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I am to find $R$ of $A$ from the following formula: $R=S\sqrt{\Lambda }S^{-1}$. I however am getting the wrong answer (not the same as the textbook answer).

The answer in the textbook is: $R=S\sqrt{\Lambda }S^{-1}=\begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}$

Here are the starting variables, I have the eigenvalues and matrix:

$\lambda_{1} = 1,\ \lambda_{2} = 9$

$A=\begin{bmatrix} 5 & 4\\ 4 & 5 \end{bmatrix}$

The first eigenvector:

$\begin{bmatrix} 5-\lambda_{1} & 4\\ 4 & 5-\lambda_{1} \end{bmatrix}=\begin{bmatrix} 4 & 4\\ 4 & 4 \end{bmatrix}$

After some row elimination: \begin{bmatrix} 1 & 1\\ 0 & 0 \end{bmatrix}

$x_{1}=1$ $x_{2}=-1$

first eigenvector $(1,-1)$

Time to get the second eigenvector:

$\begin{bmatrix} 5-\lambda_{2} & 4\\ 4 & 5-\lambda_{2} \end{bmatrix}=\begin{bmatrix} -4 & 4\\ 4 & -4 \end{bmatrix}$

After some row elimination:

\begin{bmatrix} -1 & 1\\ 0 & 0 \end{bmatrix}

$x_{1}=1$ $x_{2}=1$

second eigenvector $(1,1)$

Both eigenvectors create the following matrix:

$S=\begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix} $

$S^{-1}=\begin{bmatrix} 1/2 & 1/2\\ -1/2 & 1/2 \end{bmatrix} $

I then compute the following formula $\Lambda =S^{-1}AS$

$\Lambda =\begin{bmatrix} 1/2 & 1/2\\ -1/2 & 1/2 \end{bmatrix}\begin{bmatrix} 5 & 4\\ 4 & 5 \end{bmatrix}\begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 9 \end{bmatrix}$

Which leads to:

$R=S\sqrt{\Lambda }S^{-1}=\begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}\begin{bmatrix} \sqrt{1} & 0\\ 0 & \sqrt{9} \end{bmatrix}\begin{bmatrix} 1/2 & 1/2\\ -1/2 & 1/2 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 3 \end{bmatrix}$

This answer is however wrong, the correct answer is:

$\begin{bmatrix} 2 & 1\\ 1 & 2 \end{bmatrix}$

Where did I go wrong?

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  • $\begingroup$ You computed the inverse of $S$ wrong. Check it by multiplication. $\endgroup$ Commented Dec 16, 2014 at 15:27
  • $\begingroup$ @CameronWilliams Do you not get the inverse via having an indentity matrix beside the matrix you wish to change perform row operations until the left side is the identity matrix (leaving the right side the inverse)? $\endgroup$ Commented Dec 16, 2014 at 15:35
  • $\begingroup$ You do but you somewhere messed up. Your negative in the lower left should be in the upper right. $\endgroup$ Commented Dec 16, 2014 at 15:35
  • $\begingroup$ @KarlMorrison for $2\times 2$ matrices, just use $$S^{-1} = \frac{1}{\det S} \begin{bmatrix} d & -b \\ -c & a \\ \end{bmatrix}$$ $\endgroup$ Commented Dec 16, 2014 at 15:39

1 Answer 1

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You computed the inverse of $S$ incorrectly. The actual inverse is

$$S^{-1} = \left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{array}\right).$$

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