[محمد ابوزيد]

موقع

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اخوكم / محمد ابوزيد

[مهند الزهراني]

رااااائع جدا استاذي مبسط كثير للي يبغى الاساسيات ، اسمح لي بترك الموضوع قليلا ثم تثبيته ...

[الهَياء]

بآرك الله فيك ~

[محمد ابوزيد]

[SIZE="5"][FONT="Courier New"]محاضرة فى الجبر الخطى
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ولكن لا اعرف كيف يتم تثبيتها

[youtube]http://www.youtube.com/watch?v=ZK3O402wf1c&feature=player_embedded#![/youtube]


اخوكم / محمد ابوزيد

[محمد ابوزيد]

اشكرك اخى العزيز مهند على تثبيت الموضوع

ملاحظة:

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اخوكم / محمد ابوزيد

[محمد ابوزيد]

هذا هو نص المحاضرة باللغة الانجليزية مع الوقت الخاص بها
ان شاء الله ساحاول معرفة كيف يتم تثبيت الترجمة

0:05Hi. This is the first lecture inMIT's course 18.06, 0:11linear algebra,
and I'm Gilbert Strang.0:17The text for the course is this
book, Introduction to Linear Algebra.0:22And the course web page,
which has got a lot of exercises0:28from the past,
MatLab codes, the syllabus for the0:34course, is web.mit.edu/18.06.
And this is the first0:40lecture, lecture one.
So, and later we'll give the web0:46address for viewing these,
videotapes. Okay, so what's in the0:52first lecture?
This is my plan.0:58The fundamental problem of linear
algebra, which is to solve a system1:03of linear equations.
So let's start with a case when we1:09have some number of equations,
say n equations and n unknowns.1:14So an equal number of equations and
unknowns.1:20That's the normal,
nice case. And what I want to do is1:25-- with examples,
of course -- to describe,1:29first, what I call the Row picture.
That's the picture of one equation1:34at a time. It's the picture you've
seen before in two by two equations1:39where lines meet.
So in a minute,1:44you'll see lines meeting.
The second picture, I'll put a star1:49beside that, because that's such an
important one.1:53And maybe new to you is the picture
-- a column at a time.1:58And those are the rows and columns
of a matrix.2:02So the third --
the algebra way to look at the2:07problem is the matrix form and using
a matrix that I'll call A.2:12Okay, so can I do an example?
The whole semester will be examples2:17and then see what's going on with
the example. So,2:22take an example. Two equations,
two unknowns.2:27So let me take 2x -y =0,
let's say. And -x +2y=3.2:36Okay. let me -- I can even say
right away -- what's the matrix,2:44that is, what's the coefficient
matrix? The matrix that involves2:53these numbers --
a matrix is just a rectangular array2:59of numbers. Here it's two rows and
two columns, so 2 and --3:04minus 1 in the first row minus 1 and
2 in the second row,3:09that's the matrix. And the
right-hand -- the,3:13unknown -- well, we've got two
unknowns.3:18So we've got a vector,
with two components, x and x,3:24and we've got two right-hand sides
that go into a vector 0 3.3:30I couldn't resist writing the
matrix form, right -- even3:35before the pictures.
So I always will think of this as3:41the matrix A, the matrix of
coefficients, then there's a vector3:45of unknowns. Here we've only got
two unknowns. Later we'll have any3:49number of unknowns.
And that vector of unknowns,3:54well I'll often -- I'll make that x --
extra bold.3:58A and the right-hand side is also a
vector that I'll always call b.4:05So linear equations are A x equal b
and the idea now is to solve this4:11particular example and then step
back to see the bigger picture.4:17Okay, what's the picture for this
example, the Row picture?4:24Okay, so here comes the Row picture.
So that means I take one row at a4:31time and I'm drawing here the xy
plane and I'm going to plot all the4:38points that satisfy that first
equation. So I'm looking at all the4:46points that satisfy 2x-y =0.
It's often good to start with which4:53point on the horizontal line --
on this horizontal line,5:00y is zero. The x axis has y as zero
and that -- in this case,5:05actually, then x is zero. So the
point, the origin --5:10the point with coordinates (0,
) is on the line. It solves that5:16equation. Okay, tell me in --
well, I guess I have to tell you5:21another point that solves this same
equation. Let me suppose x is one,5:27so I'll take x to be one. Then y
should be two,5:33right? So there's the point one two
that also solves this equation.5:39And I could put in more points.
But, but let me put in all the5:44points at once,
because they all lie on a straight5:48line. This is a linear equation and
that word linear got the letters for5:52line in it. That's the equation --
