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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?

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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?

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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.

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}

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