Difference between revisions of "Determinant Transpose Proof" - Rhea

Revision as of 12:04, 8 December 2010

Statement: I am going to derive, using the cofactor expansion formula,that transposing a matrix does NOT change its determinant.

The proof is not entirely complete for the following reasons:
1) The cofactor expansion formula is never proved in the text (or in class), and
2) One key fact, namely that you can do cofactor expansion along either
rows or columns, uses (at least implicitly) the fact that det(A) = det(A^T)
hence there is some circularity in the argument.

First we will start with a 2x2 matrix as follows:

Let the 2x2 matrix A be:

$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} $

So, by calculating the determinant, we get det(A)=ad-cb, Simple enough, now lets take A^T (the transpose).

A^T=

$ \begin{bmatrix} a & c \\ b & d \end{bmatrix} $

So, det(A^T)=ad-cb.

Well, for this basic example of a 2x2 matrix, it shows that det(A)=det(A^T). Simple enough...

Now, we will use the power of induction to make some powerful assumptions, which will be proven in a bit.

Lets assume this is true for all cases of nxn...

So now assume we have a nxn matrix called B:

Then we can say that det(B)=det(B^T)

I now propose a question that is food for thought (and integral...!) for the rest of this derivation.

Question: Is it true for all (n+1)x(n+1) matrices as well??

We shall now see...

Let an (n+1)x(n+1) matrix called A be

$ \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{1m} \\ a_{21} & a_{22} & a_{23} & a_{2m} \\ a_{31} & a_{32} & a_{33} & a_{3m} \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ a_{m1} & a_{m2} & a_{m3} & a_{mm} \\ \end{bmatrix} $

So, there we have it. That is an arbitrary matrix A. Now, it is logical to extract a submatrix from within matrix A. We will call the following matrix A₁₁. It is all of the original matrix except for the 1st row, and 1st column.

A₁₁:

$ \begin{bmatrix} a_{22} & a_{23} & a_{2m} \\ a_{32} & a_{33} & a_{3m} \\ .... & .... & .... \\ .... & .... & .... \\ .... & .... & .... \\ a_{m2} & a_{m3} & a_{mm} \\ \end{bmatrix} $

Okay, now, it is time to get the transpose of A. (A^T)

Again, we have a (n+1)x(n+1) matrix.

A^T 

$ \begin{bmatrix} a_{11} & a_{21} & a_{31} & a_{m1} \\ a_{12} & a_{22} & a_{32} & a_{m2} \\ a_{13} & a_{23} & a_{33} & a_{m3} \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ a_{1m} & a_{2m} & a_{3m} & a_{mm} \\ \end{bmatrix} $

Now, it is logical to extract a submatrix from within matrix A^T. We will call the following matrix A^T₁₁. It is all of the original matrix except for the 1st row, and 1st column.

A^T₁₁:

$ \begin{bmatrix} a_{22} & a_{32} & a_{m2} \\ a_{23} & a_{33} & a_{m3} \\ .... & .... & .... \\ .... & .... & .... \\ .... & .... & .... \\ a_{2m} & a_{3m} & a_{mm} \\ \end{bmatrix} $

That covers all the matrices and the submatrices we would need to refer too for this example. Now, it is important to note that the rows of the 1st submatrix becomes the columns of the 2nd submatrix. While obvious, it can best be stated with the following relationship:

(a^T)_ij = a_ji

In addition, by using the idea of cofactors, we can also say with truth that:

(A^T)_ij = (A_ij)^T

This relationship states that i-j'th cofactor matrix of A^T is equal to the transpose of the j-i'th cofactor matrix of A, as shown in the above matrices.

Now, onto the actual gritty proof:

In the calculation of det(A), we are going to use co-factor expansion along the1st ROW of A.

Additionally, in the calculation of det(A^T), we are going to use co-factor expansion along the 1st COLUMN of A^T.

det(A)= a₁₁det(A₁₁)-a₁₂det(A₁₂)+(-1)^1+mdet(A_1m)

det(A)= a₁₁A₁₁ + a₁₂A₁₂ + .... a_1nA_1n

det(A^T)= a^T₁₁A^T₁₁ + a^T₂₁A^T₂₁ + ...+ a^T_n1A^T_n1

But:

a^T_ij = a_ji

A^T_ij= A^T with i'th row, j'th column x (-1)^ij 

= transpose of (A with j'th row and i'th column) x (-1)^ij

= (A_ji)

Therefore, det(A)=det(A^T)

However, lets keep pressing on with a more 'concrete' approach (if the above logic was too abstract)

det(A^T)=a₁₁det((A₁₁^T)-a₁₂det((A₁₂^T))+ ...(-1)^1+mdet((A_1m)^T)

So, therefore, 

det(A^T)=a₁₁det(A₁₁)-a₁₂det(A₁₂)+ ... + (-1)^1+mdet(A_1m),

so for the (n+1)x(n+1) case;

det(A)=det(A^T), assuming that its true for the nxn case.

Final Thoughts:

This proof is largely one of induction. As I had proved in the beginning 2x2 case, we could just as easily said that it would hold for any (n+1)x(n+1) matrix. By that logic, because I have shown it to be true for the nxn case, it will then be true for the 3x3 case, 4x4 case, 5x5 case, etc...you get the idea. In addition, as a disclaimer, and food for thought, it is wise in general to explain why a preliminary inductive assumption should be convincing. I mean, one could assume that 2=3, and then construct a proof that 3=4. Of course, probably not, but that is the reason behind those joke proofs such as 0=1 or -1=1, etc. Use with caution, and enjoy.