this is the line that ... of5:56solutions to 2x-y=0 my first
row, first equation.6:00So typically, maybe,
x equal a half, y equal one will6:07work. And sure enough it does.
Okay, that's the first one. Now6:13the second one is not going to go
through the origin. It's6:19always important.
Do we go through the origin or not?6:25In this case, yes, because there's
a zero over there.6:30In this case we don't go through
the origin, because if x and y are6:35zero, we don't get three.
So, let me again say suppose y is6:40zero, what x do we actually get?
If y is zero, then I get x is minus6:46three. So if y is zero,
I go along minus three. So there's6:52one point on this second line.
Now let me say, well, suppose x is6:58minus one -- just to take another x.
If x is minus one, then this is a7:05one and I think y should be a one,
because if x is minus one,7:10then I think y should be a one and
we'll get that point.7:15Is that right? If x is minus one,
that's a one. If y is a one, that's7:19a two and the one and the two make
three and that point's7:24on the equation.
Okay. Now, I should just draw the7:28line, right, connecting those two
points at -- that will give me the7:32whole line. And if I've done this
reasonably well,7:36I think it's going to happen to go
through -- well,7:39not happen -- it was arranged to go
through that point.7:43So I think that the second line is
this one, and this is the7:49all-important point that lies on
both lines. Shall we just check7:54that that point which is the point x
equal one and y was two,8:00right? That's the point there and
that, I believe, solves8:05both equations.
Let's just check this.8:12If x is one, I have a minus one
plus four equals three,8:19okay. Apologies for drawing this
picture that you've seen before.8:27But this -- seeing the row picture --
first of all, for n equal 2,8:32two equations and two unknowns,
it's the right place to start. Okay.8:36So we've got the solution.
The point that lies on both lines.8:40Now can I come to the column picture?
Pay attention,8:44this is the key point.
So the column picture.8:48I'm now going to look at the columns
of the matrix.8:52I'm going to look at this part and
this part. I'm going to say that8:57the x part is really x times --
you see, I'm putting the two -- I'm9:02kind of getting the two
equations at once --9:07that part and then I have a y and in
the first equation it's multiplying9:12a minus one and in the second
equation a two,9:17and on the right-hand side,
zero and three. You see, the9:23columns of the matrix,
the columns of A are here and the9:28right-hand side b is there.
And now what is the equation asking9:34for? It's asking us to find --
somehow to combine that vector and9:42this one in the right amounts to get
that one. It's asking us to find9:49the right linear combination --
this is called a linear combination.9:56And it's the most fundamental
operation in the whole course.10:00It's a linear combination of the
columns. That's what we're seeing10:04on the left side.
Again, I don't want to write down a10:08big definition.
You can see what it is.10:12There's column one, there's column
two.10:15I multiply by some numbers and I add.
That's a combination --10:21a linear combination and I want to
make those numbers the right numbers10:26to produce zero three.
Okay. Now I want to draw a picture10:32that, represents what this
-- this is algebra.10:37What's the geometry,
what's the picture that goes with it?10:42Okay. So again,
these vectors have two components,10:46so I better draw a picture like that.
So can I put down these columns?10:51I'll draw these columns as they are,
and then I'll do a combination of10:55them. So the first column is over
two and down one, right?10:59So there's the first column.
The first column. Column one.11:06It's the vector two minus one. The
second column is --11:12minus one is the first component and
up two. It's here.11:19There's column two.
So this, again, you see what its11:26components are.
Its components are minus one,11:32two. Good. That's this guy. Now I
have to take a combination.11:39What combination shall I take?
Why not the right combination, what11:45the hell?
Okay. So the combination I'm going11:51to take is the right one to produce
zero three and then we'll see it11:57happen in the picture.
So the right combination is to take12:03x as one of those and two of these.
It's because we already know that12:08that's the right x and y,
so why not take the correct12:13combination here and see it happen?
Okay, so how do I picture this12:17linear combination?
So I start with this vector that's12:22already here --
