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For the following symmetric matrix, I want to compute the square root.

$$A=\begin{bmatrix} D & 0 & 0 \\ 0 & D/\sin^2(\theta) & -2D \cos(\theta)/\sin^2(\theta)\\ 0 & -2D \cos(\theta)/\sin^2(\theta) & D/\sin^2(\theta) \end{bmatrix}$$

where $D > 0$ and $\theta \in[0,\pi]$. To compute the square root, I need to find its eigenvalues and eigenvectors. The eigenvalues I found are

$$\lambda_1=D \qquad \qquad \lambda_2=D/\sin^2(\theta)$$

but confused on how to obtain the eigenvectors. Any help would be appreciated.

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    $\begingroup$ Why not just multiply the matrix by itself to get the square? $\endgroup$ – Nick Alger Sep 26 '16 at 15:00
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    $\begingroup$ 1st sentence says "square root", 2nd sentence says "square". So I'm guessing Mari wants the square root. Right, Mari? $\endgroup$ – Gerry Myerson Sep 27 '16 at 9:18
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    $\begingroup$ Yes, I need the square root. $\endgroup$ – David Sep 28 '16 at 8:57
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You should go to get the basis vectors for the nullspaces of the matrices $A-\lambda I$, as the eigenvectors lay on those subspaces. This happens due to the fact that $(A-\lambda I)\nu = 0,\forall\lambda$. Using this fact, for the first eigenvalue:

$A-\lambda I =\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & \frac{D}{\sin^2\theta}-D & -2D\frac{\cos\theta}{\sin^2\theta} \\ 0 & -2D\frac{\cos\theta}{\sin^2\theta} & \frac{D}{\sin^2\theta}-D \end{array} \right)$, so now you should just compute a basis for the nullspace of this matrix, and that basis will form a group of the eignevectors of the matrix A (then you should compute the values for the others to get the complete eigenbasis. For this $\lambda_1$, it's easy to see that $\nu_1=(1,0,0)$ is the basis of the nullspace for that matrix, so $\nu_1$ is an eigenvector for A.

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  • $\begingroup$ Thank you for your response. For the second eigenvalue $\lambda=D/\sin^2\theta$, the eigenvectors I found are $\nu_2=(0,1,-1)$ and $\nu_3=(0,1,1)$. So to find the square root of the matrix, we need to compute the following $\endgroup$ – David Sep 27 '16 at 9:03
  • $\begingroup$ $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & -1 & 1\end{bmatrix}.\begin{bmatrix} D & 0 & 0\\ 0& D/\sin^2\theta & 0 \\0 & 0 & D/\sin^2\theta \end{bmatrix}.\begin{bmatrix} 1 & 0 & 0\\ 0 & 1/2 & -1/2\\ 0 & 1/2 & 1/2\end{bmatrix}$$. $\endgroup$ – David Sep 27 '16 at 9:15
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    $\begingroup$ Why does that gives you the square root? Have you tried squaring it, to see whether you get back the original matrix? $\endgroup$ – Gerry Myerson Sep 27 '16 at 9:20
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    $\begingroup$ That's the diagonalized form of A, the square root by diagonalization would be $A^{\frac{1}{2}}=V\Lambda^{\frac{1}{2}}V^{-1}$, where the square root of a diagonal matrix is the same matrix with it's component's square roots. If what you are looking for is the square of A, as you were saying in the question, then it would be $A^2=V\Lambda^2 V^{-1}$ $\endgroup$ – Josu Etxezarreta Martinez Sep 27 '16 at 9:20
  • $\begingroup$ I compute the square root of the diagonal matrix and then i multiplied the three matrices and obtained $$ \begin{bmatrix} D & 0 & 0\\ 0 & D/\sin^2\theta & 0\\ 0& 0 & D/\sin^2\theta\end{bmatrix}$$ If I multiply it by itself, clearly it does not return the original matrix. $\endgroup$ – David Sep 27 '16 at 9:51
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Solving the characteristic equation: $$ 0=det|B-\lambda I|=\begin{vmatrix} D-\lambda & 0 & 0\\ 0 & \frac{D}{\sin^2\theta}-\lambda & -2D\frac{\cos\theta}{\sin^2\theta} \\ 0 & -2D\frac{\cos\theta}{\sin^2\theta} & \frac{D}{\sin^2\theta}-\lambda \end{vmatrix}=(D-\lambda)\Big[\big(\frac{D}{\sin^2\theta}-\lambda\big)^2 -4D^2\frac{\cos^2\theta}{\sin^4\theta}\Big]. $$ So, the eigenvalues are: $$ \lambda_1=D, \quad \lambda_2=\frac{D(1+2\cos\theta)}{\sin^2\theta}, \quad \lambda_3=\frac{D(1-2\cos\theta)}{\sin^2\theta}. $$ and the eigenvectors are: $$ \mathbf{v_1}=(1,0,0). \quad \mathbf{v_2}=(0,1,-1), \quad \mathbf{v_3}=(0,1,1). $$

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