### 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 that we study
in linear algebra were developed in the 17th century, although the beginnings of
matrices and determinants can be traced back as far as the second century BC. Liebniz specifically is credited
for the introduction of the matrix determinant in Europe, whilst Gauss formalized
the term in the early 1800s. It wasnβt, however, until 1858 that
Cayley formed the concept of the matrix transpose.

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. We can then define the transpose of
π΄, which is π΄ with a superscript capital π, by using the elements of π΄. This time, the transpose of π΄ has
an πth rowβπth column element of π sub ππ. Now this definition might seem
awfully complicated, but we can simplify it somewhat. Suppose we have a two-by-two matrix
with elements π, π, π, π. The element in the first row and
first column we define as π sub one one, and thatβs equal to π. The element in the first row and
second column π sub one two is π. π sub two one is π, and π sub
two two is equal to π.

To find the corresponding elements
in the transpose of matrix π΄, we switch the values for π and π. That means that π sub one one, the
element in the first row and first column, is still π sub one one. So π remains in place. When we switch the π- and
π-values, however, the element π sub one two becomes the element π sub two one in
the matrix transpose. And that means that we place π in
the second row and first column. Continuing in this way, and the
element π sub two one becomes the element π sub one two. So the element in the first row and
second column is π. Finally, the element π sub two two
remains in this form, and so π remains in the same place.

We now see that the elements on the
main diagonal remain unchanged. But if we consider the other
elements, they appear to have reflected or flipped across that diagonal. In fact, we see that the matrix
transpose is achieved by simply swapping the rows and columns. This also means that if π΄ is an
π-by-π matrix, then its transpose must be an π-by-π matrix. Note also that whilst we looked at
how the process works for a square matrix, all matrices 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΄ equals
negative two, six, negative six, one, eight, four, find the transpose of π΄.

Remember, to find the transpose of
a matrix, we switch the rows and the columns. This means that if π΄ is an
π-by-π matrix, in other words it has π rows and π columns, then the transpose of
π΄ is an π-by-π matrix; it has π rows and π columns. We see that matrix π΄ in this case
has two rows and three columns; itβs a two-by-three matrix. This means the transpose of π΄ will
be a three-by-two matrix. In other words, it will have three
rows and two columns.

To identify the elements of the
transpose of π΄, letβs begin by inspecting the first row of π΄. It has elements negative two, six,
negative six. We know that the first row of π΄
will become the first column of the transpose of π΄. And so the first column of the
transpose of π΄ looks like this. Weβll now inspect the second row of
π΄. It has the elements one, eight,
four. And we know the second row of π΄
will become the second column of the transpose of π΄. So we put the elements one, eight,
four here. We can now indeed see that the
transposed matrix we created is a three-by-two matrix.

And so the transpose of π΄ is the
matrix with elements negative two, one, six, eight, negative six, four.

In our second example, weβll look
at 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 equals three π plus five π plus nine, find the matrix
π΄ superscript π.

Weβre given information about
matrix π΄ as a three-by-two matrix and asked to find its transpose. Thatβs represented by π΄ with a
superscript capital π. And weβre given a formula that will
allow us to generate each element in the original matrix. And so there are two ways we could
answer the question. We could use this formula to
generate matrix π΄ and then find its transpose by swapping its rows and columns. Alternatively, we can use the
definition of the transpose, and weβre going to use the latter method here.

That is, suppose π΄ is a matrix
whose πth rowβπth column element is defined by π sub ππ. The transpose of this matrix will
then be defined with πth rowβπth column elements π sub ππ. Now weβre told the matrix π΄ is a
three-by-two matrix, so it has three rows and two columns. Since the πth rowβπth column
element is defined by π sub ππ, then the element in the first row and first
column is π sub one one. The element in the first row and
second column is π sub one two, and so on.

Now that weβve defined the matrix
π΄, letβs define the matrix the transpose of π΄. Since the πth rowβπth column
element is defined by π sub ππ, then the element in the first row and first
column is still π sub one one. The element in the first row and
second column, however, will be defined by π sub two one. And the element in the first row
and third column is π sub three one. Then the second row has elements π
sub one two, π sub two two, and π sub three two.

And for clarity, we can
double-check by making sure that the rows and columns have switched places. The first row in matrix π΄ has
elements π sub one one and π sub one two. These are mapped onto the first
column in the transpose of π΄. Similarly, the second row has
elements π sub two one and π sub two two, which go into the second column of our
transpose. Finally, the third row maps onto
the third column in our transpose.

So letβs work out the element π
sub one one by using the formula. We substitute π equals one and π
equals one into the formula given. And we get three times one plus
five times one plus nine, which is equal to 17. And so the element in the first row
and first column of our transpose matrix is 17.

Next, weβll substitute π equals
two and π equals one. That gives us π sub two one equals
three times two plus five times one plus nine. And thatβs equal to 20. So the second element in our first
row is 20. Letβs repeat this one more time for
the element π sub three one, where weβre going to let π be equal to three and π
be equal to one. Thatβs three times three plus five
times one plus nine, which is equal to 23. Repeating this process for the
remaining elements, and we find that element π sub one two, that is, the element in
the second row and first column, is 22. The second element in this row is
25, and the third is 28.

And so weβve generated the matrix
transpose of π΄. Itβs the two-by-three matrix with
elements 17, 20, 23, 22, 25, and 28.