so that's one of column one,12:26that's one times column one, right
there. And now I want to add on --12:30so I'm going to hook the next vector
onto the front of the arrow will12:35start the next vector and it will go
this way. So let's see,12:39can I do it right? If I added on
one of these vectors,12:44it would go left one and up two,
so we'd go left one and up two, so12:48it would probably get us to there.
Maybe I'll do dotted line for that.12:53Okay? That's one of column two
tucked onto the end,12:57but I wanted to tuck on two of
column two. So that --13:01the second one -- we'll go up left
one and up two also.13:05It'll probably end there.
And there's another one. So what13:09I've put in here is
two of column two.13:13Added on. And where did I end up?
What are the coordinates of this13:21result? What do I get when I take
one of this plus two of that?13:30I do get that, of course. There it
is, x is zero, y is three, that's b.13:38That's the answer we wanted.
And how do I do it? You see I do13:43it just like the first component.
I have a two and a minus two that13:47produces a zero,
and in the second component I have a13:52minus one and a four,
they combine to give the three.13:57But look at this picture. So
here's our key picture.14:02I combine this column and this
column to get this guy.14:08That was the b. That's the zero
three. Okay. So that idea of14:15linear combination is crucial,
and also -- do we want to think14:22about this question? Sure, why not.
What are all the combinations?14:29If I took -- can I go back to xs
and ys? This is a question for14:36really -- it's going to come up over
and over, but why don't we see it14:42once now? If I took all the xs and
all the ys, all the combinations,14:49what would be all the results?
And, actually,14:55the result would be that I could get
any right-hand side at all.15:00The combinations of this and this
would fill the whole plane.15:06You can tuck that away. We'll,
explore it further. But this idea15:12of what linear combination gives b
and what do all the linear15:17combinations give,
what are all the possible,15:22achievable right-hand sides be --
that's going to be basic. Okay.15:27Can I move to three equations and
three unknowns?15:32Because it's easy to picture the
two by two case.15:37Let me do a three by three example.
Okay, I'll sort of start it the15:41same way,
say maybe 2x-y and maybe I'll take15:48no zs as a zero and maybe a -x+2y
and maybe a -z as a --15:56oh, let me make that a minus one and,
just for variety let me take,16:05-3z, -3ys, I should keep the ys in
that line, and 4zs is, say, 4.16:13Okay. That's three equations.
I'm in three dimensions, x, y,16:19z. And, I don't have a solution yet.
So I want to understand the16:24equations and then solve them.
Okay. So how do I you understand16:30them? The row picture one way.
The column picture is another very16:35important way.
Just let's remember the matrix form,16:41here, because that's easy. The
matrix form -- what's our matrix A?16:46Our matrix A is this right-hand
side, the two and the minus one and16:51the zero from the first row,
the minus one and the two and the16:57minus one from the second row,
the zero, the minus three and the17:02four from the third row.
So it's a three by three matrix.17:07Three equations, three unknowns.
And what's our right-hand side?17:12Of course, it's the vector, zero
minus one, four.17:17Okay. So that's the way,
well, that's the short-hand to write17:22out the three equations.
But it's the picture that I'm17:28looking for today.
Okay, so the row picture.17:34All right, so I'm in three
dimensions, x,17:41y and z. And I want to take those
equations one at a time and ask --17:48and make a picture of all the points
that satisfy --17:55let's take equation number two.
If I make a picture of all the17:59points that satisfy --
all the x, y, z points that solve18:03this equation --
well, first of all,18:07the origin is not one of them.
x, y, z -- it being 0, 0, 0 would18:11not solve that equation.
So what are some points that do18:16solve the equation?
Let's see, maybe if x is one,18:21y and z could be zero. That would
work, right? So there's one point.18:26I'm looking at this second equation,
here, just, to start18:31with. Let's see.Also, I guess,18:37if z could be one,
x and y could be zero,18:43so that would just go straight up
that axis. And,18:49probably I'd want a third point here.
Let me take x to be zero,18:55z to be zero, then y would be minus
a half, right?19:02So there's a third point,
somewhere -- oh my -- okay.19:07Let's see. I want to put in all
