So there is no basis of eigenvectors, and
2 1
0 2
is not similar to a diagonal
matrix.
6.13 Theorem
Let V be a vector space over F , let T : V V be linear, and suppose that
cfw_v1 , . . . , vk are eigenvectors of T corresponding to distinct eigenvalues
c
Sample Test 2b F 04 Page 1 of 5
MATH*2080 Sample Midterm Exam IIb
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University of Guelph
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Sample Final Exam Page 1 of 12
CALCULUS MATH*2080 SAMPLE FINAL EXAM
Prof. R.Gentry
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This exam is worth 45% of your final grade.
In Part I
you will be asked to indicate a choice for the correct
procedure. Please in
Now let cfw_w1 , . . . , wm be a spanning set. Then (2.10) tells us that m n.
Suppose now that m = n. If cfw_w1 , . . . , wm is not independent, then a1 w1 + . . . +
am wm = 0, where at least one ai = 0. But then wi is a linear combination of
the others
Let U = R2 , V = R3 and W = R4 . We dene T (x1 , x2 ) = (x1 + x2 , x1 , x2 ) and
S(y1 , y2, y3 ) = (y1 + y2 , y1 , y2 , y3).
Then ST (x1 + x2 ) = (2x1 + x2 , x1 + x2 , x1 , x2 ).
4.13 Proposition
Let U, V, W be nite-dimensional vector spaces over F with b
MATH 2080 Further Linear Algebra
Jonathan R. Partington, University of Leeds, School of Mathematics
December 8, 2010
LECTURE 1
Books:
S. Lipschutz Schaums outline of linear algebra
S.I. Grossman Elementary linear algebra
1
Vector spaces and subspaces
Vect
Note yi =
n
j=1 aij xj
for i = 1, . . . , m, where A has entries (aij )m n .
i=1 j=1
So clearly T : Rn Rm given by left multiplication by A is represented by A if
we use the standard basis.
4.4 Proposition
Let U, V, u1 , . . . , un , v1 , . . . , vm be as
We can get Q by inverting Q1 , and the answer is
1 0
.
1 2
LECTURE 13
Alternatively. From rst principles, we have T (x, y, z) = (x + y, x + y + 2z),
so the kernel consists of all vectors with z = 0 and x = y, i.e., has a basis
(1, 1, 0).
Extend to a basis
V = U W if and only if for each v V there are unique u U and w W
with v = u + w.
Proof:
The u and w are unique, since if u1 + w1 = u2 + w2 then
u1 u2 = w2 w1
U
W
and, since U W = cfw_0, we have u1 = u2 and w1 = w2 .
If v U W , then
v=
v +
0
= 0 +
v
U
W