# 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?

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Should there be an absolute value around product of singular values in (a)? – Stuart Nov 26 '12 at 4:28
@Stuart: singular values are non-negative. – Martin Argerami Nov 26 '12 at 4:30

## 2 Answers

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.

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Thank you very much for clearing that up for me, Martin! – diimension Nov 26 '12 at 5:28

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You are very welcome. – Martin Argerami Nov 26 '12 at 5:31

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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Thank you very much, Marvis! – diimension Nov 26 '12 at 5:28
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