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Usng the spectral decompostion of diagonalizable matrix

  • Thread starter syj
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  • #1
syj
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0

Homework Statement


Find the spectral decompostion of
$$
A=

\begin{matrix}
1 & 0 & 0\\
-1 & 1 & 1 \\
-1 & 0 & 2
\end{matrix}
$$
and use this to find $$ 2^{A} $$

Homework Equations





The Attempt at a Solution


I have found the eigenvalues to be : $$ \lambda_{1}=1 \text{ and } \lambda_{2}=2 $$

My textbook (matrices and linear transformations, Cullen) uses lagrange polynomials to find the spectral projectors:
$$
h_{1}(x)=\frac{x-2}{1-2}=-(x-2)\\
h_{2}(x)=\frac{x-1}{2-1}=(x-2)\\

E_{1}=h_{1}(A)\\

E_{2}=h_{2}(A)\\

\therefore A=E_{1}+2E_{2}

$$
However I now need to use this to solve for $$ 2^{A} $$
I think I need to write $$ 2^{x} $$ as an infinite series, but I am not sure how to do this.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
You know how to write [itex]e^A[/itex] as an infinite series, right? [itex]2^A=e^{A \log 2}[/itex]
 
  • #3
syj
55
0
The textbook I am using says:
[tex]
f(A)=\sum_{i=0}^{t}f(\lambda_{i})E_{i}
[/tex]

I still don't understand how to put everything together to get the answer. :(
 
  • #4
syj
55
0
After much puzzling, I believe I have the answer:

[tex]
2^{A}=2E_{1}+2^{2}E_{2}
[/tex]

where I have found
[tex]
E_{1}=\begin{bmatrix} 1&0&0\\1&1& \-1\\1&0&0\end{bmatrix}
[/tex]

[tex]
E_{2}=\begin{bmatrix}0&0&0\\-1&0&1\\-1&0&1\end{bmatrix}
[/tex]

so I have
[tex]
2^{A}=\begin{bmatrix} 2&0&0\\\-2&2&2\\\-2&0&4\end{bmatrix}
[/tex]
 
Last edited:
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
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After much puzzling, I believe I have the answer:

[tex]
2^{A}=2E_{1}+2^{2}E_{2}
[/tex]

where I have found
[tex]
E_{1}=\begin{bmatrix} 1&0&0\\1&1& \-1\\1&0&0\end{bmatrix}
[/tex]

[tex]
E_{2}=\begin{bmatrix}0&0&0\\-1&0&1\\-1&0&1\end{bmatrix}
[/tex]

so I have
[tex]
2^{A}=\begin{bmatrix} 2&0&0\\\-2&2&2\\\-2&0&4\end{bmatrix}
[/tex]
Why did it require much puzzling? You gave the answer in a previous post:
[tex] f(A) = f(\lambda_1) E_1 + f(\lambda_2) E2,[/tex]
and f(x) = 2x.

RGV
 
  • #6
syj
55
0
ha ha , it required puzzling, because I knew that
[tex]
p(x)=\sum_{i=1}^{n}p(\lambda_{i})h_{i}(x)
[/tex]
would have to be such that
[tex]
p(A)=2^{A}
[/tex]
but because of a silly mistake of writing a 0 instead of a 1, I was not getting that conclusion and so thought all my work was incorrect.
 

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