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you should use the same norme for origin and image space.
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edit approved Feb 14, 2023 at 9:24
Benoit Gaudeul
edit approved Feb 14, 2023 at 9:24
Benoit Gaudeul
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$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1/2$$1$ and the complex norm of $A$ is $1$$2$.

Indeed, if $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}\in\R^2$ with $x_1\ne0$, then $$\frac{\|Ax\|}{\|x\|}=\frac{|x_1|}{|x_1|+|x_1+ix_2|} \le\frac{|x_1|}{|x_1|+|x_1|}=\frac12,$$$$\frac{\|Ax\|}{\|x\|}=\frac{|x_1|+|x_1+0i|}{|x_1|+|x_1+ix_2|} \le\frac{2|x_1|}{|x_1|+|x_1|}=1,$$ with $\frac{\|Ax\|}{\|x\|}=\frac12$$\frac{\|Ax\|}{\|x\|}=1$ if $x_2=0$. So, the real norm of $A$ is $1/2$$1$.

If now $z=\begin{bmatrix}z_1\\z_2\end{bmatrix}\in\C^2$ with $z_1\ne0$, then $$\frac{\|Az\|}{\|z\|}=\frac{|z_1|}{|z_1|+|z_1+iz_2|} \le1,$$$$\frac{\|Az\|}{\|z\|}=\frac{2|z_1|}{|z_1|+|z_1+iz_2|} \le2,$$ with $\frac{\|Az\|}{\|z\|}=1$$\frac{\|Az\|}{\|z\|}=2$ if $z_2=iz_1$. So, the complex norm of $A$ is $1$$2$.

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1/2$ and the complex norm of $A$ is $1$.

Indeed, if $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}\in\R^2$ with $x_1\ne0$, then $$\frac{\|Ax\|}{\|x\|}=\frac{|x_1|}{|x_1|+|x_1+ix_2|} \le\frac{|x_1|}{|x_1|+|x_1|}=\frac12,$$ with $\frac{\|Ax\|}{\|x\|}=\frac12$ if $x_2=0$. So, the real norm of $A$ is $1/2$.

If now $z=\begin{bmatrix}z_1\\z_2\end{bmatrix}\in\C^2$ with $z_1\ne0$, then $$\frac{\|Az\|}{\|z\|}=\frac{|z_1|}{|z_1|+|z_1+iz_2|} \le1,$$ with $\frac{\|Az\|}{\|z\|}=1$ if $z_2=iz_1$. So, the complex norm of $A$ is $1$.

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1$ and the complex norm of $A$ is $2$.

Indeed, if $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}\in\R^2$ with $x_1\ne0$, then $$\frac{\|Ax\|}{\|x\|}=\frac{|x_1|+|x_1+0i|}{|x_1|+|x_1+ix_2|} \le\frac{2|x_1|}{|x_1|+|x_1|}=1,$$ with $\frac{\|Ax\|}{\|x\|}=1$ if $x_2=0$. So, the real norm of $A$ is $1$.

If now $z=\begin{bmatrix}z_1\\z_2\end{bmatrix}\in\C^2$ with $z_1\ne0$, then $$\frac{\|Az\|}{\|z\|}=\frac{2|z_1|}{|z_1|+|z_1+iz_2|} \le2,$$ with $\frac{\|Az\|}{\|z\|}=2$ if $z_2=iz_1$. So, the complex norm of $A$ is $2$.

added 585 characters in body
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Iosif Pinelis
Iosif Pinelis
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No$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1/2$ and the complex norm of $A$ is $1$.

Indeed, if $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}\in\R^2$ with $x_1\ne0$, then $$\frac{\|Ax\|}{\|x\|}=\frac{|x_1|}{|x_1|+|x_1+ix_2|} \le\frac{|x_1|}{|x_1|+|x_1|}=\frac12,$$ with $\frac{\|Ax\|}{\|x\|}=\frac12$ if $x_2=0$. So, the real norm of $A$ is $1/2$.

If now $z=\begin{bmatrix}z_1\\z_2\end{bmatrix}\in\C^2$ with $z_1\ne0$, then $$\frac{\|Az\|}{\|z\|}=\frac{|z_1|}{|z_1|+|z_1+iz_2|} \le1,$$ with $\frac{\|Az\|}{\|z\|}=1$ if $z_2=iz_1$. So, the complex norm of $A$ is $1$.

No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1/2$ and the complex norm of $A$ is $1$.

$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1/2$ and the complex norm of $A$ is $1$.

Indeed, if $x=\begin{bmatrix}x_1\\x_2\end{bmatrix}\in\R^2$ with $x_1\ne0$, then $$\frac{\|Ax\|}{\|x\|}=\frac{|x_1|}{|x_1|+|x_1+ix_2|} \le\frac{|x_1|}{|x_1|+|x_1|}=\frac12,$$ with $\frac{\|Ax\|}{\|x\|}=\frac12$ if $x_2=0$. So, the real norm of $A$ is $1/2$.

If now $z=\begin{bmatrix}z_1\\z_2\end{bmatrix}\in\C^2$ with $z_1\ne0$, then $$\frac{\|Az\|}{\|z\|}=\frac{|z_1|}{|z_1|+|z_1+iz_2|} \le1,$$ with $\frac{\|Az\|}{\|z\|}=1$ if $z_2=iz_1$. So, the complex norm of $A$ is $1$.

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Iosif Pinelis
Iosif Pinelis
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No. E.g., let $$A=\begin{bmatrix}1&0\\0&0 \end{bmatrix}$$ and $$\left\|\begin{bmatrix}z_1\\z_2 \end{bmatrix}\right\|=|z_1|+|z_1+iz_2|.$$ Then the real norm of $A$ is $1/2$ and the complex norm of $A$ is $1$.