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I have the following matrix :

$$ \mathbf A =\begin{bmatrix} 100 & 0 \\ 0 & 1 \\ \end{bmatrix}$$

I have to compute $ \mathbf A^{-1/2}$.

So I need spectral decomposition, $$ \mathbf A = \mathbf P \mathbf \Lambda\mathbf P',$$

$\mathbf P$ be a matrix with normalized eigenvectors and $\mathbf \Lambda$ is a diagonal matrix with diagonal elements be eigenvalues.

Eigenvalues of $ \mathbf A$ is $100$ and $1$.

But I stumbled to calculate eigenvector.

The characteristic equation is :

$$ \begin{bmatrix} 100 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}= 100\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}$$

$$\Rightarrow 100x_1 = 100x_1$$

$$x_2 = 100 x_2$$

How is $x_2 = 100 x_2$ possible ?

And is there a simpler way to calculate any power of a diagonal matrix , for example, $ \mathbf A^{-1/2}$ ?

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  • $\begingroup$ @Moo yes, do I need only $ \mathbf A^{-1/2} =\begin{bmatrix} 100^{-1/2} & 0 \\ 0 & 1^{-1/2} \\ \end{bmatrix}$ ? $\endgroup$ – user 31466 May 7 '16 at 3:00
  • $\begingroup$ One More question : The characteristic equation is : $$ \begin{bmatrix} 100 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}= 100\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}$$ $$\Rightarrow 100x_1 = 100x_1$$ $$x_2 = 100 x_2$$ How is $x_2 = 100 x_2$ possible ? $\endgroup$ – user 31466 May 7 '16 at 3:05
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    $\begingroup$ $x_2=0$, but $x_1$ is arbitrary, so $(1,0)$ is an eigenvector. $\endgroup$ – Hans Lundmark May 7 '16 at 3:06
  • $\begingroup$ Why are you diagonalizing a matrix that's already diagonal? $\endgroup$ – Erick Wong May 7 '16 at 6:20
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For diagonal matrices, the normalized eigenvectors are always standard basis vectors, i.e. vectors of all zeros except a single 1 in a particular coordinate, e.g., ${\bf{e}}_1=(1,0)$ and ${\bf{e}}_2=(0,1)$ in your example. For eigenvalue 100, the corresponding eigenvector is $(1,0)$ which means that $x_2=0$ (note that solves $x_2=100x_2$).

The power $k$ of a diagonal matrix $$ D=\left\|\begin{array}{ccc}d_1 &\cdots &0\\ \vdots & &\vdots\\ 0 &\cdots &d_n\end{array}\right\| $$ is found as
$$ D=\left\|\begin{array}{ccc}d^k_1 &\cdots &0\\ \vdots & &\vdots\\ 0 &\cdots &d^k_n\end{array}\right\| $$

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$A=SDS^{-1}$, and more generally $A^k=SD^kS^{-1}$ where D is the diagonal matrix composed from eigenvalues, S is the matrix of eigenvectors, and $S^{-1}$ is the inverse of S.

D is $ \begin{bmatrix} 100&0\\0&1 \end{bmatrix}$as you have already found.

To find eigenvectors, you need to follow this format $\left[ \begin{array}{cc|c} 100-\lambda&0&0\\0&1-\lambda&0 \end{array}\right]$
When $\lambda=100$, $\left[\begin{array}{cc|c} 0&0&0\\0&-99&0 \end{array}\right]$, $v_1=\begin{bmatrix}1\\0\end{bmatrix}$
When $\lambda=1$, $\left[\begin{array}{cc|c} 100&0&0\\0&0&0 \end{array}\right]$, $v_2=\begin{bmatrix}0\\1\end{bmatrix}$
Thus $S=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=S^{-1}$
$A^{1/2}=SD^{1/2}S^{-1}=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}\begin{bmatrix} 100^{1/2}&0\\0&1^{1/2} \end{bmatrix}\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=\begin{bmatrix} 10&0\\0&1 \end{bmatrix}$

And $A^{-1/2}=(A^{1/2})^{-1}$, $A^{-1/2}$ is the inverse of $A^{1/2}$, which is $\begin{bmatrix} \frac{1}{10}&0\\0&1 \end{bmatrix}$

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Your matrix is in Jordan Normal Form (diag) thus: $$f(A) = \begin{bmatrix}f(a_{11})& \\&f(a_{22})\end{bmatrix}$$

So yes you can use: $$A^{-1/2} = \begin{bmatrix}a_{11}^{-1/2}& \\&a_{22}^{-1/2}\end{bmatrix}$$

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