umut's blog

Theorem: Let A be nxn matrix on real numbers, then

Det(Aᵗ)=Det(A)

Proof: Let e be any elementary row operation on nxn matrices. Then, define e’ to be analogous column operation on nxn matrices. What do we mean by the word “analogous” here? Well, let’s say e multiplies r’th row by c and adds it to s’th row. Then, e’ multiplies r’th column by c and adds it to s’th column. If e multiplies r’th row by c that isn’t zero, then e’ multiplies r’th row by c. Then, there are some interesting consequences from this definition.

Proposition-1: Let A be an nxn matrix on real numbers. Also, let e be any elementary row operation. Finally, define e’ to be analogous column operation as defined above. Then,

(e(A))^t=e'(Aᵗ)

Proof:

ero-1: Let

(e(A))_(i,j)={ A_(i,j) if i≠r ; c*A_(s,j)+A_(i,j) if i=r}

then

(e(A))^t_(j,i)={ A_(i,j) if i≠r ; c*A_(s,j)+A_(i,j) if i=r}

also

(A)^t_(j,i)={A_(i,j)}

(e'(A)^t)_(j,i)={A^t_(j,i) if i≠r ; c*A^t_(j,s)+A^t_(j,i) if i=r}

this equality becomes,

(e'(A)^t)_(j,i)={A_(i,j) if i≠r ; c*A_(s,j)+A_(i,j) if i=r}

ero-2: Let

(e(A))_(i,j)={ A_(i,j) if i≠r ; c*A_(i,j) if i=r and c≠0}

then

(e(A))^t_(j,i)={ A_(i,j) if i≠r ; c*A_(i,j) if i=r and c≠0}

also

(A)^t_(j,i)={A_(i,j)}

(e'(A)^t)_(j,i)={A^t_(j,i) if i≠r ; c*A^t_(j,i) if i=r and c≠0}

this equality becomes,

(e'(A)^t)_(j,i)={A_(i,j) if i≠r ; c*A_(i,j) if i=r and c≠0 }

ero-3: Let

(e(A))_(i,j)={ A_(i,j) if i≠r,s ; A_(s,j) if i=r ; A_(r,j) if i=s}

then

(e(A))^t_(j,i)={ A_(i,j) if i≠r,s ; A_(s,j) if i=r ; A_(r,j) if i=s}

also

(A)^t_(j,i)={A_(i,j)}

(e'(A)^t)_(j,i)={A^t_(j,i) if i≠r ; A^t_(s,i) if i=r ; A^t_(r,i) if i=s}

this equality becomes,

(e'(A)^t)_(j,i)={A_(i,j) if i≠r ; A_(i,s) if i=r ; A_(i,r) if i=s}

So, proposition is proved.

Proposition-2: Let A be any nxn matrix on real numbers. Let e be any elementary row operation. Also, let e’ be analogous column operation. Then,

e'(A)=A.(e'(I))

Proof: Let A

Proposition: Let A and B be nxn matrices. Also, let

AB=I

then

BA=I

Proof: Every matrix is row equivalent to a single row-reduced echelon matrix. That is, for every matrix, there is a finite sequence of elementary row operations such that if we apply this sequence of row operations on this matrix, we get a row-reduced echelon matrix. If we denote R to be this matrix,

e_r(e_r-1(………e₁(R)…..))=A

for some elementary row operations, e_r,e_r-1,…….,e₁. Every elementary row operation on any matrix corresponds to multiplication of an elementary matrix with that matrix. This elementary matrix is obtained by doing the row operation on I_mxm where m is equal to row number of the matrix that we apply the elementary row operation. Denote E_i to be correspondent elementary matrix of e_i. Then,

E_rE_r-1…….E₁R=A

(E_rE_r-1…….E₁R)B=I

multiply both sides E_r^‘, inverse of E_r

E_r^‘(E_rE_r-1…….E₁R)B=E_r^‘I

keep multiplying,

E_r-1^‘E_r^‘(E_rE_r-1…….E₁R)B=E_r-1^‘E_r^‘I

E₁^‘……..E_r-1^‘E_r^‘(E_rE_r-1…….E₁R)B=E₁^‘……….E_r-1^‘E_r^‘I

RB=E₁^‘……….E_r-1^‘E_r^‘I

Can R contain a zero row? No, because if it were, then RB would contain a zero row and any matrix that is row equivalent to I can’t contain a zero row. Why? Because if there were such matrix, we would obtain a contradiction. Let’s assume that RB contains some zero rows and find (RB)’ (equal to RB without zero rows) matrix’s row-reduced echelon matrix, and then add zero rows at the end of this row-reduced echelon matrix. Then, this final matrix, say M, is row reduced echelon matrix of RB, and clearly it is not equal to I. This is a contradiction, since we have two distinct row reduced echelon matrix of RB; one is I and the other is M. This is not possible.

So, R must not have zero row, then it must be I (easy to verify). Then,

RB=IB=B=E₁^‘……….E_r-1^‘E_r^‘I

Is BA=I?

BA=(E₁^‘……….E_r-1^‘E_r^‘I)(E_rE_r-1…….E₁R)

=(E₁^‘……….E_r-1^‘E_r^‘I)(E_rE_r-1…….E₁I)=I

Yes, it is

Proposition: Let A and B mxn matrices. Also, let

AX=Y and BX=Y

be equivalent systems of linear equations. Then,

A’=B’

where