Videos > Geometry of Linear Algebra | 18.06SC Linear Algebra, Fall 2011
Geometry of Linear Algebra | 18.06SC Linear Algebra, Fall 2011 Video Lecture
FAQs on Geometry of Linear Algebra - 18.06SC Linear Algebra, Fall 2011 Video Lecture
| 1. What is the connection between geometry and linear algebra? |  |
Ans. Geometry and linear algebra are closely related as linear algebra provides a mathematical framework for studying geometric concepts. Linear algebra allows us to represent geometric objects such as points, lines, and planes using vectors and matrices. It also provides tools to solve geometric problems and analyze the properties of geometric transformations.
| 2. How can linear algebra be used to represent lines and planes in space? | |
Ans. Linear algebra allows us to represent lines and planes in space using vector equations. For example, a line in three-dimensional space can be represented as the set of all points that can be reached by starting from a given point and moving in the direction of a vector parallel to the line. Similarly, a plane can be represented as the set of all points that satisfy a linear equation involving the coordinates of the points.
| 3. What are eigenvectors and eigenvalues in linear algebra? | |
Ans. Eigenvectors and eigenvalues are important concepts in linear algebra. An eigenvector of a square matrix is a non-zero vector that, when multiplied by the matrix, results in a scalar multiple of the original vector. The corresponding scalar multiple is called the eigenvalue. Eigenvectors and eigenvalues have various applications, such as studying stability in dynamic systems and determining important directions in data analysis.
| 4. How can linear algebra be used to solve systems of linear equations? | |
Ans. Linear algebra provides methods for solving systems of linear equations, where multiple equations are simultaneously satisfied. One common approach is to represent the system of equations as a matrix equation, where the coefficients of the variables are arranged in a matrix and the variables themselves are represented as a vector. By performing matrix operations, such as row reduction, it is possible to find the solutions to the system of equations.
| 5. What is the significance of determinants in linear algebra? | |
Ans. Determinants are a key concept in linear algebra with various applications. A determinant of a square matrix encodes information about the matrix's invertibility, the linear dependence of its column vectors, and the scaling factor of the matrix's transformations. Additionally, determinants are used to compute eigenvalues, solve systems of linear equations, and determine the area or volume of geometric objects.
Text Transcript from Video
hello I'm Lennon welcome to recitation
of linear algebra it's my great pleasure
to guide you through the first
recitation in the first lecture we learn
some important concepts we discussed how
to view a linear system of equations
from different points and we discussed
the concepts such as row picture column
picture and the form in matrix some of
them may be new to you so today we're
going to use the simple example to
review those concepts we're going to
work with this simple system of two
equations with two unknowns so I would
like you to first solve it and then to
find out the Associated row picture and
column picture with the system after you
are done we're go also going to discuss
the matrix form of this linear system
why don't you pause the video now and
try to work them out on your own a good
suggestion would be you can try to
sketch your answer in an XY coordinate
like this okay I'll see you in a while
I hope you have just set some fun with
it let's work on it together well we're
going to solve this equations as you can
see we have two unknowns x and y and
they have to satisfy these two equations
at the same time how would you solve it
a very simple way would be you
substitute X by Y right in terms of Y so
let's do it this way so we use the
second equation to rewrite X as 2y minus
1 so 2 y -1 then you plug this into the
first equation this implies twice X
would be replaced by this 2 times 2y
minus 1 plus y is equal to 3 then you
simplify this equation here you will
arrive at that's 5y minus 2 is equal to
3 that simply tells you y is equal to 1
if Y is equal to 1 then we go back to
this formula we see that X is also equal
to 1 that's it this is the answer to
this linear system and it's easy enough
both x and y are 1 now let's try to find
out it's row picture in column picture
so I'm going to work on this XY
coordinate here first let's look at row
picture
please review what a row picture is so
by row picture I mean you have to look
at this linear system according to each
row so what is the first row well the
first row is an equation with two
unknowns so two times X plus y is equal
to three what is this equation exactly
as you may remember this equation
actually gives you a straight line in XY
plane so now let's put the line here I
want a line that satisfy satisfies 2x
plus y is 3 let's first set X to be 0 if
