- Transpose of a matrix
- Determinant of transpose
- Transpose of a matrix product
- Transposes of sums and inverses
- Transpose of a vector
- Rowspace and left nullspace
- Visualizations of left nullspace and rowspace
- rank(a) = rank(transpose of a)
- Showing that A-transpose x A is invertible
Proof by induction that transposing a matrix does not change its determinant. Created by Sal Khan.
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- These lessons are starting to run together so correct me if I'm wrong, but this also holds true when A is m x n (m =/= n) as the det(A) of an mxn is just zero. Right?(3 votes)
- Why is the last value in the determinant not (-1)ⁿ⁺¹a_1m * det(A_1m)?(2 votes)
- Two things:
1.) You're right about the "a_1m" term--Sal accidently left it out.
2.) However, since Sal is expanding along the first row (and since their are m rows), the "-1" term should be raised to the "1+m" power (and not n+1). This ensures that if their are an even number of columns, the final minor is multiplied by (-1), and if their are an odd number of rows, the final minor is multiplied by 1. (Again, assuming that we--like Sal--expand along the first row of the matrix).
Thus, the last product should read as:
- At9:42, the determinant of the transposed matrix is being calculated. Why would he be calculating it with a11, a12 (going down along the 1st column) rather than a11, a21 (going along the 1st row)?? Can you calculate a matrix like this?(2 votes)
- You can calculate the determinant by going down the rows, or going down the columns. It doesn't even have to be the first row, or first column, (as long as you put the negative sign for even rows or columns).(4 votes)
- In which video was it shown that you can do cofactor expansion down a row or column?(2 votes)
- Why can't I watch these videos??(1 vote)
- If you can't watch them on khan academy go to YouTube and search under the same title.
Try refreshing page, making sure your system or iOS is updated or it might just be your Internet connection.(3 votes)
- Do you need to prove the nxn case?(1 vote)
- That is what Sal does in this video. He accomplished this by induction on the size of the matrix. The induction works by first proving a base case, n=2 in this case. That was done first. The second step (and usually more difficult one) is proving that if we assume the theorem ( det A = det At ) is true in a particular case (n x n), then it must be the case that it's true in the next case ( n+1 x n+1 ). This second (inductive) step combined with the first (base) step, gives a proof of the theorem in all subsequent cases, sort of like a long line of dominoes set up for a chain reaction. Pushing over the first domino (base step), knocks it into the second, and the second is knocked into the third, and so on for the rest (inductive step). Hope that helps!(2 votes)
- can you illustrate using real numbers(0 votes)
- The answer depends on what you want illustrated. A 3 X 3 or 4 X 4 or whatever X whatever matrix * can be used to show that, if it has a determinant, the value is unchanged when the given matrix is tranposed. But no matter what matrix you use, you have only proven it for that matrix. If you want to prove that the determinant of a matrix and its transpose are the same, you need to use induction and for induction you can not rely on numeric values only.
* The size of the example matrix, above 2 X 2, just depends on the number of SMEs** that need fixing.
