Answer to Question #108696 in Statistics and Probability for Kentse

Question

Find an orthogonal matrix associated with matrix A D

1 2
2 4 
:

Answer 1

Find an orthogonal matrix B such that $P^TAP$ is diagonal when

$$A=\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$$

First, the find eigenvalues and eigenvectors

Start from forming a new matrix by subtracting $\lambda$ from the diagonal entries of the given matrix:

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

Find the determinant of the obtained matrix:

$$det(A-\lambda I)=\begin{vmatrix} 1-\lambda & 2 \\ 2 & 4-\lambda \end{vmatrix}=$$ $$=(1-\lambda)(4-\lambda)-2(2)=\lambda^2-5\lambda$$

This is a characteristic polynomial.

Solve the equation

$$\lambda^2-5\lambda=0$$

The roots are:

$\lambda_1=5$

$\lambda_2=0$

These are the eigenvalues.

Next, find the eigenvectors.

$\lambda_1=5$

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

Perform row operations to obtain the rref of the matrix:

$R_2=R_2+\dfrac{1}{2}R_1:$

$$\begin{bmatrix} -4 & 2 \\ 0 & 0 \end{bmatrix}$$

$R_2=-\dfrac{1}{4}R_1:$

$$\begin{bmatrix} 1& - {1 \over 2}\\ 0 & 0 \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1& - {1 \over 2}\\ 0 & 0 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$$

If we take $v_2=t,$ then $v_1={1 \over2}t, v_2=t$

Therefore

$$\text{v}=\begin{bmatrix} t/2\\ t \end{bmatrix}=\begin{bmatrix} 1/2\\ 1 \end{bmatrix}t$$

$\lambda_2=0$

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

Perform row operations to obtain the rref of the matrix:

$R_2=R_2-(2)R_1:$

$$\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$$

Now, solve the matrix equation

$$\begin{bmatrix} 1& 2\\ 0 & 0 \end{bmatrix}\begin{bmatrix} v_1\\ v_2 \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}$$

If we take $v_2=t,$ then $v_1=-2t, v_2=t$.

Therefore

$$\text{v}=\begin{bmatrix} -2t\\ t \end{bmatrix}=\begin{bmatrix} -2\\ 1 \end{bmatrix}t$$

Eigenvalue: 5, eigenvector: $\begin{bmatrix} 1/2\\ 1 \end{bmatrix}$

Eigenvalue: 0, eigenvector:$\begin{bmatrix} -2\\ 1 \end{bmatrix}$

Form the matrix $P,$ whose i-th column is the i-th eigenvector:

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

$(1/2)^2+(1)^2=5/4$

$(-2)^2+(1)2=5$

Then an orthogonal matrix $P$

$$P=\begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5}\\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix}$$