Lesson Video: Transpose of a Matrix Mathematics

In this video, we will learn how to find the transpose of a matrix and identify symmetric and skew-symmetric matrices.

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Video Transcript

in this video, we’ll 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 wasn’t, however, until 1858 that Cayley formed the concept of the matrix transpose.

Let’s take a look at the definition of the transpose of a matrix. Consider a matrix 𝐴 whose 𝑖th row and 𝑗th column element is given by π‘Ž sub 𝑖𝑗. We can then define the transpose of 𝐴, which is 𝐴 with a superscript capital 𝑇, by using the elements of 𝐴. This time, the transpose of 𝐴 has an 𝑖th row–𝑗th column element of π‘Ž sub 𝑗𝑖. Now this definition might seem awfully complicated, but we can simplify it somewhat. Suppose we have a two-by-two matrix with elements π‘Ž, 𝑏, 𝑐, 𝑑. The element in the first row and first column we define as π‘Ž sub one one, and that’s equal to π‘Ž. The element in the first row and second column π‘Ž sub one two is 𝑏. π‘Ž sub two one is 𝑐, and π‘Ž sub two two is equal to 𝑑.

To find the corresponding elements in the transpose of matrix 𝐴, we switch the values for 𝑖 and 𝑗. That means that π‘Ž sub one one, the element in the first row and first column, is still π‘Ž sub one one. So π‘Ž remains in place. When we switch the 𝑖- and 𝑗-values, however, the element π‘Ž sub one two becomes the element π‘Ž sub two one in the matrix transpose. And that means that we place 𝑏 in the second row and first column. Continuing in this way, and the element π‘Ž sub two one becomes the element π‘Ž sub one two. So the element in the first row and second column is 𝑐. Finally, the element π‘Ž sub two two remains in this form, and so 𝑑 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 𝐴 is an π‘š-by-𝑛 matrix, then its transpose must be an 𝑛-by-π‘š 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 let’s begin by looking at an example in which we’ll find the transpose of a rectangular matrix.

Given that the matrix 𝐴 equals negative two, six, negative six, one, eight, four, find the transpose of 𝐴.

Remember, to find the transpose of a matrix, we switch the rows and the columns. This means that if 𝐴 is an π‘š-by-𝑛 matrix, in other words it has π‘š rows and 𝑛 columns, then the transpose of 𝐴 is an 𝑛-by-π‘š matrix; it has 𝑛 rows and π‘š columns. We see that matrix 𝐴 in this case has two rows and three columns; it’s a two-by-three matrix. This means the transpose of 𝐴 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 𝐴, let’s begin by inspecting the first row of 𝐴. It has elements negative two, six, negative six. We know that the first row of 𝐴 will become the first column of the transpose of 𝐴. And so the first column of the transpose of 𝐴 looks like this. We’ll now inspect the second row of 𝐴. It has the elements one, eight, four. And we know the second row of 𝐴 will become the second column of the transpose of 𝐴. 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 𝐴 is the matrix with elements negative two, one, six, eight, negative six, four.

In our second example, we’ll look at how to generate a matrix transpose given a formula for the original matrix.

Given that 𝐴 is a three-by-two matrix such that π‘Ž sub 𝑖𝑗 equals three 𝑖 plus five 𝑗 plus nine, find the matrix 𝐴 superscript 𝑇.

We’re given information about matrix 𝐴 as a three-by-two matrix and asked to find its transpose. That’s represented by 𝐴 with a superscript capital 𝑇. And we’re 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 𝐴 and then find its transpose by swapping its rows and columns. Alternatively, we can use the definition of the transpose, and we’re going to use the latter method here.

That is, suppose 𝐴 is a matrix whose 𝑖th row–𝑗th column element is defined by π‘Ž sub 𝑖𝑗. The transpose of this matrix will then be defined with 𝑖th row–𝑗th column elements π‘Ž sub 𝑗𝑖. Now we’re told the matrix 𝐴 is a three-by-two matrix, so it has three rows and two columns. Since the 𝑖th row–𝑗th column element is defined by π‘Ž sub 𝑖𝑗, then the element in the first row and first column is π‘Ž sub one one. The element in the first row and second column is π‘Ž sub one two, and so on.

