1. stuck on diagonalizabilty problem

is the matrix diagonalizable, if so find an inverted matrix $\displaystyle Q$ and matrix $\displaystyle D$ such that $\displaystyle Q^{-1}AQ = D$

$\displaystyle A= \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right)$

so far I have:

$\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) - x \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} 1-x & 3 \\ \\3 & 1-x \end{array}\right)$

$\displaystyle det \left(\begin{array}{cc} 1-x & 3 \\ \\3 & 1-x \end{array}\right) = (1-x)^2-(3)(3) = 1-2x+x^2 -9 = x^2-2x- 8 = (x+2)(x-4)$

the eigenvalues are $\displaystyle \lambda_1 = -2 , \ \ \lambda_2 = 4$ with algebraic multiplicity 1 for both.

$\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) -(-2) \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} 3 & 3 \\ \\3 & 3 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\0 & 0 \end{array}\right)$

at which point I get stuck.

2. $\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) -(-2) \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} 3 & 3 \\ \\3 & 3 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\0 & 0 \end{array}\right)$

$\displaystyle \left(\begin{array}{cc} 1 \\ \\ -1 \end{array}\right) = \gamma_1$

$\displaystyle \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) -(4) \left(\begin{array}{cc} 1 & 0 \\ \\0 & 1 \end{array}\right) = \left(\begin{array}{cc} -3 & 3 \\ \\3 & -3 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\0 & 0 \end{array}\right)$

$\displaystyle \left(\begin{array}{cc} 1 \\ \\ 1 \end{array}\right) = \gamma_2$

$\displaystyle Q= \gamma_1 \bigcup \gamma_2 = \left(\begin{array}{cc} 1 \\ \\ -1 \end{array}\right) \bigcup \left(\begin{array}{cc} 1 \\ \\ 1 \end{array}\right) = \left(\begin{array}{cc} 1 & 1 \\ \\ -1 & 1 \end{array}\right)$

$\displaystyle Q^{-1} = \left(\begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ -1 & 1 & 0 & 1 \end{array}\right) = \left(\begin{array}{cc|cc} 1 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} \end{array}\right)$

$\displaystyle D = \left(\begin{array}{cc} -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{1}{2} \end{array}\right) \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 1 \\ \\ -1 & 1 \end{array}\right)$

is this right?

3. Originally Posted by lllll
$\displaystyle Q^{-1} = \left(\begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ -1 & 1 & 0 & 1 \end{array}\right) = \left(\begin{array}{cc|cc} 1 & 0 & -\frac{1}{2} & 0 \\ 0 & 1 & \frac{1}{2} & \frac{1}{2} \end{array}\right)$

$\displaystyle D = \left(\begin{array}{cc} -\frac{1}{2} & 0 \\ \frac{1}{2} & \frac{1}{2} \end{array}\right) \left(\begin{array}{cc} 1 & 3 \\ \\3 & 1 \end{array}\right) \left(\begin{array}{cc} 1 & 1 \\ \\ -1 & 1 \end{array}\right)$

is this right?
$\displaystyle Q^{-1}$ is wrong. It should be $\displaystyle \begin{pmatrix}\frac12&-\frac12\\\frac12&\frac12\end{pmatrix}$.

You can check whether your answer is right by multiplying out the matrix product for D, which ought to be a diagonal matrix with the eigenvalues as its diagonal entries.