the points that satisfy that19:12equation. Do you know what that
bunch of points will be?19:18It's a plane. If we have a linear
equation, then,19:23fortunately, the graph of the thing,
the plot of all the points that19:28solve it are a plane.
These three points determine a19:33plane, but your lecturer is not
Rembrandt and the art is going to be19:38the weak point here.
So I'm just going to draw a plane,19:43right?
There's a plane somewhere.19:48That's my plane. That plane is all
the points that solves this guy.19:53Then, what about this one? Two x
minus y plus zero z.19:58So z actually can be anything.
Again, it's going to be another20:03plane.
Each row in a three by three problem20:09gives us a plane in three dimensions.
So this one is going to be some20:14other plane -- maybe I'll try to
draw it like this.20:20And those two planes meet in a line.
So if I have two equations,20:26just the first two equations in
three dimensions,20:32those give me a line.
The line where those two planes20:39meet. And now,
the third guy is a third plane.20:45And it goes somewhere. Okay, those
three things meet in a point.20:51Now I don't know where that point is,
frankly. But --20:55linear algebra will find it.
The main point is that the three21:00planes, because they're not parallel,
they're not special.21:04They do meet in one point and
that's the solution.21:09But, maybe you can see that this row
picture is getting a little hard to21:15see. The row picture was a cinch
when we looked at two lines meeting.21:20When we look at three planes
meeting, it's not so clear and in21:26four dimensions probably
a little less clear.21:32So, can I quit on the row picture?
Or quit on the row picture before21:37I've successfully found the point
where the three planes meet?21:43All I really want to see is that
the row picture consists of three21:48planes and, if everything
works right,21:54three planes meet in one point and
that's a solution.21:59Now, you can tell I prefer the
column picture.22:04Okay, so let me take the column
picture. That's x times --22:09so there were two xs in the first
equation minus one x is,22:13and no xs in the third.
It's just the first column of that.22:19And how many ys are there? There's
minus one in the first equations,22:24two in the second and maybe minus
three in the third.22:29Just the second column of my matrix.
And z times no zs minus one zs and22:35four zs. And it's those
three columns,22:40right, that I have to combine to
produce the right-hand side,22:45which is zero minus one four.
Okay. So what have we got on this22:49left-hand side?
A linear combination.22:54It's a linear combination now of
three vectors,22:58and they happen to be --
each one is a three dimensional23:03vector,
so we want to know what combination23:07of those three vectors produces that
one. Shall I try to draw the column23:11picture, then?
So, since these vectors have three23:15components -- so it's some multiple --
let me draw in the first23:20column as before --
x is two and y is minus one.23:25Maybe there is the first column.
y -- the second column has maybe a23:32minus one and a two and the y is a
minus three, somewhere,23:39there possibly, column two.
And the third column has --23:46no zero minus one four,
so how shall I draw that?23:51So this was the first component.
The second component was a minus23:57one. Maybe up here.
That's column three,24:02that's the column zero minus one and
four.24:07This guy. So,
again, what's my problem?24:13What this equation is asking me to
do is to combine these three vectors24:19with a right combination to produce
this one. Well,24:25you can see what the right
combination is, because in24:30this special problem,
specially chosen by the lecturer,24:37that right-hand side that I'm trying
to get is actually one of these24:43columns. So I know how to get that
one. So what's the solution?24:49What combination will work? I just
want one of these and none of these.24:55So x should be zero,
y should be zero and z should be one.25:01That's the combination.
One of those is obviously the right25:06one. Column three is actually the
same as b in this particular problem.25:11I made it work that way just so we
would get an answer,25:17(0,0,1), so somehow that's the point
where those three planes met and I25:23couldn't see it before.
Of course, I won't always be able25:29to see it from the column picture,
either. It's the next lecture,25:35actually,
which is about elimination,25:41which is the systematic way that
everybody -- every bit of software,25:47too -- production, large-scale
