Description
MATH 141: Linear Analysis I |
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3. Write 1-3 sentences that interpret |
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1. Find all eigenvectors and eigenvalues of the matrix B = |
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these geometrically—in other words, what do the |
eigenvectors and eigenvalues tell you about the trans- |
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formation geometrically (in terms of stretch factors and stretch directions)?
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Construct an example for each of the following, or explain why such an example does not exist:
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a 2 2 matrix that is invertible but not diagonalizable
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a 2 2 non-diagonal matrix that is diagonalizable but not invertible
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(Strang x5.2 #11) If all eigenvalues of A are 1, 1, and 2, which of the following are certain to be true? Give a reason if true or a counterexample if false.
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A is invertible.
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A is diagonalizable.
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A is not diagonalizable.
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2 3
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4. (Strang x5.2 #12) Suppose the only eigenvectors of A are multiples of ~x = 405, which of the following are
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certain to be true? Give a reason if true or a counterexample if false.
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A is not invertible.
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A has a repeated eigenvalue.
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A is not diagonalizable.
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(Strang x5.1 #18) Suppose a 3 3 matrix A has eigenvalues 0, 3, and 5 with associated eigenvectors ~u, ~v, and w~ respectively.
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(a)
Since the eigenvalues of A are all distinct, the set f~u; ~v; w~g is
.
(b)
Write down a basis for the nullspace N(A) and the column space C(A).
(c)
Find one particular solution to A~x = ~v + w~. Find all solutions to A~x = ~v + w~.
(d)
Explain why A~x = ~u does not have a solution. (Hint: If there is a solution, then
is in C(A).
Explain why that is impossible.)
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Is A invertible? Why or why not?
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Let A be an n-by-n matrix. Suppose A~u = 2~u and A~v = 5~v for nonzero vectors ~u and ~v. Complete the following proof that f~u; ~vg is linearly independent.
Proof: In order for f~u; ~vg to be linearly independent, we need to show that
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the only solution to the vector equation
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is the trivial one.
x~u + y~v = 0
Note that in this equation, ~u and ~v are known vectors, while x and y are unknown scalars. Now assume that x = a and y = b is some solution to the above equation. That is
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~
(1)
a~u + b~v = 0:
MATH 141: Linear Analysis I Homework 13
Multiply both sides of equation (1) by matrix A from the left. Show your calculation details to explain why the following has to be true as well
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~
(2)
2a~u + 5b~v = 0:
Explain in detail how combing vectors equations (1) and (2) leads to the conclusion that a = 0 = b. There-
fore, the only solution to x~u + y~v = ~0 is the trivial solution.
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In one of the reflection questions, you saw that if A is 2 2 matrix then the product of its eigenvalues is equal to det(A) and the sum is equal to trace(A). The following shows that both claims still hold true for n n matrices.
Suppose that 1; 2; : : : ; n are the n eigenvalues of an n n matrix A. i’s are the roots of the polynomial det(A I), which means that we have a factorization
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det(A I) = ( 1 )( 2 ) ( n )
(3)
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(Strang x5.1 #8) By making a clever choice of the value for in equation (3), show that det(A) is equal to the product of eigenvalues.
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(Strang x5.1 #9) Show that trace(A) is equal to the sum of eigenvalues in three steps. First, find the
coefficient of ( )n 1 on the righthand side of equation (3). Next, find all terms on the righthand side
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of the following that involves (
)n 1
2
3
a21
a22
a2n
6
a11
a12
a1n
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det(A
I) = det
...
...
...
;
6
a
a
n2
a
nn
7
6
n1
7
4
5
where aij are the entries of A. Add up all those terms to find the coefficient of (
)n 1.
Lastly,
compare the coefficient of ( )n 1 found in these two different ways.