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## Linear algebra

### Course: Linear algebra>Unit 2

Lesson 7: Transpose of a matrix

# rank(a) = rank(transpose of a)

Rank(A) = Rank(transpose of A). Created by Sal Khan.

## Want to join the conversation?

• why number of pivot rows will always be equal to number of pivot columns?
• In both halves of the proof, he takes the rref(A). It is not (as I originally assumed) the rref(A) and rref(A^T). So, once he has rref(A), he looks for pivot entries. The pivot entries are defined by being the only non-zero entry in a column. Then the pivot column is just defined as the column containing the pivot entry. Likewise the pivot row is defined as the row containing the pivot entry. There are the same number of them because there is one of each for each pivot entry.

If you're confused, maybe it's because, like me, you assumed that the pivot row had to have only one non-zero entry (in analogy to the pivot columns). That is not the case, however, and in his rref(A) (visible at the bottom of the screen at the end of the video) he does, in fact, have multiple non-zero entries in his pivot rows. Only the pivot columns are guaranteed to have only one nonzero entry.
• at , what is a pivot row? He talks about pivot rows forming a basis for the rows in A--no discussion up to this point involved pivot rows! But now pivot rows are forming a basis for the rows in A and the columns in A transpose. Huh?
• As you may have guessed, it's a row (in rref(A)) which has a pivot element.

A pivot column is (yes, that's right) a column in rref(A) which has a pivot element.

In rref(A), all the non zero rows are pivot rows and all the pivot columns have only one non zero entry - the (you guessed it) pivot element in that row.

These row vectors (the pivot rows of rref(A)) are a l. i. set because they each have a component that doesn't occur in any of the others - (what is it?) - the pivot element. And since they are all linear combinations of the original row vectors of A (due to the row operations of Gaussian elimination), then the original row vectors must also be linear combinations of them, and so they are a basis for the row space, and therefore any basis for the row space of A has the same # vectors as there are pivots in rref(A).
• this has gotten me confused again. given a scenario with a 3x1 matrix, or simply a coolumn vector in R3. which I quite believe its rank to be 1. Now if we transpose it, we get a 1x3 matrix. How can its rank be 1. pardon mu ignorance, diss me but please sure do enlighten me.
• Note that the rank of a matrix is equal to the dimension of it's row space (so the rank of a 1x3 should also be the row space of the 1x3). And to find the dimension of a row space, one must put the matrix into echelon form, and grab the remaining non zero rows.
Well then, if you a non zero column vector (which you correctly declared has a rank of 1), then take it's transpose, you could find the rank of the transpose simply by finding the dimension of the row space.
But a single vector transposed is already in echelon form, so the dimension of the row space is 1. But we just declared the row space of the matrix transposed to be equal to the rank of the matrix transposed, therefore the rank of a non zero 1x3 is also 1.
• What is the purpose of matrixes & vectors? i dont get it be honest frends:) luv u all
(1 vote)
• I didn't think I'd ever use them when I learned basic linear algebra in middle/high school algebra 2, but it wasn't until college that I found out.

Vectors are used throughout physics, for example: you'll use them a lot when talking about position, velocity, and acceleration functions. There are many more uses of vectors, but that's just one example.

I ended up using matrices again in my circuits class (I'm an Electrical Engineering major). They come in handy when analyzing circuits (mostly mesh analysis). I got really good with solving systems of equations (made up by different variables of different circuits - real-life stuff) by setting up matrices, taking the inverse of them, and multiplying them by "b" (consider Ax = b) to get the solutions for the various x's. In mesh analysis, I was solving for the currents (and indirectly, the voltages) throughout a given circuit. I just noticed that my response is two years late, so you probably figured it out already. I guess this can be for anyone else who reads the questions and wonders the same thing.
• This is not a proof. It seems to me that the second part of the demonstration does not make sense and does not prove anything.
The only proof I know is more complex and involves "calculating" the rank of the transpose matrix and verifying that it is the same as the rank of the original matrix.
• So I accept that the # of pivot entries in rref(A) and # of pivot entries in rref(A[T]) is equal to rank(A) and rank(A[T]), respectively. I don't see how we proved that # of pivot entries in rref(A) = # of pivot entries in rref(A[T]). Can someone explain where that happened?
• We proved that the number of pivot entries in rref(A) is the rank(A[T]). This is because the rank(A[T]) is just the dim(RowSpace(A)), which is just the dim(RowSpace(rref(A))), which is just the number of pivot rows in rref(A). The number of pivot rows in rref(A) is just the number of pivot entries in the rref(A).
• At Sal uses the terms 'pivot rows' and 'pivot entries' (pivot columns) interchangeably. I've watched the whole Linear Algebra series up until this video, but this part confuses me royally. Are pivot rows and entries (columns) interchangeable? What is the relationship between a pivot entry/column and a pivot row?
• A pivot row is a row that has a pivot entry in it. A pivot entry is *not* a pivot column! A pivot column is a column with a pivot entry in it.

Let me use an analogy:

In MS Excel, you have rows, columns, and cells. Think of the cell as an entry. An entry is a specific column and row.
(1 vote)
• Hi,

Suposse we have a matrix A which is a mxn-matrix for example 3x5.

1) The maximum dimension is m right? so 3 in this example

2) Is Dim(ColSpace(A)) + Dim(NullSpace(A)) = n? so in this case 5?

3) Can we use the rule for
Dim(ColSpace(A)) = Dim(ColSpace(A^T)
and use if for NullSpaces?:
Dim(NullSpace(A)=Dim(NullSpace(A^t))?

4) if question 3 is valid than this should be correct right? (I used A^t in the NullSpace)
Dim(ColSpace(A)) + Dim(NullSpace(A^t) = n. also 5?

Thanks!
(1 vote)
• Not in this video, but what is full rank?
(1 vote)
• Full rank is when the most possible row vectors of a matrix are linearly independent.
(1 vote)
• Where can I find videos about Transposition of a subject , that is changing a subject of a equation ?
(1 vote)