# Homework Help: Usng the spectral decompostion of diagonalizable matrix

1. Oct 1, 2012

### syj

1. The problem statement, all variables and given/known data
Find the spectral decompostion of
$$A= \begin{matrix} 1 & 0 & 0\\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{matrix}$$
and use this to find $$2^{A}$$

2. Relevant equations

3. 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.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 1, 2012

### Dick

You know how to write $e^A$ as an infinite series, right? $2^A=e^{A \log 2}$

3. Oct 1, 2012

### syj

The textbook I am using says:
$$f(A)=\sum_{i=0}^{t}f(\lambda_{i})E_{i}$$

I still don't understand how to put everything together to get the answer. :(

4. Oct 1, 2012

### syj

After much puzzling, I believe I have the answer:

$$2^{A}=2E_{1}+2^{2}E_{2}$$

where I have found
$$E_{1}=\begin{bmatrix} 1&0&0\\1&1& \-1\\1&0&0\end{bmatrix}$$

$$E_{2}=\begin{bmatrix}0&0&0\\-1&0&1\\-1&0&1\end{bmatrix}$$

so I have
$$2^{A}=\begin{bmatrix} 2&0&0\\\-2&2&2\\\-2&0&4\end{bmatrix}$$

Last edited: Oct 1, 2012
5. Oct 1, 2012

### Ray Vickson

Why did it require much puzzling? You gave the answer in a previous post:
$$f(A) = f(\lambda_1) E_1 + f(\lambda_2) E2,$$
and f(x) = 2x.

RGV

6. Oct 1, 2012

### syj

ha ha , it required puzzling, because I knew that
$$p(x)=\sum_{i=1}^{n}p(\lambda_{i})h_{i}(x)$$
would have to be such that
$$p(A)=2^{A}$$
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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