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I know how to find inverse of a matrix,but am having trouble solving when a non-diagonal square matrix is raised by negative power half. $$ \begin{bmatrix} & 1 & 0.4 \\ & 0.4 & 1 \\ \end{bmatrix} $$

Can anyone please help me to find solution of the above matrix when its raised by negative power 0.5?

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3 Answers 3

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Use the Cayley-Hamilton theorem to write $A^{-1/2}=aI+bA$ for some unknown scalars $a$ and $b$. The eigenvalues of $A$ also satisfy this equation, i.e., for any eigenvalue $\lambda$, $a+b\lambda=\lambda^{-1/2}$. Compute the eigenvalues of your matrix, which are relatively “nice,” substitute into the above equation, and solve the resulting small system of linear equations for $a$ and $b$.

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Hint...one way is to assume that the required matrix has the form $$\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$$ Then $$\left(\begin{matrix}a&b\\c&d\end{matrix}\right)^2=\left(\begin{matrix}1&0.4\\0.4&1\end{matrix}\right)^{-1}$$ You can expand the LHS and evaluate the RHS and deduce the possible values of $a, b, c, d$ by solving the resulting system of equations.

The answers are not pleasant to look at. For example, $a$ takes the form $$\pm\sqrt{\frac{25\pm5\sqrt{21}}{42}}$$

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Since it is a symmetric matrix it can be diagonalized, thus we have its eigen-decomposition as $$ A = U D U^{-1} $$ where $U$ is orthogonal matrix whose columns are the eigenvectors of $A$ and $D$ is a diagonal matrix whose diagonal elements are the eigenvalues of $A$.

Let $D^{1/2}$ be the square root matrix of $D$, this is easy to compute because $D$ is diagonal, in particular: $$ \forall i: (D^{1/2})_{ii} = (D_{ii})^{1/2} $$ Now, consider $B = U D^{1/2} U^{-1}$, then $$ B^2 = BB = (U D^{1/2} U^{-1})(U D^{1/2} U^{-1}) = U D^{1/2} U^{-1} U D^{1/2} U^{-1} = U D^{1/2} D^{1/2} U^{-1} = U D U^{-1} = A $$ and thus $B$ is a square root of $A$ (in general, a matrix can have several roots). If you consider only the semi-positive definite root, then $B = A^{1/2}$.

In your case: $$ D = \begin{bmatrix} 1.4 & 0 \\ 0 & 0.6 \end{bmatrix} $$ and $$ U = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} $$ I will leave it up to you to do the multiplications and verify that $$ A^{1/2} = U D^{1/2} U^{-1} = \begin{bmatrix} 0.97890631 & 0.20430964 \\ 0.20430964 & 0.97890631 \end{bmatrix} $$

For a negative power of half you can repeat the same algorithm, but this time computing $D^{-1/2}$ instead of $D^{1/2}$, if you define $B = U D^{-1/2} U^{-1}$ (provided that $A$ and thus $D$ are positive-definite) you get that $B^2 = A^{-1}$ so $B = A^{-1/2}$.

I computed it numerically via this algorithm and got $$ A^{-1/2} = \begin{bmatrix} 1.06807435 & -0.2229201 \\ -0.2229201 & 1.06807435 \end{bmatrix} $$ so after you do the calculations manually (dragging ugly square roots) you can compare results using a calculator or Wolfram Alpha.

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