Geometrical Significance of column matrices of an $n\times n$ invertible matrix over field $F$ acting as the basis of $F^{n\times 1}\,.$

user116211

Nov 10, 2016 06:22

Anyways, I was reading about how the set of column matrices of an $n\times n$ invertible matrix over field $F$ spans all the column matrices $F^{n\times 1}\,.$

DanielSank

I'm not playing this game with you @SirCumference.

I'll just use the ignore feature, temporarily.

Sir Cumference

@DanielSank I'm not being pedantic

DanielSank

No?

Sir Cumference

I just think it's going to die out soon enough on its own

DanielSank

@MAFIA36790 Yeah ok.

user116211

Nov 10, 2016 06:23

Let $P$ be the concerned invertible $n\times n$ matrix.

DanielSank

@MAFIA36790 ok

@MAFIA36790 Do you understand geometrically why that's true?

It's actually quite intuitive!

user116211

Then $P_1, P_2,\ldots,P_n$ forms the basis of the concerned vector space.

user116211

$P_i$ is the $i$th column of $P\,.$

user116211

@DanielSank Well, the author proceeded to prove it but mentioned no geometrical significance, I fear.

user116211

Anyways, then he wrote the product $PX$ where $X$ is a column matrix as $$PX = x_1P_1 + \ldots + x_nP_n\,.$$

user116211

Nov 10, 2016 06:27

This is what I'm not getting.

user116211

Won't the product be $P_1x_1 + \ldots+ P_nx_n\,?$

Dvij

Hello. Does anyone know why people (even physicists) refer to Weak interaction as the fourth type of interaction along with Gravitational, Electromagnetic, and Strong? I mean isn't the electro-weak unification a standard established piece of today;s Physics?

user116211

What is going is this:

user116211

$$\begin{bmatrix}P_1 & \ldots& P_n\end{bmatrix}\cdot \begin{bmatrix}x_1 \\ \vdots \\ x_n\end{bmatrix}\,.$$

user116211

@DanielSank, any insight?

DanielSank

Nov 10, 2016 06:32

@MAFIA36790 How does that differ from the author's expression?

user116211

hmm.

user116211

I don't know but why he would write in that way.

DanielSank

@Dvij History. Lots of terminology used in science is very stupid if you think about it, but people use it because of history.

@MAFIA36790 How is $x_1 P_1$ different from $P_1 x_1$?

Scalar multiplication is commutative.

user116211

Okay!

user116211

I forgot $x$ is scalar.

DanielSank

Nov 10, 2016 06:36

@MAFIA36790 let me know if you want to talk about the geometrical way of thinking about matrices.

user116211

@DanielSank, Isn't every element of a matrix an element of a field? I mean field of complex numbers; they are scalars; aren't they?

DanielSank

@MAFIA36790 Yes.

user116211

Suppose $AX = Y\,.$ Then $y_i = A_{i1}x_1 + \ldots+ A_{in}x_n\,.$

user116211

You can't write $y_i = x_1A_{i1} +\ldots+ x_n A_{in},$ can you?

DanielSank

@MAFIA36790 Of course you can.

$$A = \left( \begin{array}{cc} 3 & 5 \\ 1 & -1 \end{array} \right)$$

$$X = \left( \begin{array}{cc} 2 & 4 \\ 8 & 16 \end{array} \right)$$

$$y_1 = 3 \cdot 2 + 5 \cdot 8 = 2 \cdot 3 + 8 \cdot 5$$

@MAFIA36790 Wait a second.

What does $y_i$ mean?

Your notation is inconsistent.

user116211

Nov 10, 2016 06:47

$i$th row of $Y\,.$

user116211

@DanielSank Sorry, I should have mentioned that.

DanielSank

What is x_1?

user116211

$i$th row of $X\,.$

user116211

@JohnRennie morning.

John Rennie

Morning

DanielSank

Nov 10, 2016 06:49

The statement to remember is $$Y_{nm} = \sum_i A_{ni} X_{im}$$

user116211

@DanielSank Correct.

