### Video Transcript

In this video, weβll learn how to
find the transpose of a matrix and identify symmetric and skew-symmetric
matrices. Many of the concepts you will study
in linear algebra were developed in the 17th century, although the very beginnings
of matrices and determinants can be traced back as far as the 400th century B.C. The first example of using a matrix
method to solve a system of linear equations occurred during the Han Dynasty,
somewhere between 200 and 100 B.C. Leibniz specifically is credited
for the introduction of the matrix determinant in Europe, whilst Gauss actually
formalized the term in the early 1800s. It wasnβt however until 1858 that
Cayley formed the concept of the matrix transpose. So letβs take a look at the
definition of the transpose of a matrix.

Consider a matrix π΄ whose πth row
and πth column element is given by π sub ππ. The transpose of π΄, which is
defined by π΄ with a superscript capital π, is then a matrix composed of the
elements of π΄ but in a different order. This time, the πth row πth column
is the πth row πth column of π΄. But actually, this looks awfully
complicated, and so we can simplify it somewhat. Take a two-by-two matrix with
elements π, π, π, π. In this matrix, the element in the
first row and first column is π, the element in the first row and second column π
sub one two is π, and so on.

Now, to find the corresponding
elements in the transpose of the matrix, we switch the values for π and π. But of course, switching one and
one leaves π sub one one unchanged. And so the element in the first row
and first column of the transpose of π΄ is π. Switching the one and the two
though tells us that the element in the second row and first column of the transpose
of the matrix must be π; well, thatβs here. Similarly, switching the two and
the one in our third element tells us that the element in the first row and second
column of our transpose, thatβs here, is π. Even if we switch the digits here,
we see that π sub two two remains unchanged. And the element in our second row
and second column stays as π. So the transpose of our matrix π΄
is the matrix π, π, π, π.

Notice that the elements on the
leading or main diagonal remain unchanged but that the remaining elements appear to
have sort of flipped across the diagonal. Slightly more formally, we can say
that the matrix transpose is achieved by simply swapping the rows with the columns
on our original matrix. And it follows that if π΄ itself is
an π-by-π matrix, then the transpose of π΄ must be an π-by-π matrix. And note that whilst we looked at
how the process works just now for a square matrix, all matrices will have their own
transpose regardless of how many rows or columns they might have.

So letβs begin by looking at an
example in which weβll find the transpose of a rectangular matrix.

Given that the matrix π΄ is defined
as negative two, six, negative six, one, eight, four, find the transpose of matrix
π΄.

We know that to find the transpose
of a matrix, we essentially swap the rows with the columns. The matrix π΄ has two rows and
three columns, so itβs a two-by-three matrix. Since weβre going to be switching
the rows with the columns in our transpose, it follows that that will be a
three-by-two matrix. It will have three rows and two
columns. Since the first element, negative
two, appears in the first row and the first column, that will remain unchanged. The remaining elements in this
first row though are transposed into the first column.

So once we have negative two in
place, we add a six here and a negative six here. Then, since the element four in our
original matrix is in the second row and the third column, we know that, in the
transpose of the matrix, it will be in the third row and the second column. We can then fill in the rest of
this column by using this row. We can move up from four if we
choose and put in eight here and a one here. The transpose then of matrix π΄ is
the three-by-two matrix with elements negative two, one, six, eight, negative six,
four.

In our second example, weβll
consider how to generate a matrix transpose given a formula for the original
matrix.

Given that π΄ is a three-by-two
matrix such that π sub ππ is equal to three π plus five π plus nine, find the
matrix which is the transpose of π΄.

Now, there are actually two ways in
which we could answer this question. We could use the formula to
generate matrix π΄. In other words, the element π sub
one one, the element on the first row and first column, is found by substituting π
equals one and π equals one into the formula. Then, once we performed this, we
could then find the transpose by swapping all of the rows and the columns. Alternatively, we could use the
formal definition of the transpose, and weβre going to use this latter method.

We know that if π΄ is a matrix
whose πth row and πth column element is π sub ππ, then the πth row and πth
column element of the transpose of π΄ is π sub ππ .So if we define the matrix π΄
as the three-by-two matrix with elements π sub one one, π sub one two, and so on,
then the transpose of π΄ is the firstly two-by-three matrix that its elements are π
sub one one, π sub two one, π sub three one, and so on. So letβs work out π sub one one by
using the formula. Weβre going to substitute π equals
one and π equals one in. So we get three times one plus five
times one plus nine, and thatβs 17.

Next, weβre going to substitute π
equals two and π equals one. So we get three times two plus five
times one plus nine, and thatβs 20. To find the final element on this
first row, we let π be equal to three and π be equal to one, and that gives us
23. For the element π one two, we
substitute π equals one and π equals two and we get 22. Continuing in this manner, and we
find that π sub two two is 25 and π sub three two is 28. The transpose of π΄ is therefore
the two-by-three matrix with elements 17, 20, 23, 22, 25, and 28.

In our next example, weβll look at
one of the key properties of the matrix transpose.

Given the matrix π΄ equals negative
eight, four, three, four, one, negative one, find the transpose of the transpose of
π΄.

Firstly, we know that to find the
transpose of a matrix, we simply switch the rows with the columns. Now matrix π΄ is a two-by-three
matrix. It has two rows and three columns,
so its transpose is going to have three rows and two columns. Letβs begin with the first row. That first row becomes the first
column in the transpose of our matrix. And since elements on the leading
diagonal remain unchanged, we know we have to have a negative eight here. Then we put a four here and a three
here. Next, we take elements from the
second row and we add them to the second column. Again, since elements in the
leading diagonal remain unchanged, we put negative one here. The remaining elements are one and
four. And so we have the transpose of π΄;
itβs this three-by-two matrix with elements negative eight, four, four, one, and
three, negative one.

