# I A regular matrix <=> mA isomorphism

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1. Nov 11, 2016

### Math_QED

Hello all

Let $m_A: \mathbb{K^n} \rightarrow \mathbb{K^n}: X \mapsto AX$ and $A \in M_{m,n}(\mathbb{K})$
(I already proved that this function is linear)

I want to prove that:

A regular matrix $\iff m_A$ is an isomorphism.

So, here is my approach. Can someone verify whether this is correct?

$\Rightarrow$

$A$ is a regular matrix, thus $A^{-1}$ exists
Then is $m_A$ an isomorphism, because we can find an inverse function $m_A^{-1}: X \mapsto A^{-1}X$ It's straightforward to see that this function is the inverse. ($m_A \circ m_A^{-1} = 1_{\mathbb{K^n}}$ and $m_A^{-1} \circ m_A = 1_{\mathbb{K^n}}$)

$\Leftarrow$

$m_A$ is an isomorphism, thus $m_A$ has an inverse $m_A^{-1}: X \mapsto BX$.

When we apply the definition of inverse function, we would deduce that:

$AB = I_n$ and $BA = I_n$. Then it follows, by definition of the inverse matrix, that $B = A^{-1}$. Thus $A^{-1}$ must exist, because the inverse is unique and there is only one possibility to construct this inverse. Therefore, A is a regular matrix.

2. Nov 11, 2016

### Staff: Mentor

Correct. Only "$A$ regular $\Longrightarrow \, n=m$" could be added.
Why can be assume, that $m_A^{-1}$ is of the form $m_B$? And again, why is $n=m$?

3. Nov 11, 2016

### Math_QED

n = m because $AX \in \mathbb{K^n}$, by definition of $m_A$.

Your second question made me realise that I cannot assume that. I would need to do something like:

$(m_A \circ m_A^{-1})(X) = X \iff m_A(m_A^{-1}(X)) = X \iff Am_A^{-1}(X) = X \Rightarrow m_A^{-1}(X) = BX$ such that$AB = I_n$
And analogue for $(m_A^{-1} \circ m_A)(X)$?

4. Nov 11, 2016

### Staff: Mentor

The basic difficulty here (at least mine) is to keep the parts apart. Since it is almost obviously true, I find it difficult, to separate both sides.
Isn't this the same as to start with $A \in \mathbb{M}_{n,n}(\mathbb{K})$ in the first place? Only demanding $n=m$?
I would have expected two arguments, that regular matrices as well as isomorphisms can only be established between vector spaces of the same dimension. Of course this is true (in the finite dimensional case), but is it already known or part of the proof? I would have started with "Let $m_A : \mathbb{K}^n \rightarrow \mathbb{K}^m$ given by $X \mapsto AX$ ..." because that is how you defined $A$! In this case $n=m$ would have to be shown in both directions of the statement.

Yes, this is the crucial part. That there cannot be two different inverse elements from the right and from the left should be shown separately in an auxiliary proposition because it is needed everywhere (IMO).

To do this I would consider $(Am^{-1}_A - I_n)(X)$ and apply the definitions of $0_n$ and next that of inverse elements.

5. Nov 17, 2016

### Math_QED

Hm, I have been thinking about this and I believe I might have found an easier explanation

The $\Rightarrow$ was clear. For the $\Leftarrow$ part, we know that a matrix is regular if and only of it's the matrix of an isomorphism (after we choose certain basises), and this is equivalent with saying that $A^{-1}$ exists. Well, now consider $[m_A]_{E,E}$, with E standard basis of $\mathbb{K^n}$. Then we see that $A = [m_A]_{E,E}$. Thus, A is regular and this is what we wanted to show.

Now, we can verify that $m_A^{-1}: X \mapsto A^{-1}X$ is the inverse, by applying the definition and regarding the fact that $A^{-1}$ exists and the inverse function is unique.

Last edited: Nov 17, 2016
6. Nov 17, 2016

### Staff: Mentor

I still struggle to keep assumption and implication apart. So let me proceed in small steps.
This means: Given an isomorphism $m_A$ defined by $m_A(X)=AX$.

