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I am having problem in diagonalization of \begin{bmatrix}-1&-1&-4\\-1&-4&-1\\-4&-1&-1\end{bmatrix} This is symmetric so it must be orthogonally diagonalizable. The eigen values are $-6,-3,3$. I did all the computations and found an orthogonal matrix $P$ which is \begin{bmatrix}1/\sqrt3&1/\sqrt6&-1/\sqrt2\\1/\sqrt3&-2/\sqrt6&0\\1/\sqrt3&1/\sqrt6&1/\sqrt2\end{bmatrix}

But some how $P^TAP$ is not the diagonal matrix? Can someone please find a mistake for me?

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    $\begingroup$ Look at $PAP^T$. $\endgroup$ – AlexR Mar 10 '15 at 9:08
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    $\begingroup$ looks right to me: $P$ is orthogonal and contains eigenvectors of $A$. How did you calculate $P^TAP$? By hand or with a computer? $\endgroup$ – draks ... Mar 10 '15 at 11:48
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Note: Since $P^T = P^{-1}$ for an orthogonal $P$, the equality $P^T A P = D$ is the same as $P^{-1} A P = D$.

Your answer is perfectly correct, just not in a simplified form.

We have:

$$P^T A P = \begin{bmatrix} -6 & 0 & 0 \\ \frac{2 \left(-\sqrt{\frac{2}{3}}-\frac{1}{\sqrt{6}}\right)}{\sqrt{3}}+\sqrt{2} & \sqrt{\frac{2}{3}} \left(-\sqrt{\frac{2}{3}}-\frac{1}{\sqrt{6}}\right)-2 & 0 \\ \frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{3}}+\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{3}} & \frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{6}}+\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{6}} & \frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{2}}-\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{2}} \\ \end{bmatrix}$$

For example, simplifying the bottom rightmost value yields:

$$\frac{-\frac{1}{\sqrt{2}}+2 \sqrt{2}}{\sqrt{2}}-\frac{\frac{1}{\sqrt{2}}-2 \sqrt{2}}{\sqrt{2}} = \dfrac{3}{2} + \dfrac{3}{2} = 3$$

If we simplify each value in the matrix, it reduces to:

$$P^T A P = \begin{bmatrix} -6 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix}$$

This is obviously a diagonal matrix $D$.

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