Question

Determine a real conical form of the given matrix and give a change of coordinates matrix P that brings the matrix into this form.
[ 0 -2 1 ]
[ 2 2 -1 ]
[ 0 -2 2 ]

Answer 1

The matrix P I found was singular and therefore wrong (no P inverse exists). I can give you the diagonal of the matrix, canonical form and matrix P that I calculated but they could be wrong that is why I am waiting for someone to request them instead of just typing them.

Answer 2

The matrix you listed isn't singular, right?

Answer 3

It is the matrix given in the question and is not singular.

Answer 4

Hmm, but the determinate....

Answer 5

the determinate, calculated using a graphing calculator is 4.

Answer 6

Doesn't the determinate have to be \(0\) for it to be singular?

Answer 7

yes the matrix P that I found when I attempted the problem was
[ 0.5 1 -0.5 ]
[ 0 0.5 -0.5 ]
[ 1 1 0 ]
it is singular which does not fit with what P is suppose to be.

Answer 8

Does that clear things up or should I try clarifying more?

Answer 9

I'm very confused about what matrix is what right now.

Answer 10

So the given matrix, let's call it \(A\) is some matrix.
Then the matrix you typed out is \(B\).
And \(PA=B\)? or something?

Answer 11

Or maybe \[
PAP^{-1}=B
\]

Answer 12

yes but the matrix I most recently typed out is P. This is why I know it is wrong, the matrix P that I got has no inverse.

Answer 13

I meant yes to the second formula you typed out

Answer 14

I did find the value "B" in the formula you posted to be
[ 2 0 0 ]
[ 0 1 1 ]
[ 0 -1 1 ]

Answer 15

but that could be incorrect as well.

Answer 16

B is suppose to be the real canonical form sorry I mispelt it in the original question

Answer 17

\[
A =
\begin{bmatrix}
0 & -2 & 1\\
2 & 2 & -1\\
1 & -2 & 2
\end{bmatrix}
\]\[
B =
\begin{bmatrix}
2 & 0 & 0\\
0 & 1 & 1\\
0 & -1 & 1
\end{bmatrix}
\]

Answer 18

So they give you \(A\) and \(B\)?

Answer 19

No just A. I attempted to solve for the other matrices and was incorrect.

Answer 20

What's a real canonical form? Can you use the diagonalization: \[
A=S^{-1}\Lambda S
\]?

Answer 21

Whoops, I mean \[
A = S\Lambda S^{-1}
\]

Answer 22

It uses imaginary numbers, and is similar but different to diagonalization. If you have not studied imaginary numbers I thank you for your time but you cannot help me.

Answer 23

have you studied imaginary numbers?

Answer 24

Yeah

Answer 25

http://www.wolframalpha.com/input/?i=jordan+normal+form+calculator

Answer 26

is that real canonical form? My textbook and professor have never referred to it as Jordan normal form.

Answer 27

I haven't heard real canonical form mentioned anywhere, instead it goes to jordan canonical form.

Answer 28

Different classes might have different names. This happens quite often.

Answer 29

if a and b+ci are eignevalues of A then the real canonical form is
[ a 0 0 ]
[ 0 b c ]
[ 0 -c b ]

Answer 30

*eigenvalues

Answer 31

does that fit the definition of Jordan canonical form?

Answer 32

Very similar, but no, they seem different.

Answer 33

Either way, you got eigen values of \(2, 1\pm i\)?

Answer 34

yes

Answer 35

now I need help finding the matrix P that converts matrix A into its real conical form.

Answer 36

You weren't taught any methodology for that?

Answer 37

I am suppose to find eignvectors that correspond to the eigenvalues 2 and 1+i to convert them into the matrix P.

Answer 38

http://www.wolframalpha.com/input/?i=eigen+value+calculator&f1=%7B%7B0%2C-2%2C1%7D%2C%7B2%2C2%2C-1%7D%2C%7B0%2C-2%2C2%7D%7D&f=Eigenvalue.eigvalmatrix_%7B%7B0%2C-2%2C1%7D%2C%7B2%2C2%2C-1%7D%2C%7B0%2C-2%2C2%7D%7D

Answer 39

I messed up my eigenvector corresponding to 1+i but I can't see where

Answer 40

The only issue is that these vectors have imaginary components, so I suppose they are not the \(P\) you are looking for.

Answer 41

yes they are I have another way to convert them into the matrix P that is similar to the way I stated you found the real canonical form.

