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In this video, weβll learn how to find the transpose of a matrix and identify symmetric and skew-symmetric matrices.
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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.
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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.
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Leibniz specifically is credited for the introduction of the matrix determinant in Europe, whilst Gauss actually formalized the term in the early 1800s.
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It wasnβt however until 1858 that Cayley formed the concept of the matrix transpose.
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So letβs take a look at the definition of the transpose of a matrix.
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Consider a matrix π΄ whose πth row and πth column element is given by π sub ππ.
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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.
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This time, the πth row πth column is the πth row πth column of π΄.
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But actually, this looks awfully complicated, and so we can simplify it somewhat.
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Take a two-by-two matrix with elements π, π, π, π.
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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.
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Now, to find the corresponding elements in the transpose of the matrix, we switch the values for π and π.
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But of course, switching one and one leaves π sub one one unchanged.
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And so the element in the first row and first column of the transpose of π΄ is π.
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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.
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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.
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Even if we switch the digits here, we see that π sub two two remains unchanged.
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And the element in our second row and second column stays as π.
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So the transpose of our matrix π΄ is the matrix π, π, π, π.
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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.
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Slightly more formally, we can say that the matrix transpose is achieved by simply swapping the rows with the columns on our original matrix.
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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.
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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.
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So letβs begin by looking at an example in which weβll find the transpose of a rectangular matrix.
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Given that the matrix π΄ is defined as negative two, six, negative six, one, eight, four, find the transpose of matrix π΄.
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We know that to find the transpose of a matrix, we essentially swap the rows with the columns.
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The matrix π΄ has two rows and three columns, so itβs a two-by-three matrix.
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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.
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It will have three rows and two columns.
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Since the first element, negative two, appears in the first row and the first column, that will remain unchanged.
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The remaining elements in this first row though are transposed into the first column.
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So once we have negative two in place, we add a six here and a negative six here.
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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.
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We can then fill in the rest of this column by using this row.
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We can move up from four if we choose and put in eight here and a one here.
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The transpose then of matrix π΄ is the three-by-two matrix with elements negative two, one, six, eight, negative six, four.
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In our second example, weβll consider how to generate a matrix transpose given a formula for the original matrix.
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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΄.
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Now, there are actually two ways in which we could answer this question.
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We could use the formula to generate matrix π΄.
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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.
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Then, once we performed this, we could then find the transpose by swapping all of the rows and the columns.
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Alternatively, we could use the formal definition of the transpose, and weβre going to use this latter method.
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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.
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So letβs work out π sub one one by using the formula.
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Weβre going to substitute π equals one and π equals one in.
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So we get three times one plus five times one plus nine, and thatβs 17.
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Next, weβre going to substitute π equals two and π equals one.
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So we get three times two plus five times one plus nine, and thatβs 20.
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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.
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For the element π one two, we substitute π equals one and π equals two and we get 22.
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Continuing in this manner, and we find that π sub two two is 25 and π sub three two is 28.
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The transpose of π΄ is therefore the two-by-three matrix with elements 17, 20, 23, 22, 25, and 28.
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In our next example, weβll look at one of the key properties of the matrix transpose.
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Given the matrix π΄ equals negative eight, four, three, four, one, negative one, find the transpose of the transpose of π΄.
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Firstly, we know that to find the transpose of a matrix, we simply switch the rows with the columns.
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Now matrix π΄ is a two-by-three matrix.
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It has two rows and three columns, so its transpose is going to have three rows and two columns.
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Letβs begin with the first row.
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That first row becomes the first column in the transpose of our matrix.
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And since elements on the leading diagonal remain unchanged, we know we have to have a negative eight here.
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Then we put a four here and a three here.
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Next, we take elements from the second row and we add them to the second column.
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Again, since elements in the leading diagonal remain unchanged, we put negative one here.
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The remaining elements are one and four.
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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.
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But the question wants us to find the transpose of this transpose.
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And at this point, you might be able to predict whatβs going to happen, but letβs check.
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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.
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Letβs take the row negative eight, four and weβll add it to our first column, so we get negative eight, four.
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Next, weβll take the second row and weβll add it to our second column.
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Lastly, we take the third row and we add that to our third column as shown.
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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.
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Now observe, the transpose of the transpose of a matrix is in fact the original matrix.
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Now this makes a lot of sense because we switched the rows on the columns and then we switched them back again.
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So the original matrix will be the final result.
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And this holds for all types of matrices, and this property is the first of several properties that apply to the matrix transpose.
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Given a matrix π΄, the transpose of the transpose is equal to the original matrix π΄.
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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.
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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.
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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΄.
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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.
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Now, we use the definition of the matrix transpose in our definitions for symmetric and skew-symmetric matrices.
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And these are both important concepts in linear algebra.
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Firstly, we say that a square matrix π΄ is symmetric if the transpose of that matrix is equal to the original matrix.
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For example, take the two-by-two matrix one, two, two, three.
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We switch the first row and that becomes the first column in the transpose, and we do the same with the second row.
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That becomes the second column.
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And on doing so, we notice that the matrix π΄ is the same as the transpose, and so matrix π΄ must therefore be symmetric.
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It follows of course that a nonsquare matrix cannot be symmetric.
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An π-by-π matrix will have a transpose with π rows and π columns.
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And so the size is simply going to be different.
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Next, we say that a square matrix is said to be skew-symmetric if the transpose is equal to negative π΄.
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And for this to be the case, all of the elements on the main diagonal must be equal to zero.
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For instance, take the three-by-three matrix π΄ as shown.
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The transpose of π΄ is the matrix with elements zero, four, negative nine, negative four, zero, negative one, and nine, one, zero.
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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.
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And so this matrix is skew-symmetric.
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With these definitions in mind, letβs have a look at an example of how to work with a symmetric matrix.
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Find the value of π₯ which makes the matrix π΄ equals negative one, five π₯ minus three, negative 43, negative eight symmetric.
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Remember, we say that a square matrix is symmetric if the transpose of that matrix is equal to the original matrix.
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And of course, we find the transpose by switching the rows with the columns.
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So letβs begin by finding the transpose of matrix π΄.
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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.
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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΄.
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Then we take the elements in the second row of π΄ and we add them to the second column of the transpose.
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And so the transpose of π΄ is as shown.
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We know that for the matrix to be symmetric, the transpose must be equal to the original matrix.
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And further for two matrices to be equal to one another, their individual elements must be equal to one another.
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We can already observe that the elements on the leading diagonal are already equal to one another.
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And that makes a lot of sense, since when we transpose a matrix, these elements remain unchanged.
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We do however notice that we have a pair of identical equations when we try to equate the remaining elements.
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In other words, we know that negative 43 must be equal to five π₯ minus three.
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And so we need to solve this equation for π₯.
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Weβll begin by adding three to both sides, giving us five π₯ equals negative 40.
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And finally, weβll divide through by five, which gives us π₯ equals negative eight.
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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.
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We found then the value of π₯ which makes the matrix π΄ symmetric is negative eight.
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Letβs now recap some of the key points from this lesson.
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In this video, we learned that we can find the transpose of any matrix by switching the rows with the columns.
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And we can use this formula to help us remember this.
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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.
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We also learned some of the key properties, and one of these was that the transpose of a transpose is the original matrix.
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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.
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For scalar quantities π, the transpose of π times some matrix π΄ is equal to π times the transpose of π΄.
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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.
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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.