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Notes 23: More Eigenvalues, Eigenvectors
Lecture December 8, 2010
Given a 2 2 matrix M with complex eigenvalues, our goal is to nd a basis B of R2
so that the matrix representing M with respect to the basis B is simple. Indeed we will
show how to nd a bas
Notes 22: More Eigenvalues, Eigenvectors
Lecture December 6, 2010
We apply our algorithms for nding eigenvalues and eigenvectors to 3 3 matrices. To
do this we need a dierent way of calculating the determinants of matrices. it is called
expansion by minor
Math 235 section 1 r Midterm 1 Spring 2014
psi/WU
Your Name:
Student ID:
This is a 90 minutes exam. This exam paper consists of 6 questions. It has 7 pages.
The useof calculators is not allowed on this exam. You may use one letter size page of
notes (bo
quiz 8
University of Massachusetts Amherst
Math 235 Spring 2014
April 24, 2014
name:
2
1 2
4
(1) (1 point) The matrix A = 0 2
0 0
invertible matrix S and diagonal
3
3
35 is diagonalizable, meaning that S 1 AS = D for some
3
matrix D. Give one possibility
235 Common midterm review Questions
Paul Hacking
November 3, 2013
(1) Let m and n be positive integers and v1 , v2 , . . . , vm be vectors in Rn .
What does it mean to say that a vector v is a linear combination of
v 1 , . . . , vm ?
1
(2) Is the vector
DEPARTMENT OF MATHEMATICS AND STATISTICS
UNIVERSITY OF MASSACHUSETTS
MATH 235
NAME:
Section:
MIDTERM
SOLUTIONS
Spring 2013
ID#:
Circle Instructor: Aiello, Dobson, Jennings, Kazanova, Kusner, Wilson
In this exam there are seven pages, including this one, a
Math 235
Midterm 1
Fall 2006
1. (20 points) a) Find the row reduced echelon augmented matrix of the system
x1 + x2 + x4 = 4
x2 x3 + x 4 = 3
x1 + x3 + 2x4 = 3
b) Find the general solution for the system.
1
1
1
2. (16 points) Let u1 = 0 , u2 = 2 , and u
= T)
7 x
,
=
<
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c
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s
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)
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3)

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235 Final exam review questions
December 11, 2016
(1) Find the dimension of the subspace spanned by the vectors
1
1
0
0
2 1 1 0
3 , 1 , 2 , 0 .
4 0 4 1
5
1
4
1
(2) Find the dimension of the subspace that is the solution set of the
equation Ax = 0 wi