But the question wants us to find
the transpose of this transpose. And at this point, you might be
able to predict whatβs going to happen, but letβs check. Firstly, we know that since there
are two columns and three rows in our transpose, then the transpose of the transpose
is going to have two rows and three columns. Letβs take the row negative eight,
four and weβll add it to our first column, so we get negative eight, four. Next, weβll take the second row and
weβll add it to our second column. Lastly, we take the third row and
we add that to our third column as shown. And so, given our matrix π΄, the
transpose of its transpose is the two-by-three matrix negative eight, four, three,
four, one, negative one.

Now observe, the transpose of the
transpose of a matrix is in fact the original matrix. Now this makes a lot of sense
because we switched the rows on the columns and then we switched them back
again. So the original matrix will be the
final result. And this holds for all types of
matrices, and this property is the first of several properties that apply to the
matrix transpose.

Given a matrix π΄, the transpose of
the transpose is equal to the original matrix π΄. Next, given a pair of matrices π΄
and π΅ whose sum and difference we can find, the transpose of the sum and difference
is the same as the sum and difference of their respective transposes. Take a scalar quantity π, and we
know that the transpose of π times some matrix π΄ will be equal to the transpose of
π΄ times that scalar. And finally, if we can find the
product of two matrices π΄ and π΅, the transpose of the product is equal to the
transpose of π΅ times the transpose of π΄. Now, of course, since matrix
multiplication is not commutative, itβs really important that we observe carefully
the order in which to multiply matrices for property four. Now, we use the definition of the
matrix transpose in our definitions for symmetric and skew-symmetric matrices. And these are both important
concepts in linear algebra.

Firstly, we say that a square
matrix π΄ is symmetric if the transpose of that matrix is equal to the original
matrix. For example, take the two-by-two
matrix one, two, two, three. We switch the first row and that
becomes the first column in the transpose, and we do the same with the second
row. That becomes the second column. And on doing so, we notice that the
matrix π΄ is the same as the transpose, and so matrix π΄ must therefore be
symmetric. It follows of course that a
nonsquare matrix cannot be symmetric. An π-by-π matrix will have a
transpose with π rows and π columns. And so the size is simply going to
be different.

Next, we say that a square matrix
is said to be skew-symmetric if the transpose is equal to negative π΄. And for this to be the case, all of
the elements on the main diagonal must be equal to zero. For instance, take the
three-by-three matrix π΄ as shown. The transpose of π΄ is the matrix
with elements zero, four, negative nine, negative four, zero, negative one, and
nine, one, zero. We notice that if we were to
multiply the scalar negative one across our original matrix, in other words, giving
us negative π΄, we would have the same result as the transpose. And so this matrix is
skew-symmetric.

With these definitions in mind,
letβs have a look at an example of how to work with a symmetric matrix.

Find the value of π₯ which makes
the matrix π΄ equals negative one, five π₯ minus three, negative 43, negative eight
symmetric.

Remember, we say that a square
matrix is symmetric if the transpose of that matrix is equal to the original
matrix. And of course, we find the
transpose by switching the rows with the columns. So letβs begin by finding the
transpose of matrix π΄. Since π΄ is a two-by-two matrix, it
follows that when we switch the rows and columns, we will still have a two-by-two
matrix. Then the elements on the first row
in matrix π΄ become the elements in the first column still in order of the transpose
of π΄. Then we take the elements in the
second row of π΄ and we add them to the second column of the transpose. And so the transpose of π΄ is as
shown.

We know that for the matrix to be
symmetric, the transpose must be equal to the original matrix. And further for two matrices to be
equal to one another, their individual elements must be equal to one another. We can already observe that the
elements on the leading diagonal are already equal to one another. And that makes a lot of sense,
since when we transpose a matrix, these elements remain unchanged. We do however notice that we have a
pair of identical equations when we try to equate the remaining elements. In other words, we know that
negative 43 must be equal to five π₯ minus three. And so we need to solve this
equation for π₯.

Weβll begin by adding three to both
sides, giving us five π₯ equals negative 40. And finally, weβll divide through
by five, which gives us π₯ equals negative eight. Note of course that we could go
back to the original matrix, substitute π₯ equals negative eight in, and
double-check that when we do the same with the transpose, we get the exact same
matrix. We found then the value of π₯ which
makes the matrix π΄ symmetric is negative eight.

Letβs now recap some of the key
points from this lesson. In this video, we learned that we
can find the transpose of any matrix by switching the rows with the columns. And we can use this formula to help
us remember this. And as a direct result of this
definition, we know that a matrix with π rows and π columns will have a transpose
of order π by π; it will have π rows and π columns. We also learned some of the key
properties, and one of these was that the transpose of a transpose is the original
matrix. Assuming we can add or subtract a
pair of matrices, the transpose of their sum or difference is equal to the sum or
difference of their respective transposes.

For scalar quantities π, the
transpose of π times some matrix π΄ is equal to π times the transpose of π΄. And finally being extra careful
because matrix multiplication is not commutative, we know that the transpose of π΄
times π΅ is equal to the transpose of π΅ times the transpose of π΄, assuming the two
matrices can be multiplied. Finally, we learned that a square
matrix π΄ is said to be symmetric if the transpose of π΄ is equal to π΄ and itβs
skew-symmetric if the transpose of π΄ is equal to negative π΄ and the main diagonal
entries are zero.