Answer 42

I thought the eigenvector corresponding to 1=i was
[ 1 ] [-0.5]
[0.5] + [-0.5]i
[ 1 ] [ 0 ]

Answer 43

sorry I meant corresponding to 1+i

Answer 44

For \(1+i\) they have:\[
\begin{bmatrix}
0.5 \\
0.5 \\
1
\end{bmatrix}
+
\begin{bmatrix}
0.5 \\
-0.5 \\
0
\end{bmatrix}
i
\]

Answer 45

For \(1-i\) they have: \[

\begin{bmatrix}
0.5 \\
0.5 \\
1
\end{bmatrix}
+
\begin{bmatrix}
-0.5 \\
0.5 \\
0
\end{bmatrix}
i
\]

Answer 46

what do you get for \(P\) when you use these?

Answer 47

I get the correct value of P which is (tried the operation in graphing calculator and it checked out)
[ 0.5 0.5 0.5]
[ 0 0.5 -0.5]
[ 1 1 0 ]

Answer 48

I have to use the basis for eigenvalue 2 to be
[ 0.5 ]
[ 0 ]
[ 1 ]

Answer 49

all they did was multiply it be two to get rid of the fraction but in order to get the correct value of P I cannot take that basis.

Answer 50

How in the world are you entering a 3x3 matrix in the equation area?

Answer 51

```
\begin{bmatrix}
11 & 12 & 13 \\
21 & 22 & 23 \\
31 & 32 & 33
\end{bmatrix}
```

Answer 52

I type it all up without equation editor.

Answer 53

Ooooooooooo fancy

Answer 54

\begin{bmatrix}-1-i & -2 & 1 \\2 & 1-i & -1 \\0 & -2 & 1-i\end{bmatrix}

Answer 55

http://www.wolframalpha.com/input/?i=determinate++calculator&f1=%7B%7B1%2F2%2C1%2F2%2C1%2F2%7D%2C%7B0%2C1%2F2%2C-1%2F2%7D%2C%7B1%2C1%2C0%7D%7D&f=Determinant.detmatrix_%7B%7B1%2F2%2C1%2F2%2C1%2F2%7D%2C%7B0%2C1%2F2%2C-1%2F2%7D%2C%7B1%2C1%2C0%7D%7D

The determinate is nonzero, it is definitely invertable.

Answer 56

I know the matrix P I originally had was not invertible that was how I knew it was incorrect.

Answer 57

The matrix I just entered is what I started to try and row reduce

Answer 58

then I went swapped the negative of row one with half of row two to get
\begin{bmatrix}1 & 0.5(1-i) & -0.5 \\1+i & 2 & -1 \\0 & -2 & 1-i\end{bmatrix}

Answer 59

Given the eigen vectors \(v_1, v_2, v_3\) how did you calculate \(P\)?

Answer 60

then I subtracted (1+i)row one from row two to get
\begin{bmatrix}1 & 0.5(1-i) & -0.5 \\0 & 1 & -0.5(1-i) \\0 & -2 & 1-i\end{bmatrix}

Answer 61

Since I used the eigenvalue 1+i I use the eigenvectors corresponding to that eigenvalue and the whole number eigenvalue (2)
If V1 is the eigenvector corresponding to eigenvalue 2 and
V2 is the eigenvector corresponding to eigenvalue 1+i
the I rewrite V2 and separate the real and imaginary components so that
V2=W1+W2*i where W1 and W2 are vectors
then P is the matrix with columns ( V1 W1 W2 )

Answer 62

Ah, I see!

Answer 63

thats why its so important that I messed up when calculating the eigenvector corresponding to 1+i

Answer 64

Is there an easy way of finding \(P^{-1}\)?

Answer 65

Do you just use Gauss-Jordan?

Answer 66

not that I know of but luckily the question only asks for me to find P. I don't have to find its inverse

Answer 67

I have no idea what that is but I used
\[\det(A-\lambda I)=0\] to get the eigenvalues

Answer 68

Man, the only trouble with linear algebra is that the methodology is so complicated it's easy to forget things.

Answer 69

No Idea what what is?

Answer 70

Oh, Gauss-Jordan method is just a way of inverting a matrix.

Answer 71

then row reduced\[A-\lambda I\] to get the eigenvectors (usually you do this to get a basis for the eigenspace)

Answer 72

Yeah, I have no problem with diagonalization with real eigen values.

Answer 73

I'm just a bit rusty on complex and repeated eigenvalues.

Answer 74

I don't either and getting the real canonical form is very similar but I seem to have messed up in my row reduction for eigenvalue 1+i and I don't know where thats why I started typing out my steps to see if you could spot where I went wrong in my row reduction.

Answer 75

Solving for Null space is tricky enough... doing it with complex numbers must be really tricky.

Answer 76

yes . . . and I think I may have just found my mistake one second

Answer 77

I did now I have the correct eigenvector and therefore value of P. Thanks so much, I messed up in the last step of my row reduction which had two complex numbers multiplying. I did not calculate (1-i)(1-i) properly

Answer 78

Thanks for sticking through it with me.

Answer 79

no problem