# Find an orthogonal matrix $M$ and diagonal matrix $D$ such that $M^{T}AM=D$

Hello can you help me with that. I tried all my best but I don't know how to find $$M$$ orthogonal matrix For the matrix $$A= \begin{bmatrix} 2 & 2 & -2 \\ 2 & -1 & 4 \\ -2 & 4 & -1 \end{bmatrix}$$ Find an orthogonal matrix $$M$$ and diagonal matrix $$D$$ such that $$M^{T}AM=D$$

At first I tried to find Diagonal matrix using eigenvalues

$$\begin{bmatrix} 2 - \lambda & 2 & -2 \\ 2 & -1- \lambda & 4 \\ -2 & 4 & -1- \lambda \end{bmatrix}$$

$$-\lambda^3 +27\lambda -54 = -(\lambda-3)(\lambda+6)(\lambda-3)$$

$$\lambda_1 = 3$$ $$\lambda_2=3$$ $$\lambda_3 = 6$$

$$D= \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -6 \end{bmatrix}$$

Okay I found diagonal matrix and I also know that $$A^{T} = \begin{bmatrix} 2 & 2 & -2 \\ 2 & -1 & 4 \\ -2 & 4 & -1 \end{bmatrix}$$

$$A^{-1}= \begin{bmatrix} \frac{5}{18} & \frac{1}{9} & -\frac{1}{9} \\ \frac{1}{9} & \frac{1}{9} & \frac{2}{9} \\ \frac{1}{9} & \frac{2}{9} & \frac{1}{9} \end{bmatrix}$$

• Assuming your calculation is correct, next you should find 2 orthonormal eigenvectors for $3$ and one unit eigenvector for $-6$. Then you stack them as columns to form the orthogonal matrix $M$. Dec 29, 2020 at 17:36
• thanks very much :) Dec 29, 2020 at 17:37

Look for the eigenvectors corresponding to the eigenvalue $$3$$. You will have to solve the system$$\left\{\begin{array}{l}2x+2y-2z=3x\\2x-y+4z=3y\\-2x+4y-z=3z,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}-x+2y-2z=0\\2x-4y+4z=0\\-2x+4y-4z=0.\end{array}\right.$$Each of these equations is a multiple of the other two. So, you really have only one equation to deal with, say, $$-x+2y-2z=0$$. Take two unit vectors $$(x,y,z)$$ which satisfy this condition and which are orthogonal, such as $$u=\left(\frac2{\sqrt5},\frac1{\sqrt5},0\right)$$ and $$v=\left(\frac2{3\sqrt5},-\frac4{3\sqrt5},-\frac5{3\sqrt5}\right)$$. Now, take an eigenvector corresponding to the eigenvalue $$-6$$, which is also an unit vector; you can take, say, $$w=\left(\frac13\,-\frac23,\frac23\right)$$. Then you can take the matrix$$M=\begin{bmatrix}\frac2{\sqrt5}&\frac2{3\sqrt5}&\frac13\\\frac1{\sqrt5}&-\frac4{3\sqrt5}&-\frac23\\0&-\frac5{3\sqrt5}&\frac23\end{bmatrix},$$whose colmuns are the vectors $$u$$, $$v$$; and $$w$$.