null T = (range T )?
2.

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 25
7
Operators on Inner Product Spaces
We now explore the structure of operators on inner product spaces, which
we have been building towards for quite a while now. This will lead to some
of the m

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 28
Theorem 26 (Complex Spectral Theorem). Suppose F = C and T 2 L(V ).
Then the following are equivalent:
1. T is normal
2. V has an ONB consisting of eigenvectors of T
3. T has a diagonal matrix

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 29
Warmup: If we change F = R, where does the proof of the Complex Spectral
Theorem fall apart?
Answer: To prove (1) =) (3) we used Schurs Theorem, which only applies
to complex vector spaces.
Rea

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 30
Proposition 59. If V 6= cfw_0 and T 2 L(V ) is self-adjoint, then T has an
eigenvalue (even if F = R).
To prove this, we are going to need the following proposition from Chapter
4 on polynomial

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 37
Singular Value Decomposition
Let V, W be nite dimensional inner product spaces over the eld F with
dim V = n and dim W = m.
Denition 55. Suppose T 2 L(V, W ). The Hermitian square of T is T T 2

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 33
Rigid Motions in Rn
Denition 54. A rigid motion in an inner product space V is a transformation f : V ! V that preserves distances, i.e.,
8 u, v 2 V,
kf (u)
f (v)k = ku
vk
Note, f is not assume

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 34
7.D Polar Decomposition and Singular Value Decomposition
Polar Decomposition
First we recall the polar form of a complex number z 2 C. Let z = x + iy.
Every complex number z can also be written

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 31
7.C Positive Operators and Isometries
Positive Operators
Denition 51. An operator T 2 L(V ) is positive if T is self-adjoint and
8 v 2 V, hT v, vi
0.
Examples:
1. Orthogonal projections PU (whe

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 32
Isometries
Denition 53. An operator S 2 L(V ) is an isometry if
8 v 2 V,
kSvk = kvk
Thus an isometry is an operator that preserves norms, or equivalently, preserves distances since the denition

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 38
Let T 2 L(V, W ) and recall the singular value decomposition of T :
1, . . . ,
n
the singular values of T with
1, . . . ,
r
the nonzero singular
values.
e1 , . . . , en an ONB of V consisting

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 20
Denition 43. An orthonormal basis of V is an orthonormal list of vectors
in V that is also a basis of V .
Orthonormal lists and bases are very convenient! For example:
Proposition 39. If e1 , .

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 21
Example: Lets use Gram-Schmidt nd an orthonormal basis of P2 ([ 1, 1]; R)
with the inner product:
Z 1
hp, qi =
p(x)q(x) dx.
1
Lets start with the standard basis 1, x, x2 which is linearly indep

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 3
Warmup: Is this a vector space?
1. cfw_f 2 C(0, 1); R) : f (x) = x p for some p > 0
Answer: No (all three properties fail)
2. cfw_f 2 C(R; R) : f is periodic of period
Answer: Yes (contains zer

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 5
Theorem 3. dim V < 1, U1 and U2 subspaces of V . Then
dim(U1 \ U2 )
dim(U1 + U2 ) = dim U1 + dim U2
Proof. Proof will use 3 objects:
1. B = u1 , . . . , um = basis of U1 \ U2
2. L1 = v1 , . . .

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 4
2.B Bases
span+ linear independence = basis
Denition 13. v1 , . . . , vn 2 V is a basis of V if span(v1 , . . . , vn ) = V and
v1 , . . . , vn are linearly independent.
Proposition 12. v1 , . .

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 2
Warmup: Is the empty set ; a vector space?
Answer: No since 0 2 ;
/
1.C Subspaces
A great way to nd new vector spaces is to identify subsets of an existing
vector space which are closed under ad

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 40
Singular Value Decomposition of a matrix
Suppose T 2 L(Rn , Rm ) is dened as:
T x = Ax,
8 x 2 Rn ,
where A 2 Rm,n . So in particular, M(T ) = A in the standard bases for Rn
and Rm , and we can

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 8
3.D Invertibility and Isomorphic Vector Spaces
Denition 23. A linear map that is both injective and surjective is called
bijective.
Denition 24. A linear map T 2 L(V, W ) is invertible if 9 S 2

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 6
Warmup: Let U, W be 5-dimensional subspaces of R9 . Can U \ W = cfw_0?
Answer: No. First note that dimcfw_0 = 0. Then, using Theorem 3 we have:
dim R9 = 9
dim(U1 + U2 ) = dim U1 + dim U2 dim(U1

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 7
Very important applications:
Homogeneous systems of equations
m equations and n unknowns:
n
X
a1,k xk = 0
k=1
n
X
.
.
.
(5)
am,k xk = 0
k=1
where aj,k 2 F and x = (x1 , . . . , xn ) 2 Fn .
Can

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 10
Denition 28. Suppose T 2 L(V ). A scalar
there exists v 2 V , v 6= 0, such that
2 F is an eigenvalue of T if
Tv = v
Such a v is called an eigenvector of T .
Proposition 27. T 2 L(V ) has a one

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 9
Example: Let D 2 L(P3 (R), P2 (R) be the dierentiation operator, dened
by Dp = p0 . Lets compute the matrix M(D) of D with respect to the
standard bases B3 = 1, x, x2 , x3 of P3 (R) and B2 = 1,

Fall 2015
Math 414: Linear Algebra II
Beginning of Lecture 12
Warmup: Suppose T 2 L(V ) and 6I
ble eigenvalues of T ?
5T + T 2 = 0. What are the possi-
Answer: 6I 5T + T 2 = 0 implies that (T 2I)(T 3I) = 0. Now let v 6= 0
be an eigenvector of T with eigen