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:
<br /> f(A)=\sum_{i=0}^{t}f(\lambda_{i})E_{i}<br />

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

 

[syj]

After much puzzling, I believe I have the answer:

<br /> 2^{A}=2E_{1}+2^{2}E_{2}<br />

where I have found
<br /> E_{1}=\begin{bmatrix} 1&amp;0&amp;0\\1&amp;1&amp; \-1\\1&amp;0&amp;0\end{bmatrix}<br />

<br /> E_{2}=\begin{bmatrix}0&amp;0&amp;0\\-1&amp;0&amp;1\\-1&amp;0&amp;1\end{bmatrix}<br />

so I have
<br /> 2^{A}=\begin{bmatrix} 2&amp;0&amp;0\\\-2&amp;2&amp;2\\\-2&amp;0&amp;4\end{bmatrix}<br />

 

[Ray Vickson]

After much puzzling, I believe I have the answer:

<br /> 2^{A}=2E_{1}+2^{2}E_{2}<br />

where I have found
<br /> E_{1}=\begin{bmatrix} 1&amp;0&amp;0\\1&amp;1&amp; \-1\\1&amp;0&amp;0\end{bmatrix}<br />

<br /> E_{2}=\begin{bmatrix}0&amp;0&amp;0\\-1&amp;0&amp;1\\-1&amp;0&amp;1\end{bmatrix}<br />

so I have
<br /> 2^{A}=\begin{bmatrix} 2&amp;0&amp;0\\\-2&amp;2&amp;2\\\-2&amp;0&amp;4\end{bmatrix}<br />

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
<br /> p(x)=\sum_{i=1}^{n}p(\lambda_{i})h_{i}(x)<br />
would have to be such that
<br /> p(A)=2^{A}<br />
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.