Application of Matrix Diagonalization I'm reading a book about inverse analysis and trying to figure out how the authors do the inversion.
Assume that matrix $C$ is
$$
  C
~=~
  \begin{bmatrix}
    88.53 & -33.60 & -5.33 \\
    -33.60 & 15.44 & 2.67 \\
    -5.33 & 2.67 & 0.48
  \end{bmatrix}
$$
and at some point authors diagonalize this matrix to calculate matrix $P$ using
$$
  C^{-1}
~=~
  P^{\rm t} L P
$$
where $L$ is a diagonal matrix of positive eigenvalues and the columns of $P$ are orthonormal eigenvectors.
The above equation for diagonalizing inverse of $C$ is a bit different from what is usually used and therefore I cannot calculate $P$ correctly (same as the book!).
So, that would be great if somebody can show me the way to calculate $P$.
$$
  P
~=~
  \begin{bmatrix}
    0.93 & 0.36 & -0.03 \\
    -0.36 & 0.90 & -0.23 \\
    -0.06 & 0.23 & 0.97
  \end{bmatrix}
$$
 A: The spectral theorem ensures that since $C$ is symmetric it has 3 real eigenvalues and its eigenspaces are orthogonal. Let $\lambda_1,\lambda_2,\lambda_3$ be the eigenvalues and $\vec v_1,\vec v_2,\vec v_3$ be orthonormal eigenvectors (whose existence is granted by the spectral theorem; note that $\lambda_i$ need not be distinct, since you can always orthonormalize a basis with the Gram-Schmidt process). $\vec v_i$ are such that $C\vec v_i=\lambda_iv_i$ ($i=1,2,3$): in matrix notation this means
$$
  C
  \begin{bmatrix}
  ~\\
    \vec v_1 & \vec v_2 & \vec v_3
  \\~  
  \end{bmatrix}
~=~
  \begin{bmatrix}
  ~\\
    \vec v_1 & \vec v_2 & \vec v_3
  \\~  
  \end{bmatrix}
  \begin{bmatrix}
    \lambda_1\\
    & \lambda_2\\
    && \lambda_3
  \end{bmatrix}
$$
where $\vec v\in\mathbb R^{3\times 1}$ is intended as a column vector. Therefore, setting
$$
  D
=
  \begin{bmatrix}
    \lambda_1\\
    & \lambda_2\\
    && \lambda_3
  \end{bmatrix}
\quad\text{and}\quad
  P
=
  \begin{bmatrix}
  ~\\
    \vec v_1 & \vec v_2 & \vec v_3
  \\~  
  \end{bmatrix}
$$
you have that $C=PDP^{-1}$. Since the columns of $P$ are orthonormal, it follows that $P^{-1}=P^{\rm t}$ is the transpose of $P$, therefore
$$
  C=PDP^{\rm t}
$$
Now, if you want to compute any power of $C$ (or just any integer power, if you work with real matrices and $\lambda_i$ are not all non-negative) you have that
$$
 C^n = PD^nP^{\rm t}
$$
(since $P^{\rm t}P=\rm I$, the identity matrix). Chosing $n=-1$ you have
$$
  C^{-1}
~=~
  P
  \begin{bmatrix}
    \lambda_1^{-1}\\
    & \lambda_2^{-1}\\
    && \lambda_3^{-1}
  \end{bmatrix}
  P^{\rm t}
$$
so that $L=D^{-1}$.
To sum up, $P$ is the matrix whose $i$-th column is the eigenvector $\vec v_i$ with eigenvalue $\lambda_i$, where $\vec v_1,\vec v_2,\vec v_3$ are orthonormal.
(In your case, since you require $C^{-1}=P^{\rm t}LP$ rather than $PLP^{\rm t}$, $P$ is the matrix whose $i$-th row is the eigenvector $\vec v_i$, i.e. the transpose of the $P$ above.
