Proving Det(tA)=Det(A) for Unit Matrix

• Enjoicube
In summary, the determinant of a unitary matrix can be proven to be +/-1 by using the fact that <Aw,v>=<w,(tA)v> for any symmetric bilinear form. This can be shown by proving that if U is a unitary matrix and B is a non-degenerate bilinear form, then tU commutes with B. This can be further conceptualized by picking an orthonormal basis for the form using Gram-Schmidt and showing that tU is equal to the transpose. In the complex case, the adjoint is the conjugate transpose.
Enjoicube

Homework Statement

Prove that the determinant of a unitary matrix is +/-1

Homework Equations

<Av,Aw>=<v,w>
Det(AB)=Det(A)*Det(B)

The Attempt at a Solution

Alright, I am aware that <Av,Aw>=<v,w> => A(tA)=I and (tA)A=I so Det(A(tA))=Det(I)=1 thus
Det(A)*Det(tA)=1. However, this is where I am stuck, I am aware that in some cases Det(tA)=Det(A), however this is not obvious in this case, as I cannot take tA to be the elementary defintion of switching rows and columns. How can I prove that Det(tA)=Det(A) ONLY from the fact that <Aw,v>=<w,(tA)v>. This inner product can be absolutely any symmetric bilinear form, so do not assume I mean the dot product.

Last edited:

I guess it should be |det(A)|=1, rather than det(A)=+/-1.

If you are trying to prove that the determinant of a unitary matrix is +/-1 then you MUST mean REAL unitary. It's not true for complex unitary. That means your tA is really just a plain old transpose.

Even for some random inner product, I think this is what I don't understand, why must it be the normal transpose for any completely random product, it only seems true for R^n with the standard dot product. Like for example, taking the space to be the space to be Pn(X) and <,> to be the integral of f*g.

Thank you for your responses, I think I have proved that for real unitary matrices, the transpose really is just the transpose. here is my proof if anyone can check or is interested.

Lemma: If U is unitary real, then it commutes with the matrix rep. of a nondegenerate bilinear form.
Pf: Let U be unitary, and let B be a nondegenerate bilinear form. Then

B(Uw,UV)=t(Uw)B'(Uv) for some matrix representation of B

so (tw*tU)B'(Uv)=(tw)*(tU*B'*U)*(v)

on the other hand B(Uw,Uv)=B(w,v), thus (tU*B'*U)=B',

however, since U is unitary, tU=U^-1, so (U^-1)*B'*U=B' and thus multiplying both sides by U, B'*U=U*B'. Thus U commutes with B'.

So If B is a symmetric bilinear form, tU really does equal the transpose.

Pf: Consider B(Uw,v)=B(v,Uw)=(tv)B'(Uw)=(tv)(B'U)w=(tv)(UB')w=t(tU*v)B'w=B((tU)v,w).

Sorry about the horrible notation due to my not using latex. Again, thank you Dick for pointing me towards this proof.

Ok, you're welcome! Glad you convinced yourself. I would conceptualize it by picking an orthonormal basis for the form using Gram-Schmidt, so <ei,ej>=delta_ij. So the matrix of U is U_ij=<ei,Uej>. So tU_ij=<ei,tUej>=<Uei,ej>=<ej,Uei>=U_ji. Or did I get my indices backwards? Doesn't matter, you get my point, right? So in the complex case it's pretty obvious the adjoint is the conjugate transpose.

Dick said:
Ok, you're welcome! Glad you convinced yourself. I would conceptualize it by picking an orthonormal basis for the form using Gram-Schmidt, so <ei,ej>=delta_ij. So the matrix of U is U_ij=<ei,Uej>. So tU_ij=<ei,tUej>=<Uei,ej>=<ej,Uei>=U_ji. Or did I get my indices backwards? Doesn't matter, you get my point, right? So in the complex case it's pretty obvious the adjoint is the conjugate transpose.

Yes, I think this revealed a lot to me. I think that I was just being very cautious, because the notation was very suggestive, and so I went into some sort of lockdown mode. I do like the way of conceptualizing this that you presented, and now I do understand that in the complex case it is conjugate transpose, so thank you much.

1. What is a unit matrix?

A unit matrix, also known as an identity matrix, is a square matrix with 1s along the main diagonal and 0s in all other positions. It is denoted by I or sometimes by In, where n is the dimension of the matrix.

2. How do you prove Det(tA) = Det(A) for a unit matrix?

To prove this, we need to use the properties of determinants. First, we know that the determinant of a matrix and its transpose are equal, so Det(A) = Det(AT). Next, we can use the property that multiplying a row of a matrix by a constant k multiplies the determinant by k. Since all the rows of the unit matrix are the same, we can multiply any row by t and the determinant will still be the same. Therefore, Det(tA) = tnDet(A), where n is the dimension of the matrix. Since t is equal to 1 in this case, we get Det(tA) = Det(A).

3. Why is it important to prove Det(tA) = Det(A) for a unit matrix?

This proof is important because it shows that the determinant of a unit matrix is always equal to 1. This is a fundamental property of unit matrices and is used in many mathematical and scientific applications, such as solving systems of equations and calculating areas and volumes.

4. Can this proof be extended to other matrices?

Yes, this proof can be extended to any square matrix. The determinant of a matrix and its transpose are always equal, and the property of multiplying a row by a constant also applies to any matrix. Therefore, Det(tA) = tnDet(A) for any square matrix A, where n is the dimension of the matrix.

5. Is there a visual representation of this proof?

Yes, there is a visual representation of this proof using geometric transformations. The unit matrix represents the identity transformation, which does not change the orientation or shape of a figure. When we transpose the matrix, we are essentially reflecting the figure across the line y = x. Then, when we multiply a row by t, we are stretching or shrinking the figure in the direction of that row. However, since all the rows are the same in a unit matrix, the figure is stretched or shrunk equally in all directions, resulting in the same figure. This visual representation helps to understand why the determinant remains the same in this scenario.

Replies
4
Views
1K
Replies
17
Views
2K
Replies
5
Views
2K
Replies
2
Views
6K
Replies
5
Views
2K
Replies
11
Views
2K
Replies
2
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
4
Views
6K