Usng the spectral decompostion of diagonalizable matrix

[syj]

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.

 

[Dick]

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

 

[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. :(

 

[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
$$2^{A}=\begin{bmatrix} 2&0&0\\\-2&2&2\\\-2&0&4\end{bmatrix}$$

 

[Ray Vickson]

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

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) = 2^x.

RGV

 

[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.