Question about induced matrix norm

[charlies1902]

The induced matrix norm for a square matrix $A$ is defined as:

$\lVert A \rVert= sup\frac{\lVert Ax \rVert}{\lVert x \rVert}$
where $\lVert x \rVert$ is a vector norm.
sup = supremum

My question is: is the numerator $\lVert Ax \rVert$ a vector norm?

 

[fresh_42]

The induced matrix norm for a square matrix $A$ is defined as:

$\lVert A \rVert= sup\frac{\lVert Ax \rVert}{\lVert x \rVert}$
where $\lVert x \rVert$ is a vector norm.
sup = supremum

My question is: is the numerator $\lVert Ax \rVert$ a vector norm?

The same as in the denominator.

 

[charlies1902]

The denominator is a vector norm, so are we saying that the numerator is a vector norm for all square A matrices?

 

[fresh_42]

The denominator is a vector norm, so are we saying that the numerator is a vector norm for all square A matrices?

$Ax$ is a vector, so $||Ax||$ is a vector norm. It makes sense to choose the same.

 

[pyroknife]

$Ax$ is a vector, so $||Ax||$ is a vector norm. It makes sense to choose the same.

Is this norm defined only for non-singular matrix A?

 

[fresh_42]

Is this norm defined only for non-singular matrix A?

It's defined for the whole vector space of linear transformations. It has to satisfy $||A|| = 0 ⇒ A = 0$. It can't be more singular.

 

[pyroknife]

It's defined for the whole vector space of linear transformations. It has to satisfy $||A|| = 0 ⇒ A = 0$. It can't be more singular.

I see. I have a task to prove that the norm defined by the OP is indeed a norm. In your post (the quoted), that is one of the conditions, but I am not given if A is singular/nonsingular, so essentially I have to prove that A can't be singular in order for this to satisfy the criterion of a norm.

Is this an alternative explanation why A cannot be singular?

 

[fresh_42]

The singularity, defect or rank have nothing to do with the defined norm to be one. The linear transformations may not even to be between vector spaces of the same dimension.
To start with: What do you know about your vector norm?

 

[pyroknife]

The singularity, defect or rank have nothing to do with the defined norm to be one. The linear transformations may not even to be between vector spaces of the same dimension.
To start with: What do you know about your vector norm?

I know that a vector norm has to satisfy the 3 criteria:
1.
$\lVert x \rVert \geq 0$
$\lVert x \rVert \geq 0$ = 0 IFF x = 0

2.
$\lVert constant*x \rVert = \mid constant \mid * \lVert x \rVert$

3.
$\lVert x+y \rVert \leq \lVert x \rVert+ \lVert y \rVert$

 

[fresh_42]

So $||Ax|| = 0$ implies what for which $x$? Similar the linearity. For the triangle inequality pull the factor $||x||^{-1}$ into the numerator.

 

[pyroknife]

So $||Ax|| = 0$ implies what for which $x$? Similar the linearity. For the triangle inequality pull the factor $||x||^{-1}$ into the numerator.

Well the induced matrix norm is defined for $x \neq 0$,
so $x$ can be any vector other than the zero vector.

But what if you have the matrix
A = \begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}

and

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

Here we can see that A is singular, but Ax = 0, so the vector norm of Ax is satisfied, but A is not the zero matrix?

 

[fresh_42]

Well the induced matrix norm is defined for $x \neq 0$,
so $x$ can be any vector other than the zero vector.

But what if you have the matrix
A = \begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}

and

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

Here we can see that A is singular, but Ax = 0, so the vector norm of Ax is satisfied, but A is not the zero matrix?

However, it's not the maximum / supremum choice for $x$.
If the maximum is zero, then all are. The 'iff' condition on vector norms gives you $Ax = 0$ for all $x ≠ 0$. For $x=0$ it's clearly also true. So $A$ has to be?

 

[pyroknife]

However, it's not the maximum / supremum choice for $x$.
If the maximum is zero, then all are. The 'iff' condition on vector norms gives you $Ax = 0$ for all $x ≠ 0$. For $x=0$ it's clearly also true. So $A$ has to be?

Oh I see. I disregarded the fact that Ax = 0 for $\textbf{ALL}$ $x \neq 0$
In order to satisfy Ax=0 for all non-trivial x, A must be zero.
Right?

 

[fresh_42]

Yes. That's the definition of the zero map. For the scalars you use the next condition for vector norms. For the triangle inequality the third.

 

[fresh_42]

For the triangle inequality you don't have to pull the $||x^{-1}||$ into the numerator. I first thought it would be easier but it makes no difference.