X is 0 by the first equation y should be
3 which is here now let's set X to be 1
if X is 1 then Y is also 1 so what I
have now are two points on the line and
this is enough because on the XY plane
two points will uniquely determine a
straight line and that's the line we're
looking for so all we need to do is to
connect these two points we try to draw
this line streets
okay so this is the line that is given
by 2x plus y is equal to three this is
the easy is this way to determine line
you just need to pick two points that
are on the line then you connect them
and here you have the straight line all
right so that that's the first row now
let's look at the second row the second
row is X minus 2 y is equal to negative
1 same thing we're going to locate two
points on the second line so again let's
set X to be 0 first if X is zero then Y
should be 1/2 right so this line has to
cross this point that's X 0 y 1/2 then
let's put X to be 1 again so if X is 1
then Y is also 1 so we're going to use
the same point here now we have two
points on the second line to connect
them so by connecting them I will have
my second straight line so this is the
line corresponding to X minus 2 y is
negative 1 this is the row picture
because we have separated two equations
and we look at them separately
respectively the first equation gives me
this line and the second equation gives
me this line then what do I mean by
solving this linear system well we're
putting these two lines together and
we're looking at a point which is on the
SiC first line second line at the same
time then clearly that's the point where
they intersect right we can see from the
answer over there the quantity of this
point should be 1 1
this is also clear from the construction
of the two lines we have noticed that
this point is in the first line and the
second line at the same time okay two
lines meeting at the point one one
that's the row picture of this linear
system now let's move on to the column
picture again I will need an XY
coordinate okay so where can I find my
columns if you look at that two
equations then you focus on the
coordinates in front of X sorry the
coefficient in front of X in both
equations okay what would that be well
in the first equation I have a 2 in
front of X in the second equation I have
a 1 in front of X I want to put them
together as a column vector let me call
it v1 okay and I'm going to do the same
thing to the coefficients in front of Y
in the first equation I have a 1 in
front of Y in the second equation I have
a negative 2 in front of Y put them
together call it a column vector v2
these are the columns I am looking for
then what does that linear system say
well I have extracted the coefficients
in front of X now I can consider X to be
the coefficient of this vector in same
thing I'm going to view Y as the
coefficient of this vector then what
that linear system is doing is just to
sum them up that gives you the left-hand
side of the linear system which is the
right-hand side again you put the two
constants as a column vector which is 3
negative 1 that is the right-hand side
okay so what I'm doing here is I'm
taking the linear combination of V 1 and
V 2 and the coefficients are given by x
and y respectively and I want the result
of this combination to be 3 negative 1
now let's incorporate this into this
picture I have a V 1 so I'm going to
draw a vector V 1 X is 2 y is 1 so V 1
is here that's the one and then neither
V 2 X is 1 Y is negative 2 so that's my
V 2
I want to take the sum of X multiple of
V 1 and Y multiple of V 2 and I want the
result to be 3 negative 1 well taking a
hint from the previous consideration we
know that both x and y should be 1 so
I'm actually some a one copy of V one
and one copy of V two okay so how do you
indicate the sum of these two vectors
you complete the parallelogram spanned
by these two vectors then the vector
given by the diagonal is the sum of V 1
and V 2 is this vector going to be 3
negative 1 well we can check the x
coordinate will be 2 plus 1 which is 3
that's 2 plus 1 3 and the y coordinate
will be 1 minus 2 so 1 minus 2 which
should be negative 1 okay that's it
that's one multiple of V one and one
multiple of V 2 the sum will be 3
negative 1 and that's the row picture
where does that X is equal to Y is equal
to 1 come from it comes from solving the
linear system it comes from the row
picture ok so here we have found out the
row picture and the column picture of
this linear system what I would like to
mention is the matrix form of this
linear system so what is the matrix form
what if I put these two column vectors
together so I want to put them back to
back V 1 and V 2 and I call this matrix
to be a so if you write it out a should
be given by 2 1 1 negative 2 matrix a
has V 1 and V 2 as its column vectors
and it's a 2 by 2 matrix if I consider
if I take this into account then what
will be the right-hand side sorry the
left-hand side of the linear system in
other words what will be the left-hand
side of this equation this is actually
matrix a multiplying a vector given by X