So if we denote the matrix of $m_A$ according to the basis $E=\{E_1,\ldots,E_N\}$ by $[m_A]$, we have $[m_A](E_i)=AE_i$ by definition of $m_A$ and thus $[m_A]=A$.
Now, why isn't this exactly what we have to show: $m_A$ isomorphic $\Longrightarrow A = [m_A]$ regular? Somewhere the existence of $m_A^{-1}$ has to be used and we cannot assume, yet, that $[m_A^{-1}]=A^{-1}$ or that $A^{-1}$ even exists. Only $[m_A^{-1}]$.

(This were my thoughts, and I hope I didn't confuse something, because the longer I think about it, the more I get confused ... The entire exercise sounds like the question about hen and egg.)

7. Nov 17, 2016

### Math_QED

Well, let me write out everything more clearly.

Definitions:

1) An $n$ x $n$ matrix $A$ is regular if it's the matrix of an isomorphism, or equivalent: if $A^{-1}$ exists.
2) $m_A: \mathbb{K^n} \rightarrow \mathbb{K^n}: X \mapsto AX$

Theorem: A regular matrix $\iff m_A$ is an isomorphism.

Proof
:

$\Rightarrow$ If $A$ is a regular matrix, then $m_A$ is an isomorphism.

If A is regular, then we know, by definition, that $A^{-1}$ exists. I now claim that $m_A^{-1}: X \mapsto A^{-1}X$

This claim is true, because $m_A \circ m_A^{-1} = 1_{\mathbb{K^n}}$ and $m_A^{-1} \circ m_A = 1_{\mathbb{K^n}}$
and thus this function is the (unique) inverse. Because $m_A$ has an inverse, we know that $m_A$ is bijective and thus an isomorphism (It's clear that $m_A$ is a linear function)

$\Leftarrow$ If $m_A$ is an isomorphism, then $A$ is a regular matrix.

$m_A$ is an isomorphism, thus every matrix associated with it is regular. $[m_A]_{E,E} = A$ when we choose that $E$ is the standard basis of $\mathbb{K^n}$. Thus, because $A$ is the matrix of an isomorphism, $A$ must be regular (by definition) and this ends the proof.

QED.

8. Nov 17, 2016

### Staff: Mentor

Why? And why isn't this exactly what has to be shown? $m_A$ is an isomorphism only guarantees another isomorphism $\varphi$ such that $\varphi m_A = 1$. If you already know, that these belong to regular matrices, why to prove something at all?

9. Nov 17, 2016

### Math_QED

By definition! And you don't know that A is a matrix associated with $m_A$ until you have found a basis $B$ such that $[m_A]_{B,B}$ = A. But such a basis exists, as it is the standard basis.

10. Nov 17, 2016

### Staff: Mentor

We know by construction of $m_A$ as $m_A(X)=AX$ that $[m_A]_{B,B} = A$ by applying unit vectors. But how do you get from $\varphi m_A = 1$ to $\varphi(X) = A^{-1}X$? In this direction we only have $\varphi$ and know nothing about a matrix.
(why?) and
is what I don't see. By definition we only get $\varphi$, no matrix of $\varphi$. If we automatically have matrices, then there is nothing left to show, because it's clear then that:
$$m_A \text{ isomorphism with matrix } A \Leftrightarrow m_A^{-1} \text{ isomorphism with matrix } B \Leftrightarrow AB=1$$
To me the bridge between regular matrices and isomorphisms is what it's all about, i.e. why is $[\varphi] = A^{-1}$.

11. Nov 17, 2016

### Math_QED

In the $\Leftarrow$ part I never use the existence of $m_A^{-1}$ (check my new proof that I wrote 2 posts ago). We know from the definition that every matrix of an isomorphism is regular. That's the only thing we use. We choose the standard basis to show that the isomorphism $m_A$ can be represented by $A$ and we know from the definition that, when this is possible, A is regular. And, yes we have shown it's possible.

12. Nov 17, 2016

### Staff: Mentor

But that's all you have. How can you not use it?
In this case I fold. If you have such a definition, then all of the above is trivial since $m_A(X)=AX$ implies that there is already a basis (to express $A$), and all which is left to show is, that there cannot be two different matrices.