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# 电子代写|数字信号处理代写Digital Signal Processing代考|EEE5502 The Eigenvalue Decomposition

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## 电子代写|数字信号处理代写Digital Signal Processing代考|The Eigenvalue Decomposition

To demonstrate the eigenvalue decomposition, we continue our earlier example, which calculated the eigenvalues and eigenvectors of the transformation matrix, $A$, shown below. We find the eigenvalues by solving for $\lambda$ in the polynomial $p(\lambda)=\operatorname{det}(A-\lambda I)$.
$$\begin{gathered} A=\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \ p(\lambda)=\operatorname{det}\left[\begin{array}{cc} 2-\lambda & -4 \ -1 & -1-\lambda \end{array}\right]=(2-\lambda)(-1-\lambda)-(-4)(-1)=\lambda^2-\lambda-6=(\lambda-3)(\lambda+2) \end{gathered}$$
Recall the original requirement of $A v_n=\lambda_n v_n$ for each eigenpair.
$$\begin{gathered} A \cdot v_n=\lambda_n \cdot v_n \ {\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \cdot\left[\begin{array}{c} -4 \ 1 \end{array}\right]=3 \cdot\left[\begin{array}{c} -4 \ 1 \end{array}\right]} \ {\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \cdot\left[\begin{array}{l} 1 \ 1 \end{array}\right]=-2 \cdot\left[\begin{array}{c} 1 \ 1 \end{array}\right]} \end{gathered}$$
We can easily combine these two systems of equations into one using notation $V$ to represent the matrix of eigenvectors and $A$ for the eigenvalues.
$A \quad v_1\left|v_2 \quad 3 \cdot v_1\right|-2 \cdot v_2$
$\left[\begin{array}{cc}2 & -4 \ -1 & -1\end{array}\right] \cdot\left[\begin{array}{c|c}-4 & 1 \ 1 & 1\end{array}\right]=\left[3 \cdot\left[\begin{array}{c}-4 \ 1\end{array}\right] \mid-2 \cdot\left[\begin{array}{l}1 \ 1\end{array}\right]\right]$
We now reformulate the expression.
$$\begin{gathered} A \ {\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \cdot\left[\begin{array}{c|c} -4 & 1 \ 1 & 1 \end{array}\right]=\left[\begin{array}{c|c} -4 & 1 \ 1 & 1 \end{array}\right] \cdot\left[\begin{array}{cc} \lambda_1=3 & 0 \ 0 & \lambda_2=-2 \end{array}\right]} \end{gathered}$$

## 电子代写|数字信号处理代写Digital Signal Processing代考|The Power of the Eigenvalue Decomposition

There are mathematical challenges which require that we compute an equation that takes a matrix as an argument. Can you image computing $A^{1 / 2}$, or $e^4$ ? Nothing that we have learned up to now allows us to even guess at how to compute the quantities above. To get a handle on these challenges, we will use the eigenvalue decomposition to compute the quantity $A^2$.
$$A \cdot A=\left(V \cdot \Lambda \cdot V^{-1}\right) \cdot\left(V \cdot \Lambda \cdot V^{-1}\right)$$
Note that the quantities $V^{-1}$ and $V$ appear next to one another and their product reduces to the identity matrix, $I$.
\begin{aligned} &A^2=V \cdot \Lambda \cdot I \cdot \Lambda \cdot V^{-1} \ &A^2=V \cdot \Lambda \cdot \Lambda \cdot V^{-1} \ &A^2=V \cdot \Lambda^2 \cdot V^{-1} \end{aligned}
This rather interesting manipulation proves that we can square a matrix A by firsts computing its eigenvalue decomposition, squaring the diagonal entries in $\Lambda$ and multiplying out $V \cdot \Lambda^{2 \cdot} V^{-1}$. Without a doubt, it would have been easier to just compute $A^2$ directly. But this example is pointing us into an interesting direction. Could we compute the square root of $A$, or equivalently $A^{1 / 2}$, in a similar manner?

## 电子代写|数字信号处理代写数字信号处理代考|特征值分解

$$\begin{gathered} A=\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \ p(\lambda)=\operatorname{det}\left[\begin{array}{cc} 2-\lambda & -4 \ -1 & -1-\lambda \end{array}\right]=(2-\lambda)(-1-\lambda)-(-4)(-1)=\lambda^2-\lambda-6=(\lambda-3)(\lambda+2) \end{gathered}$$

$$\begin{gathered} A \cdot v_n=\lambda_n \cdot v_n \ {\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \cdot\left[\begin{array}{c} -4 \ 1 \end{array}\right]=3 \cdot\left[\begin{array}{c} -4 \ 1 \end{array}\right]} \ {\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \cdot\left[\begin{array}{l} 1 \ 1 \end{array}\right]=-2 \cdot\left[\begin{array}{c} 1 \ 1 \end{array}\right]} \end{gathered}$$

$A \quad v_1\left|v_2 \quad 3 \cdot v_1\right|-2 \cdot v_2$
$\left[\begin{array}{cc}2 & -4 \ -1 & -1\end{array}\right] \cdot\left[\begin{array}{c|c}-4 & 1 \ 1 & 1\end{array}\right]=\left[3 \cdot\left[\begin{array}{c}-4 \ 1\end{array}\right] \mid-2 \cdot\left[\begin{array}{l}1 \ 1\end{array}\right]\right]$

$$\begin{gathered} A \ {\left[\begin{array}{cc} 2 & -4 \ -1 & -1 \end{array}\right] \cdot\left[\begin{array}{c|c} -4 & 1 \ 1 & 1 \end{array}\right]=\left[\begin{array}{c|c} -4 & 1 \ 1 & 1 \end{array}\right] \cdot\left[\begin{array}{cc} \lambda_1=3 & 0 \ 0 & \lambda_2=-2 \end{array}\right]} \end{gathered}$$

## 电子代写|数字信号处理代写数字信号处理代考|特征值分解的力量

.

$$A \cdot A=\left(V \cdot \Lambda \cdot V^{-1}\right) \cdot\left(V \cdot \Lambda \cdot V^{-1}\right)$$

\begin{aligned} &A^2=V \cdot \Lambda \cdot I \cdot \Lambda \cdot V^{-1} \ &A^2=V \cdot \Lambda \cdot \Lambda \cdot V^{-1} \ &A^2=V \cdot \Lambda^2 \cdot V^{-1} \end{aligned}

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