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 ei
Sample Test 2b F 04 Page 1 of 5
MATH*2080 Sample Midterm Exam IIb
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Sample Final Exam Page 1 of 12
CALCULUS MATH*2080 SAMPLE FINAL EXAM
Prof. R.Gentry
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In Part I
you will be asked to
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
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 Eleme
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 Propos
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 an
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 =
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