1 so that's 2 1 1 negative 2 multiplying
x and y so you put both unknowns
together as a column vector that's the
left-hand side of the equation and again
the right-hand side is given by this
column vector 3 negative 1 this is the
matrix form of this linear system we can
actually solve this directly in other
words we can get this unknown vector at
once both x in both x coordinate and y
coordinate okay let's recall if you have
a scalar equation like this let's say a
is some constant times X is unknown is
equal to B if a is nonzero what would be
X so clearly X should be B over a I can
also write it as a inverse times B
that's what we do when we have members
so here what we want to apply is a
similar idea but to matrix what you want
to find is a matrix a inverse such that
a inverse times a is equal to an
identity matrix which is 1 0 0 1 this
may be new to you but as you as you go
further into this course this idea will
become more and more natural if such an
inverse matrix matrix exists then what
would be this vector here then X Y will
simply be a inverse times 3 negative 1
that will give you the answer here I'm
not going to go into detail but we will
return to this later in this course I
hope this simple example is helpful to
you in reviewing what you've learned in
the lecture thank you for watching and
I'm looking forward to seeing you again
of linear algebra it's my great pleasure
to guide you through the first
recitation in the first lecture we learn
some important concepts we discussed how
to view a linear system of equations
from different points and we discussed
the concepts such as row picture column
picture and the form in matrix some of
them may be new to you so today we're
going to use the simple example to
review those concepts we're going to
work with this simple system of two
equations with two unknowns so I would
like you to first solve it and then to
find out the Associated row picture and
column picture with the system after you
are done we're go also going to discuss
the matrix form of this linear system
why don't you pause the video now and
try to work them out on your own a good
suggestion would be you can try to
sketch your answer in an XY coordinate
like this okay I'll see you in a while
I hope you have just set some fun with
it let's work on it together well we're
going to solve this equations as you can
see we have two unknowns x and y and
they have to satisfy these two equations
at the same time how would you solve it
a very simple way would be you
substitute X by Y right in terms of Y so
let's do it this way so we use the
second equation to rewrite X as 2y minus
1 so 2 y -1 then you plug this into the
first equation this implies twice X
would be replaced by this 2 times 2y
minus 1 plus y is equal to 3 then you
simplify this equation here you will
arrive at that's 5y minus 2 is equal to
3 that simply tells you y is equal to 1
if Y is equal to 1 then we go back to
this formula we see that X is also equal
to 1 that's it this is the answer to
this linear system and it's easy enough
both x and y are 1 now let's try to find
out it's row picture in column picture
so I'm going to work on this XY
coordinate here first let's look at row
picture
please review what a row picture is so
by row picture I mean you have to look
at this linear system according to each
row so what is the first row well the
first row is an equation with two
unknowns so two times X plus y is equal
to three what is this equation exactly
as you may remember this equation
actually gives you a straight line in XY
plane so now let's put the line here I
want a line that satisfy satisfies 2x
plus y is 3 let's first set X to be 0 if
X is 0 by the first equation y should be
3 which is here now let's set X to be 1
if X is 1 then Y is also 1 so what I
have now are two points on the line and
this is enough because on the XY plane
two points will uniquely determine a
straight line and that's the line we're
looking for so all we need to do is to
connect these two points we try to draw
this line streets
okay so this is the line that is given
by 2x plus y is equal to three this is
the easy is this way to determine line
you just need to pick two points that
are on the line then you connect them
and here you have the straight line all
right so that that's the first row now
let's look at the second row the second
row is X minus 2 y is equal to negative
1 same thing we're going to locate two
points on the second line so again let's
set X to be 0 first if X is zero then Y
should be 1/2 right so this line has to
cross this point that's X 0 y 1/2 then
let's put X to be 1 again so if X is 1
then Y is also 1 so we're going to use
the same point here now we have two
points on the second line to connect
them so by connecting them I will have
my second straight line so this is the
line corresponding to X minus 2 y is
negative 1 this is the row picture
because we have separated two equations
and we look at them separately
respectively the first equation gives me
this line and the second equation gives
me this line then what do I mean by
solving this linear system well we're
