According to the spectral decomposition theorem, every real symmetric matrix has a spectrum (it can be diagonalised by an orthonormal matrix). Moreover, it can be shown that the diagnolising matrix is its eigenmatrix. I need to use use this to diagnolise the matrix $C$, and I must elaborate the calculations.
I have this matrix:
$$ C = \begin{bmatrix} 5.6 & 7.2\\ 7.2 & 10.8 \end{bmatrix} $$
So $C$ is symmetric. If there is an invertible $n\times n$ matrix $Q$, and a diagonal matrix $D$ such that $C = QDQ^{-1}$, then an $n\times n$ matrix $C$ is diagonalisable.
First, I need to detect the eigenvalues of the matrix, so:
$$\det(C – \lambda I) = 0$$
$$\det\left(\left[\begin{matrix}
5.6&7.2\\\\
7.2&10.8\\
\end{matrix}\right] – \lambda \left[\begin{matrix}
1&0\\\\
0&1\\
\end{matrix}\right]\right)=0$$
$$\det\left[\begin{matrix}
5.6 – \lambda&7.2\\\\
7.2&10.8 – \lambda\\
\end{matrix}\right] = 0$$
Now solving the determinant:
$$(5.6 – \lambda)(10.8 – \lambda) – (7.2)^2 = 0$$
$$60.48 – 5.6\lambda – 10.8 \lambda + \lambda^2 -51.84 = 0$$
$$8.64 – 16.4\lambda + \lambda^2 = 0$$
We get $\lambda_1 = 15.8551$ and $\lambda_2 = 0.544936$
Hence, we got this diagonal matrix $D =\left[\begin{matrix}
15.8551&0\\\\
0&0.544936\\
\end{matrix}\right]$
The next step is to find the corresponding eigenvectors which we will use as the columns of matrix $Q$. So, we start with $\lambda_1=15.8551$ , and we assign vector $x$, and this vector is an eigenvector:
$$(C – \lambda I)x = 0$$
$$\left(\left[\begin{matrix}
5.6&7.2\\\\
7.2&10.8\\
\end{matrix}\right] – (15.8551) \left[\begin{matrix}
1&0\\\\
0&1\\
\end{matrix}\right]\right)x = 0$$
After calculations, we get:
$$\left[\begin{matrix}
-10.2551&7.2\\\\
7.2&-5.0551\\
\end{matrix}\right] \left[\begin{matrix}
x_1\\\\
x_2\\
\end{matrix}\right] = \left[\begin{matrix}
0\\\\
0\\
\end{matrix}\right]$$
Write the equations which results in multiplying the matrix with vector $x$:
$$-10.2551x_1 + 7.2x_2 = 0$$
$$7.2x_1 – 5.0551x_2 = 0$$
After solving I get $x_1 = 0$ and $x_2 = 0$
And those are the eigervectors for $\lambda_1 = 15.8551$, but I get the same result for $\lambda_2 = 0.544936$. Which means that $Q$ is an all-zero matrix, and that matrix is non-invertible, which is not good, cause this way I can't diagonalise $C$.
So there it is, this is my problem, I'm stuck at this $Q$ matrix, I must be doing something wrong but I don't know what it is exactly. $Q$ must be invertible, how can I find it?
Thank you!
Best Answer
For a $2 \times 2$ symmetric matrix $C$, there is a guaranteed method to diagnolize it, and write it as $ C = R D R^T $, where $D$ is a diagonal matrix of the eigenvalues and $R$ is a unitary matrix whose columns are the corresponding eigenvectors. The method goes as follows
Find $\theta = \dfrac{1}{2} \tan^{-1} \dfrac{2 C_{12} } { C_{11} - C_{22} } $
Then
$ R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
And the diagonal elements of $D$ (whare are the eigenvalues) are given by
$ D_{11} = C_{11} \cos^2 \theta + C_{22} \sin^2 \theta + 2 C_{12} \sin \theta \cos \theta $
$D_{22} = C_{11} \sin^2 \theta + C_{22} \cos^2 \theta - 2 C_{12} \sin \theta \cos \theta $
Applying this to your matrix
$ C = \begin{bmatrix} 5.6 && 7.2 \\ 7.2 && 10.8 \end{bmatrix} $
Then
$ \theta = \dfrac{1}{2} \tan^{-1} \dfrac{ 2(7.2) }{ 5.6 - 10.8} =\dfrac{1}{2} \tan^{-1} (-2.76923076923) = -0.61212873118 $
Therefore,
$ R = \begin{bmatrix} 0.81842668 && 0.5746109711 \\ -0.5746109711 && 0.81842668 \end{bmatrix} $
And
$ D = \begin{bmatrix} 0.5449363164 && 0 \\ 0 && 15.8550636836 \end{bmatrix} $