I have computed the singular value decomposition of the following matrix
$$A= \begin{bmatrix}1&2\\0&1\\-1&0\\\end{bmatrix}$$

here are the important findings below.

$$\Sigma=\left[\begin{matrix}1 & 0 \\ 0 & \sqrt 6 \\ 0 & 0\end{matrix}\right]$$

$$V=\left[\begin{matrix}-\frac{2}{\sqrt 5} & \frac{1}{\sqrt 5} \\ \frac{1}{\sqrt 5} & \frac{2}{\sqrt 5}\end{matrix}\right]$$

$$U=\left[\begin{matrix}0 & \sqrt{\frac 5 6} & \frac{1}{\sqrt 6} \\ \frac{1}{\sqrt 5} & \sqrt{\frac 2 {15}} & -\sqrt{\frac 2 3} \\ \frac{2}{\sqrt 5} & -\frac{1}{\sqrt 30} & \frac{1}{\sqrt 6}\end{matrix}\right]$$

Now here is the question I am trying to solve now,

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> Use the singular value decomposition of $A$ to prove that $\|A\|\_2 = \sigma\_1 = \sigma\_{\max}(A)$. Show further that if $A$ is invertible $\|A^{-1}\|\_2 = \sigma\_n = \sigma\_{\min}(A)$ and thus that the condition number based on the spectral norm is $k(A) = \frac{\sigma\_{\max}(A)}{\sigma\_{\min}(A)}$
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here are my workings for the first part. To find $\|A\|\_2=\sigma\_1$ use that $\|A\|\_2 = \sqrt{\lambda \_{\max}(A^TA)}$

$$A^TA =\left[\begin{matrix}2 & 2 \\ 2 & 5 \\\end{matrix}\right]$$ which gives the $$\det(A^TA-\lambda I) = \left[\begin{matrix}2- \lambda & 2 \\ 2 & 5-\lambda \\\end{matrix}\right]$$ resulting in $\lambda = 6,1$ then $\|A\|\_2 = \sqrt6 = \sigma\_1 = \sigma\_{\max}$

**I am wondering how I do the next two parts, is the next part the same as I have done above?**