这是一份leeds利兹大学MATH260001作业代写的成功案例
We shall verify the three axioms for a norm. First, if $A \neq 0$, then $A$ has at least one nonzero column; say, $A^{(j)} \neq 0$. Consider the vector in which 1 is the $j$ th component – that is, $x=(0, \ldots, 0,1,0, \ldots, 0)^{T}$. Obviously, $x \neq 0$ and the vector $v=x /|x|$ is of norm 1 . Hence by the definition of $|A|$,
$$
|A| \geq|A v|=\frac{|A x|}{|x|}=\frac{\left|A^{(\jmath)}\right|}{|x|}>0
$$
Next, the vector norm, we have
$$
|\lambda A|=\sup {|\lambda A u|:|u|=1}=|\lambda| \sup {|A u|:|u|=1}=|\lambda||A|
$$
For the triangle inequality, we use the analogous property of the vector norm and Problem 4 to write
$$
|A+B|=\sup {|(A+B) u|:|u|=1}
$$
MATH260001 COURSE NOTES ：
If $A$ is an $n \times n$ matrix such that $|A|<1$, then $I-A$ is invertible, and
$$
(I-A)^{-1}=\sum_{k=0}^{\infty} A^{k}
$$
First, we shall show that $I-A$ is invertible. If it is not invertible then it is singular, and there exists a vector $x$ satisfying $|x|=1$ and $(I-A) x=0$. From this we have
$$
1=|x|=|A x| \leqq|A||x|=|A|
$$
which contradicts the hypothesis that $|A|<1$.