software would solve the equations.25:53So the lecture that's coming up. If
I was to add that to the syllabus,25:59will be about how to find x, y, z in
all cases.26:05Can I just think again,
though, about the big picture?26:11By the big picture I mean let's
keep this same matrix on the left26:17but imagine that we have a different
right-hand side.26:23Oh, let me take a different
right-hand side.26:29So I'll change that right-hand side
to something that actually is also26:33pretty special.
Let me change it to --26:37if I add those first two columns,
that would give me a one and a one26:41and a minus three.
There's a very special right-hand26:45side. I just ****ed it up by adding
this one to this one.26:48Now, what's the solution with this
new right-hand side?26:54The solution with this new
right-hand side is clear.27:00took one of these and none of those.
So actually, it just changed around27:05to this when I took this
new right-hand side.27:11Okay. So in the row picture,
I have three different planes,27:17three new planes meeting now at this
point. In the column picture,27:23I have the same three columns, but
now I'm combining them27:29to produce this guy,
and it turned out that column one27:35plus column two which would be
somewhere -- there is the right27:41column -- one of this and one of
this would give me the new b.27:46Okay. So we squeezed in an extra
example. But now think about all bs,27:52all right-hand sides.
Can I solve these equations for27:58every right-hand side?
Can I ask that question?28:05So that's the algebra question.
Can I solve A x=b for every b? Let28:13me write that down.
Can I solve A x =b for every28:20right-hand side b?
I mean, is there a solution?28:26And then, if there is, elimination
will give me a way to find it.28:31I really wanted to ask, is there a
solution for every right-hand side?28:36So now, can I put that in different
words -- in this linear28:41combination words?
So in linear combination words,28:49do the linear combinations of the
columns fill three dimensional space?29:00Every b means all the bs in three
dimensional space.29:11Do you see that I'm just asking the
same question in different words?29:22Solving A x -- A x --
that's very important.29:28A times x -- when I multiply a
matrix by a vector,29:34I get a combination of the columns.
I'll write that down in a moment.29:40But in my column picture,
that's really what I'm doing.29:46I'm taking linear combinations of
these three columns and I'm trying29:55to find b. And,
actually, the answer for this matrix30:05will be yes. For this matrix A --
for these columns, the answer is yes.30:14This matrix -- that I chose for an
example is a good matrix.30:19A non-singular matrix. An
invertible matrix.30:24Those will be the matrices that we
like best. There could be other --30:29and we will see other matrices where
the answer becomes,30:34no -- oh, actually,
you can see when it would become no.30:40What could go wrong? How could it
go wrong that out of these --30:45out of three columns and all their
combinations --30:51when would I not be able to produce
some b off here?30:56When could it go wrong?
Do you see that the combinations --31:02let me say when it goes wrong.
If these three columns all lie in31:08the same plane,
then their combinations will lie in31:14that same plane.
So then we're in trouble.31:21If the three columns of my matrix --
if those three vectors happen to lie31:27in the same plane -- for example,
if column three is just the sum of31:33column one and column two,
I would be in trouble. That would31:39be a matrix A where the answer would
be no, because the combinations --31:45if column three is in the same plane
as column one and two,31:50I don't get anything new from that.
All the combinations are in the31:55plane and only right-hand sides b
that I could get would be the ones32:00in that plane.
So I could solve it for some32:04right-hand sides,
when b is in the plane,32:08but most right-hand sides would be
out of the plane and unreachable.32:13So that would be a singular case.
The matrix would be not invertible.32:17There would not be a solution for
every b. The answer would become no32:24for that. Okay.
I don't know -- shall we take just32:30a little shot at thinking about nine
dimensions? Imagine that we have32:36vectors with nine components.
Well, it's going to be hard to32:42visualize those.
I don't pretend to do it.32:48But somehow, pretend you do.
Pretend we have -- if this was nine32:53equations and nine unknowns,
then we would have nine columns,32:59and each one would be a vector in