** Stupid Math Errors(1 vote)
Let's see if taking the transpose of a matrix does anything to its determinant. So a good place to start is just with the 2-by-2 scenario. So let's do the 2-by-2 scenario. So if I start with some matrix here. Let me just take its determinant. So a, b, c, d. And then let me take its determinant. So this is going to be equal to ad minus bc. Let me take the transpose of this and then take its determinant. So that would be the determinant of ac, the columns turn into the rows, and then bd, the rows turn into the columns. What is this going to be equal to? This is going to be equal to ad minus bc again. The only thing that happened is these two guys got swapped and they multiply times each other anyway. So these two things are equivalent. So at least for the 2-by-2 case, the determinant of some matrix is equal to the determinant of the transpose of that matrix. Now, that's just the 2-by-2 case. Now, I'm going to make an inductive argument, or I could just say an argument by induction, to show that this works for all n by n, for all cases. And the way you construct an argument by induction is you assume that it's true for the n-by-n case. So let's assume that for n by n, and let's say I have some matrix. Let me call it matrix B, and let's say it's an n-by-n matrix, we assume that the determinant of any matrix B that's n by n is equal to the determinant of B's transpose. That's where we started off with our inductive argument. And then we see if given this, if given that, is it true of n plus 1 by n plus 1 matrix? Because if we can, if we can say, look, given that it's true for the n-by-n case, it's going to be true for the n plus 1 by n plus 1 case, then we're done because we know it's true for the base case, the 2-by-2 case, which you could say, well, that's your first n by n. So if it's true for the 2 by 2 case, then it'll be true for the 3-by-3 case, because that's just one increment. But then if it's true for the 3-by-3 case, then it'll be true for the 4-by-4 case. And if it's true for the 4-by-4 case, it'll be true for the 5-by-5 case, and you just keep going up like that. So when you do a proof by induction, you prove a base case, and then you prove that if it's true for n, or in this case an n-by-n determinant, if you can prove that given it's true for an n-by-n determinant, it's going to be true for an n plus 1 by n plus 1 determinant or an n plus 1 by n plus 1 matrix, then you have completed your proof. So let's see if this is the case. So let me construct an n plus 1 by n plus 1 matrix. So let's say I have my matrix A, my favorite letter to use for matrices, I think the entirely linear algebra's favorite letter to use for matrices. And let's say it's an n plus 1 by n plus 1 matrix. And just to simplify my notation let's just say that m is equal to n plus 1. So we could call it an m-by-m matrix. And what is it going to look like? Let's draw its entries right here. I'm going to draw more than the normal amount of entries. a11, this is its first row, a12, all the way to a1m. We have m columns, which is the same thing as m plus 1. That's not m times 1 columns. This is m plus 1 as well. And then we do our second row right here. a21, a22, a23, all the way to a2m. Then you have your third row right here: a31, a32, a33, all the way to a3m. And then you go all the way down here. At the end you have your mth row, which you could also say is your m plus 1 row. So it's your mth row, first column, and then you have a sub m2, and then a sub m3, all the way to a sub mm. Fair enough. Now, let me draw the transpose of A. So a transpose is also going to be an n plus 1 by n plus 1 matrix, which you could also write as an m-by-m matrix. I'm just going to have to take the transpose of this. So the transpose of that, this row becomes a column, so it becomes a11, and this entry right here is a12. It's this entry right there. Then you can go all the way down to a1m. Then this pink row becomes a pink column here, a21-- I wanted to do it in pink. You have a21, a22, you have an a23, and it goes all the way down to a2m. You have your green row right here, it's your third one, so it's a31, a32, a33, all the way down to a3m. And then we can just skip a bunch of rows in this case, but it's columns in this case. So you just draw some dots and you have am1, am2. I'm just going down this guy, but this row is now going to become-- which was the last row. It's now going to become the last column. am3, all the way down to amm. And I have my transpose. Now, let's take the determinant of A. Let me do it in purple. So the determinant of my matrix A, we could just go down this first row up here. It's going to be equal to a11 times the determinant of its submatrix, so it's the determinant of this submatrix right here. We could call that the submatrix a sub 11. We've seen this notation before. So it's the determinant of a sub 11, and then it is minus a12 times the determinant of its submatrix, so you cross out that row and that column. So that is going to be a sub 12, and you're going to go all the way to-- and I don't what the sign on that is, so we could call it negative 1 to the 1 plus m-- that'll give us the right sign for the checkerboard pattern-- times the determinant of the submatrix for this guy. So we call it a1m, where you cross out that guy's row, that guy's column, and you're