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?
Best Answer
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.