DanielSank

That defines matrix multiplication.

user116211

yes.

DanielSank

soooo

I don't understand your question.

user116211

In a nutshell, I'm surprised as the author switched the position like saying $Y_{nm} =\displaystyle \sum_i X_{im}A_{ni}\,.$

user116211

Nov 10, 2016 06:57

Matrix multiplication is not commutative, as everyone knows.

DanielSank

@MAFIA36790 If that's all the author did, it's just silly, but not wrong.

$$X_{im} A_{ni} = A_{ni} X_{im}$$

user228700

Man, I do not get springs >.<

Secret

[Example of circular logic] I want the feeling of hopelessness to feel the pain of the feeling of hopelessness itself, so it will stop bothering me by making me to feel hopeless, which is painful.

user116211

@DanielSank okay.

DanielSank

@Kaumudi Hi.

Bonjour, Monsieur LeRennie.

That "R" sound is just delicious.

user116211

Nov 10, 2016 07:12

@DanielSank yes, I would surely love to; but let me complete what the author has to say.

user228700

@DanielSank Ello :-)

user116211

Got it; that's easy.

user116211

If you have time @DanielSank, you can go.

user116211

They first proved, the set is linearly independent; then they showed why it spans $F^{n\times 1}\,.$

John Rennie

@Kaumudi Hookes law? You've done much harder stuff than that ...

DanielSank

Nov 10, 2016 07:21

@MAFIA36790 Imagine you live in a 2D plane.

user116211

done.

DanielSank

On that plane there is a set of coordinate axes.

user116211

okay.

user228700

@JohnRennie Hooke's law isn't so bad. It's when I've to deal with springs attached to blocks, performing S.H.M >.<

DanielSank

Those axes could be thought of as a set of vectors $$\left( \begin{array}{c} 1 \\ 0 \end{array} \right) \qquad \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$$

user116211

Nov 10, 2016 07:23

okay.

John Rennie

@Kaumudi SHM is easy, and the spring is just an excuse to introduce the type of force required for SHM.

DanielSank

Now, suppose you have a matrix $A$ with components $A_{ij}$.

What do you get if you act $A$ on the first coordinate (basis) vector?

user228700

@JohnRennie Yeah, but I'm finding it difficult to internalize everything that's happening when there are multiple springs and blocks and what not.

DanielSank

@MAFIA36790 you dead?

user228700

It's very interesting and all, but it's a little difficult :-|

user228700

Nov 10, 2016 07:31

@JohnRennie: Are u especially busy at the moment? (Or are u just drinking coffee and chilling? :-P)

user116211

@DanielSank sorry, connectivity interruption.

user116211

@DanielSank Order of $A\,?$

John Rennie

@Kaumudi no. Are you still perplexed by springs? :-)

DanielSank

@MAFIA36790 huh?

user116211

I mean $m\times n\,.$

user228700

Nov 10, 2016 07:33

Yep yep yep. I've an actual problem (and a crappy solution :-|) Maybe it will help to see how it should be approached..? (By superhuman geniuses like urself :-P)

DanielSank

@MAFIA36790 $2 \times 2$.

user116211

okay.

John Rennie

Ok let's see the problem ...

Secret

0

Q: Is it possible for $\nabla f=y\mathbf{i}-x\mathbf{j}$

is it possible for a function $f(x,y)$ satisfying $\nabla f=y\mathbf{i}-x\mathbf{j}$ ? Since $\frac{\partial^2 f}{\partial x\partial y}=-1$ and $\frac{\partial^2 f}{\partial y\partial x}=1$, thus $f$ should be singular. I find a similar answer in wikipedia "Symmetry of second derivatives", $f(...

