Singular value proofs 
1.) Let $A$ be a nonsingular square matrix.
  
  
*
  
*a.) Prove that the product of the singular values of $A$ equals the
  absolute value of its determinant:
  $\sigma_1\sigma_2...\sigma_n=|detA|$. 
  
*b.) Does the sum equal the absolute value of the trace? And a matrix
  whose determinant is very small is ill-conditioned?
  
*c.) Show that if $|detA|< 10^{-k}$, then its minimal singular value
  satisfies $\sigma_n <10^{-\frac{k}{n}}$. Can you construct an
  ill-conditioned matrix with $detA = 1$?
2.) Let A be a square matrix. Prove that its maximum eigenvalue is smaller than its maximal singular value.

My attempt:
a.) I know that the equation for singular values is $A = U \dot\ E \dot\ V$ so $|det(A)| = |det(U) \dot\ det(E) \dot\ det(V)|
= |\pm1 \dot\ $ (product of singular values) |
= product of singular values. Is that correct?
b.) True for the first part. For the second part, do they mean a matrix can have ill-conditioned determinant if its determinant is small? Not sure what they are asking.
c.) I do not know how to do.
2.) Not sure how to do because I thought that the singular value should be smaller than the maximum eigenvalue?
 A: a) is correct
b) No, it's not true. If $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, then the sum of the singular values is $2$, while the absolute value of the trace is zero. 
To discuss whether your matrix is ill-conditioned, you need to say which norm you are talking about. Assuming we are talking about the operator norm (=largest singular value), if the determinant is small it means that some singular values are small; then the inverse will have big singular values and the condition number will be large. 
c) You have $\sigma_1\sigma_2\cdots\sigma_n<10^{-k}$; if all $\sigma_j\geq 10^{-k/n}$, then $$|\det A|=\sigma_1\cdots\sigma_n\geq(10^{-k/n})^n=10^{-k};$$
so at least one singular value is less than $10^{-k/n}$.
2) This is not well phrased, because they can be equal. The maximum singular value is $\|A^TA\|^{1/2}$. Now let $\lambda$ be
an eigenvalue of $A$ with unit eigenvector $v$. Then
$$
|\lambda|=\|\lambda v\|=\|Av\|=(v^TA^TAv)^{1/2}\leq\|A^TA\|^{1/2}(v^Tv)^{1/2}=\|A^TA\|^{1/2}.
$$
So every eigenvalue is smaller in absolute value than the biggest singular value.
A: For $1(c)$, make use of the fact that $\sigma_1 \geq \sigma_2 \geq \cdots \geq \sigma_n$. Hence, we get that $$\sigma_n^n \leq \sigma_1 \sigma_2 \cdots \sigma_n = \det(A) < 10^{-k}$$ Hence, we get that $$\sigma_n < 10^{-k/n}$$
The maximal eigen value is given by
\begin{align}
\left \vert \lambda_{\max} \right \vert & = \max_{\Vert x \Vert_2 = 1} \left \vert x^T A x \right \vert\\
 & = \underbrace{\max_{\Vert x \Vert_2 = 1} \left \vert x^T U \Sigma V^T x \right \vert \leq \max_{\Vert x \Vert_2 = 1} \Vert x^T U \Vert_2 \Vert \Sigma \Vert_2 \max_{\Vert x \Vert_2 = 1}  \Vert V^T x \Vert_2}_{\text{By sub-multiplicativity of matrix norm}}\\
& = 1 \times \Vert \Sigma \Vert_2 \times 1 = \sigma_1
\end{align}
