Given $$A=\left[\begin{array}{rr}2&2\\-1&1\end{array}\right],$$ find and compare $$A=S\Lambda S^T\quad \text{and} \quad A=U\Sigma V^T.$$
I can find $U$ and $V$, beginning with finding eigenvalues of $A^T T$.
However, I cannot find $S$ such that $S\Lambda S^T =A$.
From $\det (A-\lambda{I})=0$, I got $\lambda_1=\frac{3+i\sqrt7}{2}$ and $\lambda_2 = \frac{3-i\sqrt7}{2}$.
From $(A-\lambda_1 I)\mathbf{x}_1=(A-\lambda_1 I)\begin{bmatrix}a\\b\end{bmatrix}=\mathbf{0}$, I could find $\mathbf{x}_1=\left[\begin{array}{r}1+i\sqrt7\\-2\end{array}\right]$.
Similarly, I found $\mathbf{x}_2=\left[\begin{array}{r}-4\\1+i\sqrt7\end{array}\right]$.
Then, letting $S=\begin{bmatrix}\mathbf{x}_1&\mathbf{x}_2\end{bmatrix}$, A can be decomposed to $S\Lambda S^{-1}$, however $S^{-1}$ is not equal to $S^T$.
Is there any method to find $S$ such that $A=S\Lambda S^T$?
If $A=S\Lambda S^T$, and if you assume that $\Lambda$ is a diagonal matrix (which is not clear from your post), thus symmetrical, then let's see what is $A^T$:
$A^T=S^{TT}\Lambda^TS^T=S\Lambda S^T=A$
Thus $A$ itself is symmetrical (which it isn't). This shows us that $A$ cannot be written as a product $S\Lambda S^T$ for any diagonal (or, for that matter, symmetrical) matrix $\Lambda$.
Note if you can choose $\Lambda$ more freely, you can trivially choose $\Lambda=A, S=I$, but my guess is that this was not the point of your exercise.