Off to MSE with you

DanielSank

@MAFIA36790 I'm going to sleep, unless you want to go through this now.

user116211

Nov 10, 2016 07:37

$(A\textrm{Basis}_1)_{11} = A_{11}\cdot 1 + A_{12} \cdot 0 $

user228700

John Rennie

That looks straightforward ...

user116211

$\begin{bmatrix}A_{11}\\ A_{21}\end{bmatrix}$

user228700

Of course :-P

John Rennie

The system is symmetrical about the mid point, so you need only consider the motion of one block.

user228700

Nov 10, 2016 07:39

( ^ Sarcasm)

Anonymous

@JohnRennie did you mean centre of mass ?

Anonymous

or literally mid point ?

John Rennie

The mid point is the centre of mass

user116211

Am I right @DanielSank?

user228700

But $m_1≠m_2$

Anonymous

Nov 10, 2016 07:40

@JohnRennie But the masses are different

John Rennie

@S007 Oops, well spotted, I assumed the masses were identical :-)

Anonymous

@JohnRennie :-)

John Rennie

@Kaumudi as S007 says, find the centre of mass and put your origin at that point.

DanielSank

@MAFIA36790 Yeah

Right. What about for the second basis vector?

John Rennie

@Kaumudi I have something to do for the next ten minutes or so, but if you don't mind waiting I can go through the detail.

user228700

Nov 10, 2016 07:43

@JohnRennie I'll try and wrap my head around it while I wait...

user116211

$\begin{bmatrix}A_{12}\\ A_{22}\end{bmatrix}$

DanielSank

Very good.

user228700

Perhaps we should continue later...I need to do something else in 15 mins :-|

DanielSank

Notice this simple fact: The $i^\text{th}$ column of a matrix is simply the vector you get by acting that matrix on the $i^\text{th}$ basis vector.

user116211

yes!

Anonymous

Nov 10, 2016 07:45

@Kaumudi The question is simple....see this youtube.com/… will get it

DanielSank

Ok, so, think about what happens if we act $A$ on our pair of basis vectors.

We get two new vectors.

user116211

The sum of which is ....

DanielSank

Don't worry about the sum.

user116211

okay.

DanielSank

The point is that we start with two orthogonal vectors, and we wind up with two new vectors.

user116211

Nov 10, 2016 07:45

yes.

DanielSank

Now, what does it mean for $A$ to be invertible?

user116211

$A$ is row-equivalent to $2\times 2$ identity matrix?

DanielSank

$A$ is arbitrary.

If $A$ is invertible, then given any vector $v$ in the 2D plane, there is some vector $u$ such that $Au = v$.

@MAFIA36790 No, $A$ is not the identity.

user116211

yes.

DanielSank

$A$ is something arbitrary.

user116211

Nov 10, 2016 07:47

gotcha.

DanielSank

Ok, now...

Let's call the two basis vectors $e_1$ and $e_2$.

user116211

okay.

DanielSank

Suppose $v$ can be written as a sum of $Ae_1$ and $Ae_2$.

So, e.g. $v = \alpha A e_1 + \beta A e_2$.

user116211

yeh.

DanielSank

In that case, can you tell me what the inverse of $v$ is?

Uh, not the inverse... I mean the thing that $A$ turns into $v$.

"Inverse" is the wrong word.

user116211

Nov 10, 2016 07:51

$\alpha e_1 + \beta e_2\,?$

DanielSank

Exactly

So, it seems like we can always find some $u$ such that $Au = v$. All we do is write $v$ in terms of $Ae_1$ and $Ae_2$, and then the value of $u$ is obvious.

Right?

user116211

yeh, sure.

DanielSank

Ok! Now if that were true that we could always do this, then every matrix $A$ would be invertible.

user116211

yes, there is always a $u$ for which the relation, above, is true.

DanielSank

However, what happens if, for some $v$ we cannot write $v$ as a sum of $Ae_1$ and $Ae_2$?

For example, what if $Ae_1$ and $Ae_2$ are parallel?

Then there is a direction in which we could point $v$ such that $v$ cannot be expressed as a sum of $Ae_1$ and $Ae_2$.

...and so our strategy fails.

user116211

Nov 10, 2016 07:55

Then our assumption should be wrong.

user116211

@DanielSank yes.

