Transposes and Inverses
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We introduce the notion of the transpose of a matrix, give some properties, and relate it to the dot product of vectors.
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hello everyone I'm Norman Weber today we
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talk about transposes and inverses so
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these are important operations on
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matrices so it transposes act on any
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sized matrix while inverses are only
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defined for square matrices that means
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ie matrices that are n by n matrices
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okay so we want to talk about these two
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different kinds of transformations the
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inverses are more complicated and
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important but transposes act in some
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ways a little bit like inverses so it's
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good to start with them all right so
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what is the transpose of matrix well
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it's a kind of a reflection so the
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transpose of the matrix is described in
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terms of its size if a is an M by n
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matrix then it's transpose and usually
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denoted a with a superscript capital T
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is an N by M matrix the rows of a become
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the columns of a transpose and
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vice-versa so let's give an example so
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if a is the matrix site 2 1 3 4 - 6 17
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then a transpose is the matrix - 1 3 4 -
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6:17 so you can see that the rows of a
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becomes the columns of a transpose while
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the columns of a become the rows of a
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transpose and a is a two by three matrix
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while a transpose is a three by two
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matrix that's quite simple and if we
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wanted to describe it in terms of
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entries so an official definition might
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be is that if a equals say the matrix
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with entries little a IJ okay is M by n
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matrix then a transpose is the N by M
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matrix where the IJ entry of a transpose
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is a ji in other words it's the J entry
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of a alright that's a concise way of
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describing the relationship between the
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entries of the matrix a and the entries
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of the transpose basically interchanging
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properties first of all that it's linear
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if a and B are both matrices of the same
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size then a plus B transpose is a
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transpose plus B transpose so here if a
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and B are both the same size say m by n
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and secondly then if we take lambda
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times a and we take the transpose of
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that so if we multiply all the entries
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by five well then kind of obviously the
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transpose of that is going to be lambda
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times the transpose of a
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but a more subtle property is the
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following that if a is M by N and B is n
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by P then in that case we're allowed to
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multiply them because this end and that
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end are the same okay so then if we
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multiply a times B and take the
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transpose of that what we get is the
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transpose of B times the transpose of a
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so when we take transpose as the order
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interchanges and that makes sense at the
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level of sizes because if we look at the
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sizes of things a B itself is a B is
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going to be an M by P so a B transpose
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is P by M and on the right hand side B
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transpose is P by n just reverse these
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ones and a transpose is n by M reverse
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these ones so B transpose times a
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transpose is now these ends here also
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agree but the other elements so we get P
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by M so this dimension of a B transpose
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agrees with the dimension of B transpose
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times a transpose okay so let's see if
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we can prove this we improve the third
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with the first two almost like exercises
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but the third one is a little bit
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trickier so let's prove this third
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property prove three so a is M by N and
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B is n by P so let's have a look at the
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IJ entry
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of a B transpose in IJ entry of a B
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transpose by definition that's equal to
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the J I entry of the product a B that's
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the definition of the transpose now what
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is the ji entry of a B so I remind you
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that a is we're assuming it's M by N and
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B is n by P so this is jit entry it's
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the sum from k equals 1 to N of the the
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J caithe entry of a times the K I entry
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of b-but this is then also equal to sum
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from k equals 1 to n I will write it
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with the B's first be ki a JK and then
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I'm going to write this as the ki
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element of B is the same as the I cave
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element of B transpose and a sub J k is
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the KJ element of a transpose and when
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we write it that way now we see that
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this is the altogether it's equal to the
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IJ entry of B transpose times a
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transpose
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because there's a common index there and
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there and the outside indices are I and
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J and we're talking about the matrices B
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transpose and a transpose so that's the
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computation that shows that the
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transpose satisfies this pleasant
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doesn't condition up here now when we
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take the transpose of a product we get
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the product of the transpose is but in
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the opposite order now another role for
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transposes is to have a look at the dot
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product in matrix language so the dot
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product of vectors in matrix language so
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we now have an opportunity of sort of
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rethinking what the dot product of
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vectors is so I remind you that if you
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have two vectors say if a is an element
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in n-dimensional space that means it's a
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column vector of length or size N and B
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is also in RN then we can define the dot
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product then a dot B is is a number okay
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and more specifically if a is equal to a
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one down to a n and B is be one down to
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BN
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then I remind you that the dot product
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of these two vectors is the sum a 1 B 1
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plus all the way up to a and BN alright
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coming from vector geometry now we can
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interpret column vectors as matrices so
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about a column vector
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in say RN can be viewed as an n by 1
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matrix right this vector here is
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essentially a matrix with n rows and one
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column now similarly a row vector a row
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vector which typically we usually write
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may be something like the one comma V 2
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comma V n so this is what a row vector
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looks like this can be interpreted or
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recast as 1 by n matrix okay in that
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case we write rather write without
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commas when we write matrices we
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generally do not have commas so we would
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write this row vector rather as V 1 V 2
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V 2 V 2 up to VN without any columns so
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this way both row vectors and column
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vectors candies are brought into the
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realm of matrices with column vectors
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being n by 1 matrices and row vectors as
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being 1 by n matrices so now if we look
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at our two original column vectors so a
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equals a 1 down to a n and B equals B 1
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through B n then let's have a look at
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this following product when we take a
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transpose times B
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so the transpose of a will be the row
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vector a 1 up to a an ax B is the column
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vector B 1 down to BN and when we
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multiply them we're getting a 1 B 1 plus
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all the way up to a n BN actually we're
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getting that as a 1 by 1 matrix but if
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we don't bother with the brackets denote
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the 1 by 1 matrix then we can interpret
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this as just the dot product of a and B
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so we've make established proposition is
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that if a and B are column vectors both
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n-dimensional then the dot product
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between them he is just a transpose
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times B