So there is no basis of eigenvectors, and
is not similar to a diagonal
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
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
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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
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 ).
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
S. Lipschutz Schaums outline of linear algebra
S.I. Grossman Elementary linear algebra
Vector spaces and subspaces
Note yi =
j=1 aij xj
for i = 1, . . . , m, where A has entries (aij )m n .
So clearly T : Rn Rm given by left multiplication by A is represented by A if
we use the standard basis.
Let U, V, u1 , . . . , un , v1 , . . . , vm be as
We can get Q by inverting Q1 , and the answer is
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.
The u and w are unique, since if u1 + w1 = u2 + w2 then
u1 u2 = w2 w1
and, since U W = cfw_0, we have u1 = u2 and w1 = w2 .
If v U W , then
= 0 +