Question

Let A=|-39 28|
|-60 43|
has an eigenvalue of lambda = 1.
An eigenvector for lambda = 1
has the form | a |
|-10|
What is the value of a? (If the values in A are too large or unwieldy, you may be able to regenerate a new problem depending on the settings of this assignment.)

Answer 1

if it has a an Evalue of 1, subtract it from the main diag and row reduce

Answer 2

looks like it turns it into even numbers that can be scaled down if we are lucky

Answer 3

and then??

Answer 4

and then pull out the nulspace

Answer 5

what is that mean?

Answer 6

I don't understand that part?

Answer 7

its easier to show, but essentially you write up a parametric form of the row reduction equal to zero

Answer 8

1 -0.7; 0 0

Answer 9

1 -.7
0 0

is your row reduced form right?

Answer 10

yea

Answer 11

you see your free variable column in that? the one without a pivot point

Answer 12

the first column?

Answer 13

the first column has a pivot point, its the top entry; the second column has no pivot point (no 1 in it)

Answer 14

\[\begin{pmatrix}
x_1&x_2\\
1&-.7\\
0&0
\end{pmatrix}\]

the x2 is your free variable column and we need to define x1 and x2 in terms of that column

Answer 15

\[\binom{x_1}{x_2}=x_2\binom{.7}{1}\]

Answer 16

is n't it negative?

Answer 17

no, read it as equations:

x1 - .7x2 = 0

x1 = .7x2

Answer 18

and x2 always equals itself and nothing else

Answer 19

So what is the final answer?

Answer 20

pull off the vector; [.7 1] and scale it by whatever you want other than 0

Answer 21

[7 10] works as a pretty integer setup

Answer 22

So what is the (a) value?

Answer 23

well, since the vector can be scaled by any value; and we have it in the form [7 10], but we want to view it in the form [a -10] , what do you propose we do to it?

Answer 24

multiple it by -1?

Answer 25

7s = a
10s = -10

yes

Answer 26

so a = -7?

Answer 27

if you did your row reduction correctly, then that is what we would get

Answer 28

Thanks it's the right answer

Answer 29

yay!! :)