just left with all of this stuff over here. Fair enough. Now, let's look at the determinant of a transpose. We learned earlier, you don't have to go down the first row or you don't even go down a row. You could go down a column. Let me be clear. So for our determinant of A, we went down this row, and our submatrices, this was my first submatrix. My second submatrix, you know what it looks like. You would cross out the second column and that row, and whatever's left over would be the second submatrix and so on and so forth. But for our determinant of A transpose, let's just go down this first column and get the submatrices like that. So this is going to be equal to-- let's get our first term right here. a sub 11 times the determinant of its submatrix. So what's the determinant of its submatrix? It's going to be-- its submatrix, you cross out its row and its column, and you're going to be left with this thing right here. Now, an interesting question is how does this thing that I've just squared off, this guy's submatrix, relate to this guy's submatrix? Well, if you look at it carefully, this row from a22 to a2m has now become a column from a22 to a2m. This row, which is the next one, from a32 to a3m has now become a column from a32 to a3m. If you keep going down this last row, it has now become this column. So this guy's submatrix or the thing we're going to have to take the determinant of right here is equal to the transpose of this guy. So this is equal to a sub 11 transpose. And if you go through it, we go minus this guy, minus a12 times the determinant of his submatrix. And if we cross out this guy's row and that guy's column, what is that going to look like? His submatrix is going to look like this. It's going to have that there and it's going to have that right there. How does that compare to a12? So a12 is if you crossed out this and this, you're left with all of this right here. So once again, you see that this row is the same thing as this column, that this row is the same thing as this column, that that row is the same thing as that column. So once again, the submatrix we have to take the determinant of is equal to the transpose of this thing over here. So it's equal to a12 transpose. This thing-- I could draw it shaded in-- is equal to the transpose of this thing, is equal to the transpose of that thing right there. So, in general, each of these submatrices when we go down this row is equal to the transpose of each of these. So you keep going, so then you're going to go all the way to plus minus 1. We're going to go all the way down to the minus 1, 1 plus m times the determinant of-- it's going to be this guy transposed. You can even do it. If you go all the way to-- if you cross that guy out and that guy out, you're left with everything else on this matrix, and that's equal to the transpose of if you cross this guy out and that guy out. This row turns into that column, that row turns into that column. I think you see the point. I don't want to beat a dead horse. So that's going to be equal to a sub 1m transpose. Now, remember, going into this inductive proof, or proof by induction, I assume that for-- remember, this is an n plus 1 by n plus 1 matrix. But going into it, I assumed that for an n-by-n matrix, the determinant of B is equal to the determinant of B transpose. Well, these guys right here, these are n-by-n matrices, right? This guy right here is an n plus 1 by n plus 1. Same thing for this guy right here. But these guys right here are n by n. So if we assume for the n-by-n case that the determinant of a matrix is equal to the determinant of a transpose-- this is the determinant of the matrix, this is the determinant of its transpose-- these two things have to be equal. So we can then say that the determinant of A transpose is equal to this term A sub 11 times this, but this is equal to this for the n-by-n case. Remember, we're doing the n plus 1 by n plus 1 case. But these submatrices are one dimension smaller in each direction. It has one less row and one less column. So these two things are equal. So instead of writing this, I can just write this, so times the determinant of a sub 11. Then you keep going. Minus a sub 12 times the determinant. Instead of writing this, I could write that because they're equal. Determinant a sub 12, all the way to plus minus 1 to the 1 plus m times the determinant of this. These two things are equal. That is equal to that. That was our assumption in this inductive proof, a sub 1m. And then you see that, of course, this line, this blue line here, is equivalent to this blue line there. So we get that the determinant of A, which is an n plus 1 by n plus 1, so this is the n plus 1 by n plus 1 case. We get the determinant of A is equal to the determinant of A transpose. And we got this assuming that it is true-- let me write it-- assuming that it's true for n-by-n case. And then we're done. We've now proved that this is true in general, because we've proven the base case. We've proven it for the 2-by-2 case, and then we showed that if it's true for the n case, it's true for the n plus 1 case. So if it's true for the 2 case, it's going to be true for the 3-by-3 case. If it's true for the 3-by-3 case, it's true for the 4-by-4 case, so on and so forth. But the takeaway is pretty neat. You can take the transpose, the determinant doesn't change.