Induced norm form with Nilpotent Matrices

Question

1)Is $||A||=\sum |a_{ii}|$ a norm?

2)Is $||x||=min |x_i|$ a norm?

3) Is $||A||=\sum_{i,j} |a_{ij}|$ an induced norm?

1,2) I believe no since it is easy to prove that a nonzero matrix's trance (even with abs value) =0.

3) I believe it is not. (cuz of identity)

Any hint would be appreciated.

Answer 1

A norm is a mapping $\Vert \cdot \Vert : X \to [0, \infty)$ and a $\mathbb K$-vector space $X$ with the following properties:

1. $x = 0 \Leftrightarrow \Vert x \Vert = 0.$
2. $\Vert \lambda x \Vert = \vert \lambda \vert \Vert x \Vert$ for all $\lambda \in \mathbb K$, $x \in X$.
3. $\Vert x + y \Vert \leq \Vert x \Vert + \Vert y \Vert$ for all $x,y \in X$.

Now to your questions:

1) Notice that $\mathbb K^{d \times d}$ is a $\mathbb K$-vector space. Consider the matrix $A = \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$. Then you have $\Vert A \Vert = 0$ but $A \neq 0$. Hence you don't have a norm (since 1. isn't true).

2) Notice that $\mathbb K^d$ is a $\mathbb K$-vector space. Consider the vector $x = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$. Then you have $\Vert x \Vert = 0$ but $x \neq 0$. Hence you don't have a norm (since 1. isn't true).

Now let $X = \mathbb K^{d \times d}$. A norm $\Vert \cdot \Vert_{\operatorname{op}} : \mathbb K^{d \times d} \to [0, \infty)$ is called induced, if there exists a vector norm $\Vert \cdot \Vert : \mathbb K^d \to [0, \infty)$ such that $$ \Vert A \Vert_{\operatorname{op}} = \sup_{\Vert x \Vert = 1} \Vert Ax \Vert \qquad \text{for all } A \in \mathbb K^{d \times d}.$$ From that it follows directly that $\Vert I \Vert_{\operatorname{op}} = 1$.

3) For the map above you have $\Vert I \Vert = d$. Thus you have no induced norm in the case $d > 1$. In the case $d = 1$ it is induced (by $1$-norm or $\infty$-norm for example).

Is absolutve value of norm of a matrix smaller than one mean that is it a nilpotent matrix?

No it doesn't mean that. For example scaling a given norm by a positive constant gives you another equivalent norm. You can choose the constant as small as you want and thus it doesn't say anything about nilpotency of a matrix. But now let $A \in \mathbb K^{d \times d}$ with $\Vert A \Vert < 1$ for some induced norm. We want to show that $I - A$ is invertible. We use a Neumann-series argument for that. Since $\Vert A \Vert < 1$ we know that the series $\sum_{n = 0}^\infty A^n $ converges absolutely since $\sum_{n = 0}^\infty \Vert A ^ n\Vert \leq \sum_{n = 0}^\infty \Vert A\Vert^n.$ converges since the geometric series converges (notice that induced norms are submultiplicative). Now notice the following $$ (I - A) \sum_{n = 0}^N A ^ n = \sum_{n = 0}^N A ^ n - \sum_{n = 0}^N A ^ {n+1} = -A^{N + 1}.$$ Taking norms we arrive at $$ \Vert (I - A) \sum_{n = 0}^N A ^ n \Vert = \Vert A^{N + 1} \Vert \leq \Vert A \Vert^{N + 1} \to 0$$ for $N \to \infty$. Since $\sum_{n = 0}^\infty A^n $ converges absolutely we have $\sum_{n = 0}^\infty A^n \in \mathbb K^{d \times d}$ and the inequality above leads in the limit that $(I - A)^{-1} = \sum_{n = 0}^\infty A^n$. Hence $I - A$ is invertible if $\Vert A \Vert < 1$.

I hope that helps you :)