Now that we’ve defined the matrix 𝐴, let’s define the matrix the transpose of 𝐴. Since the 𝑖th row–𝑗th column element is defined by π‘Ž sub 𝑗𝑖, then the element in the first row and first column is still π‘Ž sub one one. The element in the first row and second column, however, will be defined by π‘Ž sub two one. And the element in the first row and third column is π‘Ž sub three one. Then the second row has elements π‘Ž sub one two, π‘Ž sub two two, and π‘Ž 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 𝐴 has elements π‘Ž sub one one and π‘Ž sub one two. These are mapped onto the first column in the transpose of 𝐴. Similarly, the second row has elements π‘Ž sub two one and π‘Ž 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 let’s work out the element π‘Ž sub one one by using the formula. We substitute 𝑖 equals one and 𝑗 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, we’ll substitute 𝑖 equals two and 𝑗 equals one. That gives us π‘Ž sub two one equals three times two plus five times one plus nine. And that’s equal to 20. So the second element in our first row is 20. Let’s repeat this one more time for the element π‘Ž sub three one, where we’re going to let 𝑖 be equal to three and 𝑗 be equal to one. That’s 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‘Ž 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 we’ve generated the matrix transpose of 𝐴. It’s the two-by-three matrix with elements 17, 20, 23, 22, 25, and 28.

We’ve now demonstrated how to generate a matrix transpose given a matrix and given a formula that describes a matrix. In our next example, we’ll consider a key property of the matrix transpose.

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

Remember, if we have a matrix 𝐴, the transpose of 𝐴, which is defined using the superscript capital 𝑇, is found by switching the rows and columns of matrix 𝐴. What this actually means is that if matrix 𝐴 is an π‘š-by-𝑛 matrix, then the transpose is an 𝑛-by-π‘š matrix. Now, of course, our matrix 𝐴 has two rows and three columns. So it’s 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. We’re going to take the elements in the first row of matrix 𝐴 and put them in the first column of the transpose of 𝐴.

So the first column has elements negative eight, four, three. Then we take the elements in the second row of 𝐴. 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 𝐴, let’s find the transport of this new matrix. It follows that since the transpose of 𝐴 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. We’re 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 𝐴. It’s 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.

Let’s look at these, take matrices 𝐴 and 𝐡 so that the following operations can be performed. The transpose of the transpose of 𝐴 is simply 𝐴. The transpose of the sum or difference of 𝐴 and 𝐡 is equal to the sum or difference of the transpose of 𝐴 and the transpose of 𝐡. For a constant 𝐾, the transpose of 𝐾 times 𝐴 is equal to 𝐾 times the transpose of 𝐴. And finally, the transpose of 𝐴 times 𝐡 is equal to the transpose of 𝐡 times the transpose of 𝐴.

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 𝐴 is symmetric if it’s equal to its transpose. For instance, the matrix 𝐴 with elements one, two, two, three is symmetric. That’s because its transpose is also the two-by-two matrix with elements one, two, two, three. They are equal, so 𝐴 is symmetric.

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

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

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

Remember, a square matrix 𝐴 is symmetric if its transpose is equal to the original matrix. We see that 𝐴 is a square matrix; it’s two by two. And so we’ll 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 𝐴 is a two-by-two matrix, its transpose will also be two by two.

The first row of 𝐴 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. Let’s 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, that’s negative 43. Since these are equal, we form and solve an equation for π‘₯.

It’s 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‘₯, let’s 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 𝐴 symmetric is negative eight.

Let’s 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 𝑖th row–𝑗th column element of 𝐴 is π‘Ž sub 𝑖𝑗, then the 𝑖th row–𝑗th column element of the transpose of 𝐴 is given by π‘Ž sub 𝑗𝑖. We also learned that a matrix of order π‘š by 𝑛 will have a transpose of order 𝑛 by π‘š.

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

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

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