Orthogonal Vectors and Subspaces | 18.06SC Linear Algebra, Fall 2011 Video Lecture

hi everyone I'm Dave now today I'd like
to tackle
in orthogonal subspaces so the problem
we'd like to tackle given a subspace s
and suppose s is spanned by two vectors
1 2 2 3 and 1 3 3 2 we have a question
here which is to find a basis for s perp
s perp is another subspace which is
orthogonal to s and then secondly can
every vector in R 4 be uniquely written
in terms of s and s perp so I'll let you
think about this for now and I'll come
back in a minute
hi everyone welcome back okay so why
don't we tackle this problem okay so
first off what does it mean for a vector
to be an S perp well if I have a vector
X and s perp and X is in s perp what
this means is X is going to be
orthogonal to every vector in s now
specifically s is spanned by these two
vectors so it's sufficient that X be
perpendicular to the two basis vectors
in s so specifically I can take one two
two three and dot it with X and it's
going to be zero so I'm writing I'm
treating X as a column vector here in
addition X must also be orthogonal to
one three three two so any vector X
that's an S perp must be orthogonal to
both of these vectors so what we can do
is we can write this as a matrix
equation and we do this by combining
these two vectors as rows of the matrix
so if we step back and take a look at
this equation we see that what we're
really asking is to find all X that are
in the null space of this matrix so how
do we find X in the null space of a
matrix well what we can do is we can row
reduce this matrix and try and find a
basis for the null space so I'm going to
just row reduce this matrix and notice
that by row reduction we don't actually
change the null space of a matrix so if
I'm only interested in the null space
this system is going to be equivalent to
I can keep the top row the same and then
just to simplify our lives we can take
the second row and subtract one copy of
the first row now if I do that I obtain
0 1 1 minus 1 ok now to parameterize the
null space what I'm going to do is I'm
going to write X out as components so if
I write X with components x1 x2 x3 and
x4 we see here that this matrix has a
rank of 2 now we were looking at vectors
which live in our 4 so we know that the
null space is going to have a dimension
which is 4 minus 2 so that means there
should be two vectors in the null space
of this matrix to parameterize these two
dimensional vectors what I'm going to do
is I'm going to let X 4 equal some
constant and X 3 equal another constant
so specifically I'm going to let X 4
equal B and X 3 equal a now what we do
is we take a look at these two equations
and this bottom equation will say that X
2
is equal to negative x3 plus x4
which is going to equal negative a x4
plus B and then the top equation says
that x1 is equal to negative 2 X 2 minus
2 X 3 plus 3 x4 sorry minus 3 X 4 and if
I substitute in X 2 is minus a plus B X
3 is a and X 4 is B so when the dust
settles the A's cancel and I'm left with
minus 5 B ok so we can combine
everything together and we end up
obtaining X 1 X 2 X 3 X 4 equals minus 5
B x2 is minus a plus B X 3 is a and X 4
is B and now what we can do is we can
take this vector and we can decompose it
into pieces which are a multiplied by a
vector and B multiplied by a vector so
you'll note that this is actually a
times 0 minus 1 1 0 plus B times minus 5
1 0 1 ok
so we have successfully achieved a
parameterization of the null space of
this matrix as some constant a times a
vector 0 minus 1 1 0 plus B times the
vector minus 5 1 0 1 and now we claim
that this is the entire space s perp so
s perp is going to be spanned by this
vector and this vector now notice how if
I were to take either of these two
vectors in s and dot it with any vector
in the null space by construction it
automatically vanishes ok so this
concludes part 1 now for part 2 can
every vector V and r4 be written
uniquely in terms of s and s perp the
answer is yes so how do we see this well
if I have a vector V what I can do is I
can try and write it as some constant C
1 times the vector 1 2 2 3 plus C 2
times the vector 1 3 3 2 plus the vector
C 4 sorry C 3 0 minus 1 1 0 plus C 4
minus 5 1 0 1 ok so C 1 and C 2 are
multiplying the vectors in s and C 3 and
C 4 are multiplying the vectors in s
perp so the question is given any V can
I find constants C 1 C 2 C 3 C 4 such
that this equation holds and the answer
is yes now notice how just to see just
to see why it's yes what we can do is we
can rewrite this in a matrix notation
and there
kind of a handy trick what I can do is I
can take these columns and write them as
columns of the matrix and this whole
expression is actually equivalent to
this matrix multiplied by the constant C
1 C 2 C 3 C 4 and on the right hand side
we have the vector V now by construction
these vectors are linearly independent
and we know from linear algebra that if
we have a matrix with linearly
independent columns the matrix is
invertible what this means is for any V
on the right hand side we can invert
this matrix and obtain unique
coefficients C 1 C 2 C 3 C 4 this then
gives us a unique decomposition for V in
terms of a piece which is in s and a
piece which is in s perp and in general
this can be done for any vector space
well I'd like to conclude this problem
now and I hope you had a good time