# Find the power of a square matrix

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?

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 $$b$$.

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 $$\pm\sqrt{\frac{25\pm5\sqrt{21}}{42}}$$

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.