Transformation of coordinate axis to make matrix diagonal

Question

Consider the matrix

$$ A= \begin{bmatrix}1/8 & \frac{-5}{8\sqrt{3}} \\ \frac{-5}{8\sqrt{3}} & 11/8 \end{bmatrix} $$ which of the following transformations of the coordinate axis will make the matrix $A$ diagonal?

Rotation in $-60$ counter clock wise, $-30$ ccw, $30$ ccw, $60$ ccw or $45$ ccw?

-I tried graphing out the matrix for better visual grasp on the question. I multiplied $\cos(\theta) -\sin(\theta)$ to random vectors in $A$, replacing $\sin(60)$ with sqrt3/2 $\sin(\theta)$ $\cos(\theta)$ ,$\cos(60)$ with $1/2$ and etc. I tried all of the rotations but I can't get my end matrix to be diagonal and my results always come out clunky with weird random numbers. Can anyone tell me what I am doing wrong or give a brief explanation on what I should be doing?

Answer 1

You can use eigen value decomposition. We know that the eigen values and eigen vectors are computed by solving this equation: $$Av_i=\lambda_i v_i \, , \quad i=1,\ldots,n $$ So if we re-write the equation for all $i$ in a matrix form, we will have: $$AP=P\Lambda \Rightarrow P^{-1}AP=\Lambda$$ where $P=(v_1,\ldots,v_n)$ and $$\Lambda=\left[\begin{matrix} \lambda_1 &&\\& \ddots \\& &\lambda_n \end{matrix}\right]$$

In this problem: $$P=\left[ \begin{matrix} -0.9659&-0.2588 \\ -0.2588 &0.9659 \end{matrix} \right]$$ and $$\Lambda=\left[ \begin{matrix} 0.0283&0 \\ 0 &1.4717 \end{matrix} \right]$$

Answer 2

Hint: We want $\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}A\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$ to be diagonal. So we need the $(1,2)$ and $(2,1)$ entries of the product to be zero.

Second approach: diagonalize the matrix, by computing the characteristic polynomial, getting the eigenvalues, and a basis of (two) eigenvectors.