putting these two lines together and
we're looking at a point which is on the
SiC first line second line at the same
time then clearly that's the point where
they intersect right we can see from the
answer over there the quantity of this
point should be 1 1
this is also clear from the construction
of the two lines we have noticed that
this point is in the first line and the
second line at the same time okay two
lines meeting at the point one one
that's the row picture of this linear
system now let's move on to the column
picture again I will need an XY
coordinate okay so where can I find my
columns if you look at that two
equations then you focus on the
coordinates in front of X sorry the
coefficient in front of X in both
equations okay what would that be well
in the first equation I have a 2 in
front of X in the second equation I have
a 1 in front of X I want to put them
together as a column vector let me call
it v1 okay and I'm going to do the same
thing to the coefficients in front of Y
in the first equation I have a 1 in
front of Y in the second equation I have
a negative 2 in front of Y put them
together call it a column vector v2
these are the columns I am looking for
then what does that linear system say
well I have extracted the coefficients
in front of X now I can consider X to be
the coefficient of this vector in same
thing I'm going to view Y as the
coefficient of this vector then what
that linear system is doing is just to
sum them up that gives you the left-hand
side of the linear system which is the
right-hand side again you put the two
constants as a column vector which is 3
negative 1 that is the right-hand side
okay so what I'm doing here is I'm
taking the linear combination of V 1 and
V 2 and the coefficients are given by x
and y respectively and I want the result
of this combination to be 3 negative 1
now let's incorporate this into this
picture I have a V 1 so I'm going to
draw a vector V 1 X is 2 y is 1 so V 1
is here that's the one and then neither
V 2 X is 1 Y is negative 2 so that's my
V 2
I want to take the sum of X multiple of
V 1 and Y multiple of V 2 and I want the
result to be 3 negative 1 well taking a
hint from the previous consideration we
know that both x and y should be 1 so
I'm actually some a one copy of V one
and one copy of V two okay so how do you
indicate the sum of these two vectors
you complete the parallelogram spanned
by these two vectors then the vector
given by the diagonal is the sum of V 1
and V 2 is this vector going to be 3
negative 1 well we can check the x
coordinate will be 2 plus 1 which is 3
that's 2 plus 1 3 and the y coordinate
will be 1 minus 2 so 1 minus 2 which
should be negative 1 okay that's it
that's one multiple of V one and one
multiple of V 2 the sum will be 3
negative 1 and that's the row picture
where does that X is equal to Y is equal
to 1 come from it comes from solving the
linear system it comes from the row
picture ok so here we have found out the
row picture and the column picture of
this linear system what I would like to
mention is the matrix form of this
linear system so what is the matrix form
what if I put these two column vectors
together so I want to put them back to
back V 1 and V 2 and I call this matrix
to be a so if you write it out a should
be given by 2 1 1 negative 2 matrix a
has V 1 and V 2 as its column vectors
and it's a 2 by 2 matrix if I consider
if I take this into account then what
will be the right-hand side sorry the
left-hand side of the linear system in
other words what will be the left-hand
side of this equation this is actually
matrix a multiplying a vector given by X
1 so that's 2 1 1 negative 2 multiplying
x and y so you put both unknowns
together as a column vector that's the
left-hand side of the equation and again
the right-hand side is given by this
column vector 3 negative 1 this is the
matrix form of this linear system we can
actually solve this directly in other
words we can get this unknown vector at
once both x in both x coordinate and y
coordinate okay let's recall if you have
a scalar equation like this let's say a
is some constant times X is unknown is
equal to B if a is nonzero what would be
X so clearly X should be B over a I can
also write it as a inverse times B
that's what we do when we have members
so here what we want to apply is a
similar idea but to matrix what you want
to find is a matrix a inverse such that
a inverse times a is equal to an
identity matrix which is 1 0 0 1 this
may be new to you but as you as you go
further into this course this idea will
become more and more natural if such an
inverse matrix matrix exists then what
would be this vector here then X Y will
simply be a inverse times 3 negative 1
that will give you the answer here I'm
not going to go into detail but we will
return to this later in this course I
hope this simple example is helpful to
you in reviewing what you've learned in
the lecture thank you for watching and
I'm looking forward to seeing you again
About this Video
Oct 17, 2025
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