nine-dimensional space and we would33:04be looking at their linear
combinations.33:09So we would be having the linear
combinations of nine vectors in33:13nine-dimensional space,
and we would be trying to find the33:17combination that hit the correct
right-hand side b.33:20And we might also ask the question
can we always do it?33:24Can we get every right-hand side b?
And certainly it will depend on33:28those nine columns.
Sometimes the answer will be yes --33:32if I picked a random matrix, it
would be yes, actually.33:36If I used MatLab and just used the
random command,33:40picked out a nine by nine matrix,
I guarantee it would be good. It33:44would be non-singular,
it would be invertible,33:48all beautiful.
But if I choose those columns so33:54that they're not independent,
so that the ninth column is the same34:01as the eighth column,
then it contributes nothing new and34:09there would be right-hand sides b
that I couldn't get.34:16Can you sort of think about nine
vectors in nine-dimensional space an34:21take their combinations?
That's really the central thought --34:25that you get kind of used to in
linear algebra.34:30Even though you can't really
visualize it, you sort of think you34:34can after a while.
Those nine columns and all their34:39combinations may very well fill out
the whole nine-dimensional space.34:45But if the ninth column happened to
be the same as the eighth column and34:50gave nothing new,
then probably what it would fill out34:56would be -- I hesitate
even to say this --35:01it would be a sort of a plane --
an eight dimensional plane inside35:06nine-dimensional space.
And it's those eight dimensional35:11planes inside nine-dimensional space
that we have to work with eventually.35:16For now, let's stay with a nice
case where the matrices work,35:21we can get every right-hand side b
and here we see how to35:26do it with columns.
Okay. There was one step which I35:33realized I was saying in words that
I now want to write in letters.35:41Because I'm coming back to the
matrix form of the equation,35:48so let me write it here.
The matrix form of my equation,35:55of my system is some matrix A times
some vector x equals some right-hand36:00side b. Okay.
So this is a multiplication.36:05A times x. Matrix times vector,
and I just want to say how do you36:11multiply a matrix by a vector?
Okay, so I'm just going to create a36:16matrix --
let me take two five one three --36:21and let me take a vector x to be,
say, 1and 2. How do I multiply a36:27matrix by a vector?
But just think a little bit about36:33matrix notation and how to do that
in multiplication.36:39So let me say how I multiply a
matrix by a vector.36:45Actually, there are two ways to do
it. Let me tell you my favorite way.36:50It's columns again.
It's a column at a time.36:56For me, this matrix multiplication
says I take one of that column and37:02two of that column and add.
So this is the way I would think of37:07it is one of the first column and
two of the second column and let's37:11just see what we get.
So in the first component I'm37:15getting a two and a ten.
I'm getting a twelve there.37:20In the second component I'm getting
a one and a six, I'm37:24getting a seven.
So that matrix times that vector is37:29twelve seven. Now,
you could do that another way.37:35You could do it a row at a time.
And you would get this twelve --37:41and actually I pretty much did it
here -- this way.37:47Two -- I could take that row times
my vector. This is the idea of a37:52dot product. This vector times this
vector, two times one plus five37:56times two is the twelve.
This vector times this vector --38:01one times one plus three times two
is the seven.38:05So I can do it by rows,
and in each row times my x is what38:11I'll later call a dot product.
But I also like to see it by38:17columns. I see this as a linear
combination of a column.38:23So here's my point.
A times x is a combination of the38:29columns of A. That's how I hope you
will think of A times x when we need38:37it. Right now we've got --
with small ones, we can always do it38:45in different ways,
but later, think of it that way.38:53Okay. So that's the picture for a
two by two system.39:01And if the right-hand side B
happened to be twelve seven,39:07then of course the correct solution
would be one two.39:14Okay. So let me come back next
time to a systematic way,39:21using elimination, to find the
solution, if there is one,39:28to a system of any size and find out
-- because if elimination fails,39:33find out when there isn't a solution.
Okay, thanks.



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