Weβve now demonstrated how to
generate a matrix transpose given a matrix and given a formula that describes a
matrix. In our next example, weβll consider
a key property of the matrix transpose.

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

Remember, if we have a matrix π΄,
the transpose of π΄, which is defined using the superscript capital π, is found by
switching the rows and columns of matrix π΄. What this actually means is that if
matrix π΄ is an π-by-π matrix, then the transpose is an π-by-π matrix. Now, of course, our matrix π΄ has
two rows and three columns. So itβs a two-by-three matrix. That means its transpose is a
three-by-two matrix. In other words, it has three rows
and two columns. Weβre going to take the elements in
the first row of matrix π΄ and put them in the first column of the transpose of
π΄.

So the first column has elements
negative eight, four, three. Then we take the elements in the
second row of π΄. And we move them into the second
column of the transpose. So this column has elements four,
one, negative one. Now that we have the transpose of
π΄, letβs find the transport of this new matrix. It follows that since the transpose
of π΄ is a three-by-two matrix, the transpose of this matrix will be two-by-three
once again; it will have two rows and three columns. Weβre going to take the elements in
the first row and put them into the first column. So that contains the elements
negative eight and four.

Then we take the elements in the
second row and put them in the second column. So the second column has elements
four and one. And finally, we take the elements
in the third row, and we put them in the third column. And so we found the transpose of
the transpose of π΄. Itβs the two-by-three matrix with
elements negative eight, four, three, four, one, negative one.

Now, if we look really carefully,
we see that the transpose of the transpose of the matrix was in fact the original
matrix. And this holds for all
matrices. This property is in fact the first
of several properties that apply to the matrix transpose.

Letβs look at these, take matrices
π΄ and π΅ so that the following operations can be performed. The transpose of the transpose of
π΄ is simply π΄. The transpose of the sum or
difference of π΄ and π΅ is equal to the sum or difference of the transpose of π΄ and
the transpose of π΅. For a constant πΎ, the transpose of
πΎ times π΄ is equal to πΎ times the transpose of π΄. And finally, the transpose of π΄
times π΅ is equal to the transpose of π΅ times the transpose of π΄.

We also use the definition of the
matrix transpose in the definitions for symmetric and skew-symmetric matrices, which
are both important concepts in linear algebra. For instance, we say that a square
matrix π΄ is symmetric if itβs equal to its transpose. For instance, the matrix π΄ with
elements one, two, two, three is symmetric. Thatβs because its transpose is
also the two-by-two matrix with elements one, two, two, three. They are equal, so π΄ is
symmetric.

The next definition is that of a
skew-symmetric matrix. A square matrix π΄ is said to be
skew-symmetric if its transpose is equal to negative π΄. Now, in fact, in this case, the
main diagonal entries end up being equal to zero. For instance, this time, suppose π΄
is the square matrix with elements zero, negative four, nine, four, zero, one,
negative nine, negative one, zero. The transpose is the matrix with
elements zero, four, negative nine, negative four, zero, negative one, nine, one,
zero. This is quite clearly equal to the
negative of π΄. If we take each element in the
original matrix and multiply it by negative one, we get the transpose of π΄. And so matrix π΄ here is
skew-symmetric.

In our final example, weβll use the
definition of a symmetric matrix to find missing values.

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

Remember, a square matrix π΄ is
symmetric if its transpose is equal to the original matrix. We see that π΄ is a square matrix;
itβs two by two. And so weβll be able to identify
the value of π₯ which makes the matrix symmetric by first finding its transpose. The transpose is found by switching
the rows and columns. And so since π΄ is a two-by-two
matrix, its transpose will also be two by two.

The first row of π΄ has elements
negative one and five π₯ minus three. So this is the first column of the
transpose. The second row has elements
negative 43 and negative eight. So this is the second column. So we need to find the values of π₯
which make these two matrices equal. Now, of course, for two matrices to
be equal, we know their individual elements must also be equal. Letβs consider in particular the
element in the first row and second column. In our first matrix, that is five
π₯ minus three. And in the transpose, thatβs
negative 43. Since these are equal, we form and
solve an equation for π₯.

Itβs worth also noting that had we
instead equated the element in the second row and first column, we would have ended
up with the same equation. To solve this equation for π₯,
letβs add three to both sides. So five π₯ equals negative 40. Then we divide through by five, so
π₯ is equal to negative eight. The value of π₯ which makes the
matrix π΄ symmetric is negative eight.

Letβs now recap the key points from
this lesson. In this video, we learned that to
find the transpose of a matrix, we switch the rows and the columns. So if the πth rowβπth column
element of π΄ is π sub ππ, then the πth rowβπth column element of the transpose
of π΄ is given by π sub ππ. We also learned that a matrix of
order π by π will have a transpose of order π by π.

And we saw that for matrices π΄ and
π΅ where the following operations are possible, the transpose of the transpose of π΄
is matrix π΄. The transpose of the sum or
difference of π΄ and π΅ is equal to the sum or difference of the transpose of π΄ and
the transpose of π΅. For a constant πΎ, the transpose of
πΎ times π΄ is πΎ times the transpose of π΄. And the transpose of the product of
π΄ and π΅ is equal to the transpose of π΅ times the transpose of π΄.

Finally, we learned that a square
matrix is said to be symmetric if its transpose is equal to the original matrix. And itβs said to be skew-symmetric
if its transpose is equal to the negative of the original matrix.