## condition number of a matrix

What is the Condition Number of a Matrix? » Cleve’s Corner

Condition Number for Inversion. A condition number for a matrix and computational task measures how sensitive the answer is to perturbations in the input data and to roundoff errors made during the solution process. When we simply say a matrix is “ill-conditioned”, we are usually just thinking of the sensitivity of its inverse and not of all the other condition numbers.

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Condition number of matrix – MATLAB cond – MathWorks

Compute Different Condition Numbers of Matrix. Compute the 1-norm condition number, the Frobenius condition number, and the infinity condition number of the inverse of the 3-by-3 magic square A. A = inv(sym(magic(3))); condN1 = cond(A, 1) condNf = cond(A, ‘fro’) condNi = cond(A, inf)

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7.4 Matrix Norms and Condition Numbers

7.4 Matrix Norms and Condition Numbers Thus the accuracy of the solution depends on the conditioning number of the matrix 10. let x be the solution of Ax = b, We need condition number for that: condition number: cond(A) = kAkkA 1k 2. where the kkabove could be any of …

Condition Number — from Wolfram MathWorld

Condition Number. An estimate of the -norm condition number of a matrix can be computed in the Wolfram Language using LinearAlgebra`MatrixConditionNumber [ m , p] for , 2, or , where omitting the is equivalent to specifying Infinity . A similar approximation for the condition number is given by LUDecomposition [ mat ][[-1]].

Condition number of a product of two matrices – Stack Exchange

Condition number of a product of two matrices. The definition of condition number of matrix is as follows: $$\operatorname{cond}(A)=||A|| \cdot ||A^{-1}||$$ linear-algebra matrices norm condition-number. share | cite | improve this question. edited Jan 19 ’18 at 13:52. Rodrigo de Azevedo.

When $A$ and $B$ are square matrices, the inequality is true for every matrix norm (one that satisfies $\|AB\|\le \|A\|\,\|B\|$.) Indeed,

$$

\operatorname{cond}(AB)=\|AB\|\,\|(AB)^{-1} | \le \|A\|\,\|B\|\,\|B^{-1}\|\,\|A^{-1} \| =\operatorname{cond}(A)\,\operatorname{cond}(B)

$$

If $A$ and $B$ are non-square, then $A^{-1}$ is not meaningful, and the condition number has to be defined differently. The one definition I know for this case (which agrees with the above when the operator norm is used), is

$$

\operatorname{cond}(A)=\frac{\sigma_1(A)}{\sigma_n(A)} = \frac{\max\{|Ax|:|x|=1\}}{\min \{|Ax| : |x|=1\}}

$$

(Here $\sigma_1$ and $\sigma_n$ are the greatest and smallest singular values of $A$, defined in the quotient on the right).

This definition is of interest only when the kernel is trivial. The submultiplicative inequality still holds, because $\sigma_1(AB)\le \sigma_1(A)\sigma_1(B)$ and $\sigma_n(AB)\ge \sigma_n(A)\sigma_n(B) $.Best answer · 3The inequality is false for non-square matrices. $||.||$ denotes the matricial $2$-norm; if $U\in M_{p,n}\setminus \{0\}$, then we put $cond(U)=||U||||U^+||$ where $U^+$ is the Moore-Penrose inverse of $U$. More precisely, let $\sigma_1\geq \cdots\geq \sigma_k > 0,\cdots,0$ be the singular values of $U$. Then $||U||=\sigma_1,||U^+||=1/\sigma_k$ and (with respect to the definition above) $cond(U)=\sigma_1/\sigma_k$. Counter-example to $n\not= p,A\in M_{n,p},B\in M_{p,n},cond(AB)\leq cond(A)cond(B)$. Take $n=2,p=4$ and $A=\begin{pmatrix}99&-95.001&-25&76\\99&-95&-25&76\end{pmatrix},B=\begin{pmatrix}10&-62\\-44&-83\\26&9\\-3&88\end{pmatrix}$. Note that $A$ is ill conditioned. For more details, see my answer in Counter example or proof that $\kappa(AB) \leq \kappa(A)\kappa(B) $0I’d like to clarify this slightly: the reasoning in user103402’s answer isn’t quite clear in the case where the matrices are not square, and it can make you think that something false is true as shown in loup blanc’s answer . $

\newcommand{\sig}{\operatorname{\sigma}}

\newcommand{\R}{\mathbb R}

\newcommand{\norm}[1]{\lVert #1 \rVert}

$ Let $A \in \R^{m \times n}, B \in \R^{n \times p}$ .