DanielSank

So we see the following: we can't invert $A$ if $A$ takes the two basis vectors and makes them parallel.

Think of $e_1$ and $e_2$ as cutting out a square in the 2D plane.

user116211

yes.

DanielSank

Another way to say what I said above just now is "If the shape cut out by $Ae_1$ and $Ae_2$ has nonzero area, then $A$ is invertible".

Mew

@heather, what is it?

@S007, let me know when you're going to behave sensibly and we can chat

DanielSank

Nov 10, 2016 07:58

@Mew Yeah, that's a useful tone which is likely to encourage everyone to get along and make progress.

Mew

@DanielSank, he spammed the new site with foul language and offensive posts

most of it has been hidden by moderators

DanielSank

@Mew ok.

Mew

I don't think I need to be polite to him when he's showing this behaviour

DanielSank

Still not seeing how sarcasm is going to help.

@Mew I strongly disagree. This site has a "be nice" policy.

Mew

@DanielSank, even to those who are trying to destroy it?

DanielSank

Nov 10, 2016 08:00

This site's rules don't care about users' activities on other sites.

user116211

@DanielSank cutting a square? I'm failing to visualise :(

DanielSank

@Mew Yes.

Mew

well we aren't on the site are we and i'm not being mean

DanielSank

Mew

I'm just saying when he starts being sensibly, we can chat

behaving*

DanielSank

Nov 10, 2016 08:00

@MAFIA36790 see picture

Mew

and clearly he isn't behaving sensibly

user116211

@DanielSank got it.

DanielSank

Ok, so we've established that if the shape cut out by the transformed basis vectors has zero area, then the matrix is not invertible.

user116211

@DanielSank Yes, when they are parallel.

DanielSank

Another way to see why is like this: If the transformation enacted by the matrix squashes the basis vectors together, then there's no way to invert because a point on the squashed vectors could have come from either $e_1$ or $e_2$.

Let me know if that makes sense. It's a critical thing to understand.

user116211

Nov 10, 2016 08:02

Okay, I'm seeing what you wanted to mean by the geometrical significance.

DanielSank

Ok, @MAFIA36790 now there's one last point to make that's very interesting.

user116211

tell tell.

DanielSank

This whole reasoning works in any dimension. If we take a 3x3 matrix, then we could act it on the x, y, and z basis vectors.

user116211

yes.

user116211

That's very true.

DanielSank

Nov 10, 2016 08:04

If the action of the matrix turns the xyz cube into something with zero volume, then the matrix is not invertible.

Ok so far?

user116211

@DanielSank okay.

DanielSank

Now to the final point: do you know what a determinant is?

user116211

yes from high school; but haven't read it rigorously now.

user116211

It is in chapter 5 in Hoffman, Kunze.

user116211

And I'm in chapter 2: Vector Spaces.

DanielSank

Nov 10, 2016 08:05

ok ok

There is a particular number associated to each matrix called its determinant. Never mind how you compute it for now.

Here's the important thing to remember:

user116211

okay.

DanielSank

The determinant of a matrix is the volume/area/whatever of the geometrical object produced by acting that matrix on the set of basis vectors.

user116211

okay!

user116211

This is new to me!

DanielSank

SO! If the determinant of a matrix is zero, the matrix is not invertible!

If the determinant of the matrix is not zero, the matrix is invertible.

user116211

Nov 10, 2016 08:07

Yes, I know that.

DanielSank

But now you know why. It's because the determinant is the volume of the transformation.

user116211

Singular matrices has no inverse matrix.

DanielSank

If the transformation squashes the volume, then you can't find an inverse for any point in the squashed area.

user116211

@DanielSank yes, I didn't know about the geometrical motivation behind it; now I got the whole point.

DanielSank

cool

Geometrical Significance of column matrices of an $n\times n$ invertible matrix over field $F$ acting as the basis of $F^{n\times 1}\,.$

The h Bar

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