Question

For each matrix, determine if the matrix is diagonalizable, nilpotent or neither.

If the matrix is diagonalizable, find a matrix P that diagonalizes the matrix. If the matrix is nilpotent, find a string basis associated with the matrix. If he matrix is neither, find the Jordan canonical form of the matrix.

A = [5 4 2]
[4 5 2]
[2 2 2]

Answer 1

I do not know the meaning of those terms.

Answer 2

I can't find the eigenvalues!

Answer 3

@Calcmathlete

Answer 4

@apoorvk

Answer 5

what is your characteristic polynomial?

Answer 6

What do you mean?

Answer 7

http://en.wikipedia.org/wiki/Characteristic_polynomial

Answer 8

Do you have any idea about what the Jordan canonical form is? I am familiar with other terms but not this one...

Answer 9

No but we can find other terms

Answer 10

the eigenvalues are easy to find...I'm not sure what your issue is.

Answer 11

I don't know I got this equation: x^3 - 17x^2 + 25x - 26 = 0
I don't how to solve from here

Answer 12

that is not what I get

Answer 13

what do you get?

Answer 14

\[-\lambda^3+12\lambda^2-21\lambda+10\]

Answer 15

the above equation is the characteristic polynomial

Answer 16

equal to zero that is

Answer 17

I don't know how did you get this equation and how to solve it?

Answer 18

I computed

\[\det\left(A-\lambda I_3\right)\]

Answer 19

I did the samething

Answer 20

then you must have made a mistake in your calculation

Answer 21

ok how to solve this equation?

Answer 22

easy...it factors ...you get 1,1,10 as the roots

Answer 23

it's cubic equation!

Answer 24

so?

Answer 25

how to solve it?!

Answer 26

factor it...use the rational root test

Answer 27

can you show me please?

Answer 28

test \(\pm10,\pm5,\pm2,\pm1\) into the above equation

you will see that 10 and 1 are the only ones that work...and in fact 1 is a repeated root.

Answer 29

ok how did you get these numbers?

Answer 30

http://en.wikipedia.org/wiki/Rational_root_theorem

Answer 31

but you said it's 1, 1 and 10

Answer 32

yes..that is correct

Answer 33

that is what I said both time I gave you the roots

Answer 34

ok sorry

Answer 35

So it's gonna be 1 and 10?

Answer 36

yes...where 1 is a repeated root

Answer 37

ok
So it's not diagonalizable right?

Answer 38

it is ;)

Answer 39

how come ? it's not equal to n

Answer 40

this is one of those weird cases where you have a repeated root but it is still diagonalize

Answer 41

so it's considered 3 not 2?

Answer 42

have you tried to find the eigenvectors?

Answer 43

no

Answer 44

Why do I want to find the eigenvector?

Answer 45

have you ever diagonalized a matrix before?

Answer 46

ok sorry
got it! to find P right?

Answer 47

yes

Answer 48

ok

Answer 49

ok I got 4b1 + 4b2 + 2b3 = 0

Answer 50

for lambda = 1

Answer 51

I get \[p=\left[\begin{matrix}-1 & 2 & -1 \\ 1 & 2 & 0\\ 0 & 1 & 2\end{matrix}\right]\]

Answer 52

Can you explain how to get this?

Answer 53

ok...can you solve your equation in terms of the free variables

Answer 54

do you mean do it in rref?

Answer 55

I assume you looked at

A-I

and put it in rref

Answer 56

I don't how to get P

Answer 57

that is what we are working on...there are several steps to go through

Answer 58

Put \(A-I_3\) into rref


I get

\[\left[\begin{matrix}1 & 1 & 1/2 \\ 0 & 0 & 0\\ 0 & 0 & 0\end{matrix}\right]\]

Answer 59

A-1 = [ 1 , 1 , 0.5 ; 0 , 0 , 0 ; 0 , 0 , 0]

Answer 60

yes

Answer 61

now can you solve this system in terms of the free variables

Answer 62

to get what?

Answer 63

the eigenspace (the eigenvectors)

Answer 64

i don't know?

Answer 65

have you ever found a Null space before?

Answer 66

yes, I got b1 = -b2 and b3 = -2 b1

Answer 67

if we use x,y,z then [1 , 1 , 0.5]

turns into

x+y+.5z=0

let z=t and y=r

they are free variables

then we get the solution x=-r-.5t,y=y, z=t

so we have

\[\left(\begin{matrix}-r-.5t \\ r\\t\end{matrix}\right)=\left(\begin{matrix}-.5t \\ 0\\t\end{matrix}\right)+\left(\begin{matrix}-r \\ r\\0\end{matrix}\right)\]

\[=\left(\begin{matrix}-.5 \\ 0\\1\end{matrix}\right)t+\left(\begin{matrix}-1 \\ 1\\0\end{matrix}\right)r\]

Answer 68

typo
*then we get the solution x=-r-.5t,y=r, z=t

Answer 69

ok got it. This is the eignvector? right?

Answer 70

\[\left(\begin{matrix}-.5 \\ 0\\1\end{matrix}\right)\text{ and }\left(\begin{matrix}-1 \\ 1\\0\end{matrix}\right)\] are


though I used \[\left(\begin{matrix}-1 \\ 0\\2\end{matrix}\right)\text{ and }\left(\begin{matrix}-1 \\ 1\\0\end{matrix}\right)\]

Answer 71

notice these two vectors are in the matrix P that I wrote above

Answer 72

from where you got 2 2 1 ?

Answer 73

that is the eigenvector corresponding to the eigenvalue 10

Answer 74

are we working on 1 or 10 ?

Answer 75

we just did 1...since it was a repeated root we got 2 vectors

Answer 76

I don't get this part, so this is the eignvector for lambda = 1 right?

Answer 77

there are two vectors for lambda =1. the two vectors I wrote above

Answer 78

ok where is the third one comes from?

Answer 79

from lambda =10

Answer 80

so for lambda = 1 we have 2 vectors and for lambda = 10 we have 3 vectors?

Answer 81

no...we get one vector from lambda =10. which gives us 3 total

Answer 82

ohh ok got it
Can you tell how to get the nilpoten?

Answer 83

multiply the matrix to get the zero vector?

Answer 84

it is not nilpotent

Answer 85

So I don't have to find it?

Answer 86

becasue the det(A) is not = 0 right?

Answer 87

if A was nilpotent then \(A^n=0\) for some \(n\)

what would imply that the determinant of A was zero. But the determinant of A is not zero. Thus A is not nilpotient

Answer 88

correct

Answer 89

ok so for this matrix the answer is done?

Answer 90

because I have two more matrices and I will try to do it by myself

Answer 91

you have what you need

Answer 92

ok Thank you so much for your help and your time

I really appreciate that

Answer 93

No problem ... I just hope you undrstand some of it now

Answer 94

yea I do