MHB Uniqueness of Inverse Matrices: Proof and Explanation

[ognik]

I have an exercise which says to show that for vectors, $ A \cdot A^{-1} = A^{-1} \cdot A = I $ does NOT define $ A^{-1}$ uniquely.

But, let's assume there are at least 2 of $ A^{-1} = B, C$

Then $ A \cdot B = I = A \cdot C , \therefore BAB = BAC, \therefore B=C$, therefore $ A^{-1}$ is unique? (got lazy with dots)

 

[Sudharaka]

I have an exercise which says to show that for vectors, $ A \cdot A^{-1} = A^{-1} \cdot A = I $ does NOT define $ A^{-1}$ uniquely.

But, let's assume there are at least 2 of $ A^{-1} = B, C$

Then $ A \cdot B = I = A \cdot C , \therefore BAB = BAC, \therefore B=C$, therefore $ A^{-1}$ is unique? (got lazy with dots)

Hi ognik, :)

I am not getting your question exactly. Do you mean that you have to show that $A^{-1}$ is unique? Could you please elaborate a bit.

 

[ognik]

I'm not sure to be honest, I wonder if it is more that the identity does not define the inverse uniquely, ie there could be more than 1 inverse ...

 

[Deveno]

You'll have to be a bit more explicit in exactly what you're asking.

If, for a matrix $A$, there exists a matrix $B$ such that $AB = BA = I$, then $B$ is unique (invertible $n \times n$ matrices form a *group*, called the general linear group of degree $n$ over whatever field you're working with, and inverses in a group are unique).

However, if $B$ is merely a one-sided inverse, that is $AB = I$ or $BA = I$, but not both, $B$ may not be unique. This only happens when $A$ is non-square.

Furthermore, your question begins: "For vectors..." -it is hard to imagine what the (multiplicative) inverse of a vector might be.

 

[I like Serena]

It seems to me that a multiplicative inverse of a vector, is a vector such that the dot-product is equal to $1$.
So for instance:
$$(^2_3) \cdot (^{1/2}_0) = 1$$

 

[ognik]

I had what ILS suggests in mind, but I found the question hard to follow myself, hoping this is helpful - it is exactly as follows:

View attachment 4980

 

[Ackbach]

As the dot product is commutative, the fact that $A\cdot A^{-1}=A^{-1}\cdot A$ is superfluous. Choose
$$A=\begin{bmatrix}1 \\ -1\end{bmatrix}.$$
Choose
$$B=\begin{bmatrix}1 \\ 0\end{bmatrix} \qquad \text{and} \qquad C=\begin{bmatrix}2 \\ 1\end{bmatrix}.$$
Then $A\cdot B=1$ and $A\cdot C=1$, but $B\not=C$.

 

[ognik]

I see that thanks, but then what is wrong with my original 'proof'? It applies to matrices and a vector is a type of matrix?

 

[I like Serena]

I see that thanks, but then what is wrong with my original 'proof'? It applies to matrices and a vector is a type of matrix?

I presume you're referring to $BAB=BAC∴B=C$.

It appears you're removing $BA$ from the left side.
But that would correspond to multiplying on the left with $(BA)^{-1}$.
This can only be done if $BA$ is invertible and we cannot assume any such thing, and in fact we know that it's not.

 

[Deveno]

You may or may not know this, but any vector $v$ in a finite-dimensional inner product space determines (uniquely) a *hyperplane* (a subspace of dimension: $\dim(V) - 1$):

$E_v = \{w \in V: \langle v,w\rangle = 0\}$.

The vector $\dfrac{1}{\|v\|}v$ serves as a (unit) *normal* to $E_v$, if we choose an orientation for $V$, we have that either:

$\dfrac{1}{\|v\|}v$ or $-\dfrac{1}{\|v\|}v$ can be called *the* normal to the hyperplane $E_v$ (the "usual" orientation for $\Bbb R^3$ is set so that $\mathbf{i} \times \mathbf{j} = \mathbf{k}$, often called the "right-hand" orientation, since it corresponds to forming the (positive) $x$-axis with the right-hand index finger, the (positive) $y$-axis with the right-hand middle finger, and the thumb (pointing up), is the (positive) $z$-axis. This is a purely arbitrary convention, which is why cross-products are often called "pseudo-vectors", their sign isn't independent of axis orientation).

So...where was I?

Pick any vector in $E_v$, say, $w$, and consider $w + v$.

Then $\langle w+v,v\rangle = \langle w,v\rangle + \langle v,v\rangle = 0 + \|v\|^2 \neq 0$ (unless $v = 0$).

Thus $\langle\dfrac{1}{\|v\|^2}(w+v),v\rangle = 1$, and it is clear we have just as many such vectors as we have elements of $E_v$.

Why does your result not hold when $\dim(V) = 1$?

 

[ognik]

I presume you're referring to $BAB=BAC∴B=C$.

It appears you're removing $BA$ from the left side.
But that would correspond to multiplying on the left with $(BA)^{-1}$.
This can only be done if $BA$ is invertible and we cannot assume any such thing, and in fact we know that it's not.

That's a surprise, I thought we effectively had (BA)B = (BA)C, therefore B must = C?

 

[Ackbach]

I see that thanks, but then what is wrong with my original 'proof'? It applies to matrices and a vector is a type of matrix?

I presume you're referring to $BAB=BAC∴B=C$.

It appears you're removing $BA$ from the left side.
But that would correspond to multiplying on the left with $(BA)^{-1}$.
This can only be done if $BA$ is invertible and we cannot assume any such thing, and in fact we know that it's not.

I would also add that it's not at all clear that $BAB$ is well-defined. If the multiplication is a dot product, then the result of $A\cdot B$ is a number, not a vector; in that case, it's unclear what $B\cdot A \cdot B$ would even be. How would you define that?

 

[I like Serena]

I would also add that it's not at all clear that $BAB$ is well-defined. If the multiplication is a dot product, then the result of $A\cdot B$ is a number, not a vector; in that case, it's unclear what $B\cdot A \cdot B$ would even be. How would you define that?

Indeed.

That's a surprise, I thought we effectively had (BA)B = (BA)C, therefore B must = C?

If $(BA)$ is for instance the zero matrix, we cannot conclude that $B=C$.

Just for fun:
$$a^2-a^2=a^2-a^2 \Rightarrow a(a-a) =(a+a)(a-a) \Rightarrow a=a+a \Rightarrow a=2a$$
This holds for any $a$, therefore $1=2$.

 

[ognik]

Indeed.
If $(BA)$ is for instance the zero matrix, we cannot conclude that $B=C$.
.

I had postulated both B and C = $A^{-1}$ where we are a looking at A non-singular, but I need to remember that BA could be 0, even if B and A aren't.

Is $AA^{-1}=1$ by definition only?