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First in the case of unitary matrix $U^*=U^{-1}$. and $\det U=1$.

See more here: https://en.wikipedia.org/wiki/Unitary_matrix

However, notice $A=\begin{pmatrix} 0.1&0.1\\ 0.1&0.1\\ \end{pmatrix}$, $B=\begin{pmatrix} 10&10\\ 10&10\\ \end{pmatrix}$

Thus $\lim_{n\rightarrow\infty}A^n=\begin{pmatrix} 0&0\\ 0&0\\ \end{pmatrix}$, $\lim_{n\rightarrow\infty}B^n=\begin{pmatrix} \infty&\infty\\ \infty&\infty\\ \end{pmatrix}$, where $\lim_{n\rightarrow\infty}U^n$ was still a unitary matrix.

Obviously, $A$ converge to $0$ and $B$ blow up. In the sense that $|B|>|U|$ and $|U|>|A|$.

My question was that, is there any way to measure the magnitude of a matrix? (Notice $\det A=\det B=0$)

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    I give a brief explanation of how matrix norms work here – Ben Grossmann May 25 '18 at 23:55
  • @Omnomnomnom Thank you. I had an other example. Consider a matrix $A=\begin{pmatrix} 1.01&0\ 0&0.1 \end{pmatrix}$

    $B=\begin{pmatrix} 0.99&0\ 0&0.99 \end{pmatrix}$. Then if we set the norm to be $(a_1^2+a_2^2+a_3^2+a_4^2)^{1/2}$ then $||B||>||A||$ yet $A$ still blow up and $B$ converge to $0$.

     –  May 26 '18 at 00:04
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    You might also be interested in the spectral radius. Do you know what eigenvalues are? – Ben Grossmann May 26 '18 at 00:07
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    It is notable that, while the particular norm you have chosen fails to capture the phenomenon as you hoped, there always exists a (multiplicative) matrix norm such that $|A| > 1$ if $A^n \to \infty$ and there always exists a (multiplicative) matrix norm such that $|A|<1$ if $A^n \to 0$ – Ben Grossmann May 26 '18 at 00:10
  • In your particular example, the spectral norm (the operator norm associated with the usual Euclidean difference) satisfies $|A| = 1.01$ and $|B| = 0.99$. – Ben Grossmann May 26 '18 at 00:12

1 Answers1

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Yes. Take a look at the operator norm or in the case of matrices, the equivelant concept of matrix norm.

Wraith1995
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  • Thank you. That helped a lot, however, consider Consider a matrix $A=\begin{pmatrix} 1.01&0\ 0&0.1 \end{pmatrix}$

    $B=\begin{pmatrix} 0.99&0\ 0&0.99 \end{pmatrix}$. Then if we set the norm to be $(a_1^2+a_2^2+a_3^2+a_4^2)^{1/2}$ then $||B||>||A||$ yet $A$ still blow up and $B$ converge to $0$.

     –  May 26 '18 at 00:00