Then the question becomes

$$

\operatorname{cond}(AB) = \frac{\sig_1(A B)}{\sig_{\min(m,p)}(A B)}

\stackrel{?}{\le} \frac{\sig_1(A)}{\sig_{\min(m,n)}(A)} \frac{\sig_1(B)}{\sig_{\min(n,p)}(B)} = \operatorname{cond}(A) \operatorname{cond}(B)

.$$ $\sig_1(AB) \le \sig_1(A) \sig_1(B)$ always holds; the question is whether

$$

\sig_{\min(m,p)}(AB) \stackrel{?}{\ge} \sig_{\min(m,n)}(A) \sig_{\min(n,p)}(B)

.$$ If we assume that $m \ge n \ge p$ , it holds. We can see this because: If $B$ has a nontrivial null space, $\sigma_p(B) = 0$ and the inequality holds trivially. So assume it doesn’t. In general note that, if $C \in \mathbb R^{q \times r}$ with $q \ge r$ , then

$$

\sig_r(C)

= \sqrt{\lambda_{\min}(C^T C)}

= \sqrt{\inf_{x \in \R^r_*} \frac{x^T C^T C x}{x^T x}}

= \inf_{x \in \R^r_*} \frac{\norm{C x}}{\norm x}

$$

where $\R^r_* = \R^r \setminus \{0\}$ . Then we can do, because $B x \ne 0$ for $x \ne 0$ ,

\begin{align*}

\sig_p(A B)

& = \inf_{x \in \R^p_*} \frac{\norm{A B x}}{\norm x}

\\& = \inf_{x \in \R^p_*} \frac{\norm{A B x} \norm{B x}}{\norm{B x} \norm{x}}

\\&\ge \left( \inf_{x \in \R^p_*} \frac{\norm{A B x}}{\norm{B x}} \right)

\left( \inf_{x \in \R^p_*} \frac{\norm{B x}}{\norm{x}} \right)

\\&\ge \left( \inf_{y \in \R^n_*} \frac{\norm{A y}}{\norm{y}} \right)

\left( \inf_{x \in \R^p_*} \frac{\norm{B x}}{\norm{x}} \right)

\\& = \sig_n(A) \sig_p(B)

.\end{align*} It holds for $m \le n \le p$ as well by just transposing everything. But

$$

A = \begin{bmatrix}1 & 0\end{bmatrix}

\qquad

B = \begin{bmatrix}0 \\ 1\end{bmatrix}

\qquad

A B = \begin{bmatrix}0\end{bmatrix}

\qquad

A^T B^T = \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}

$$

gives us a counterexample for both $m \le n \ge p$ and $m \ge n \le p$ .0

linear algebra – Condition number of a rectangular matrix |

What is the practical impact of a matrix’s condition number? |

MATH2071: LAB #2: Norms, Errors and Condition Numbers

From the definitions of norms and errors, we can define the condition number of a matrix, which will give us an objective way of measuring how “bad” a matrix is, and how many digits of accuracy we can expect when solving a particular linear system.

Matrix Condition Number and Matrix Norms – Well

Mar 04, 2017 · Multiply the norm of the inverse by the norm of the original matrix and this is the condition number STEP 4 If the condition number is quite large then we have an ill conditioned matrix. If the

Author: Blake Tabian

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Lecture 7 Norms and Condition Numbers – ERNET

The condition number a measure of how close a matrix is to being singular: a matrix with large condition number is nearly singular, whereas a matrix with condition number close to 1 is far from being singular. It is obvious from the definition that a nonsingular matrix and its inverse have the same condition number.