2x2 Fibonacci matrix singular value decomposition $A = \left[\begin{array}[c]{rr}1 & 1\\1 & 0\end{array}\right]$
I am supposed to find all the eigenvalues and vectors for this matrix so that $Av=σu$ and then form a singular value decomposition $UΣVᵀ$ and show that $A=UΣVᵀ$. 
I have attempted this so many times and just keep getting so confused with what the question really wants, since I get something very complicated, as I get when I try to solve it. However, when I look at a similar markscheme it's quite simply put, but I don't understand how. 
I know that I can use $det(AᵀA)$ to get $σ^2$ or I can get σ with $det(A-λI)$. So those give me the same, which are the diagonals of $Σ$ and the rest of $Σ$ consists of zeros.
So I get
$$\Sigma = \left(
\begin{array}{cc}
 \sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)} & 0 \\
 0 & \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)} \\
\end{array}
\right)$$
and $v_1$ and $v_2$ I calculated to be as follows
$v_1$$=$$\left[\begin{array}[c]{r}(1+√5)/2\\1\end{array}\right]$
$v_2$$=$$\left[\begin{array}[c]{r}(1-√5)/2\\1\end{array}\right]$
After that am I not supposed to convert it to the unit vector? That comes out really messy...
I just feel like I'm doing something wrong because it's not supposed to be a very complicated question, and if I continue from here it gets way ugly. I'm not sure how I am supposed to show what they're asking for. 
Also, I believe $u_1=v_1$ and $u_2=-v_2$, but how are those calculated?
Another note - can this be done by keeping the eigenvectors as variables instead of substituting the values? Maybe that would not come out as complicated. 
 A: We are given:
$$A = \begin{bmatrix}1 & 1\\1 & 0 \end{bmatrix}$$
We have:
$$W = A^T A = \begin{bmatrix}2 & 1\\1 & 1 \end{bmatrix}$$
The characteristic polynomial and eigenvalues of W are:
$$\lambda^2 +3 \lambda -1 = 0 \implies \lambda_1 = \frac{1}{2} \left(3-\sqrt{5}\right), ~ \lambda_2 = \frac{1}{2} \left(3+\sqrt{5}\right)$$
This gives us the singular values:
$$\sigma_1= \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)}, ~ \sigma_2= \sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)}$$
The eigenvectors of $W$ are:
$$v_1 = \begin{bmatrix} \frac{1}{2} \left(1-\sqrt{5}\right) \\ 1 \end{bmatrix}, ~ v_2 = \begin{bmatrix} \frac{1}{2} \left(1+\sqrt{5}\right) \\ 1 \end{bmatrix}$$
Normalizing these eigenvectors, we have:
$$v_1 = \begin{bmatrix} \frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \\ \frac{1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \end{bmatrix}, ~ v_2 = \begin{bmatrix} \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} \\ \frac{1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \end{bmatrix}$$
Now, we can write $W = U \Sigma V^T$ as:
$$U = \left(
\begin{array}{cc}
 \frac{\frac{1}{2} \left(1+\sqrt{5}\right)+1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1+\sqrt{5}\right)+1\right)^2}} & \frac{\frac{1}{2} \left(1-\sqrt{5}\right)+1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1-\sqrt{5}\right)+1\right)^2}} \\
 \frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1+\sqrt{5}\right)+1\right)^2}} & \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+\left(\frac{1}{2} \left(1-\sqrt{5}\right)+1\right)^2}} \\
\end{array}
\right)$$
$$\Sigma = \left(
\begin{array}{cc}
 \sqrt{\frac{1}{2} \left(3+\sqrt{5}\right)} & 0 \\
 0 & \sqrt{\frac{1}{2} \left(3-\sqrt{5}\right)} \\
\end{array}
\right)$$
$$V = \left(
\begin{array}{cc}
 \frac{1+\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} & \frac{1-\sqrt{5}}{2 \sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \\
 \frac{1}{\sqrt{\frac{1}{4} \left(1+\sqrt{5}\right)^2+1}} & \frac{1}{\sqrt{\frac{1}{4} \left(1-\sqrt{5}\right)^2+1}} \\
\end{array}
\right)$$
Notes: 


*

*Recall: The columns of V are called the right singular vectors. The columns of U are called the left singular vectors. To find $U$, we calculate the columns as:


$$u_1 = \dfrac{1}{\sigma_1} A v_1, ~ u_2 = \dfrac{1}{\sigma_2} A v_2$$
Care needs to be taken when one or both of the eigenvalues are zero!
This SVD can also be written as:
$$\sigma_1~u_1~v_1^T + \sigma_2~u_2~v_2^T$$


*

*Recall we need to form $V^T$ when writing this out.

