To prove this, we replace the columns of F one at a time with columns taken from A and show that we can do this for all of A's columns.

• Define F^[k] = [D^[k] a_1:k] consisting of r-k columns selected from F followed by the first k columns of A. We prove by induction that, for each k in the range 0:n, we can choose D^[k] and B^[k] so that F=F^[k]B^[k].
• For k=0, the proposition is true with D^[0]=F and B^[0]=I.
• Suppose F=F^[k]B^[k] is true for some k<n. Then, A=F^[k]B^[k]G and if we define x = (B^[k]G)_k+1, the (k+1)^th column of B^[k]G, we have a_k+1 = F^[k]x.
• Suppose x_i is the first non-zero element of x. We must have i <= r-k for otherwise a_k+1 = [0 a_1:k]x or equivalently A[x_r-k+1:r; -1; 0_[n-k-1#1]] = 0. This would contradict the linear independence of A.
• If we define D^[k+1] to be D^[k] with its i^th column removed then F^[k] = F^[k+1]S where S=[I_[i-1#i-1] 0;0 -x_i^-1x_i+1:r I_[r-i#r-i];0_[1#i-1] x_i^-1 0_[1#r-i]]. It follows that if B^[k+1]=SB^[k] we have F=F^[k+1]B^[k+1] as required.
• Finally, we have F=F^[n]B^[n] = [D^[n] A] where D^[n] consists of r-n columns selected from F. In order for D^[n] to exist, we must therefore have r>=n.

We prove this by induction on n. If n=1, set F=A and G=1.

Now suppose A, F and G satisfy the theorem and consider the m#n+1 matrix A⁺=[A x]. If x=Fy for some y then set F⁺=F and G⁺=[G y], otherwise set F⁺=[F x] and G⁺=[G 0; 0T 1]. It is straightforward to show in both cases that A⁺=F⁺G⁺ that G⁺ is upper triangular and that r⁺<=n+1.

We must also show that in the second case (i.e. if x!=Fy for any y) that the columns of [F x] are linearly independent. If [F x][a; b] = 0, then Fa=-bx. We must have b=0 since otherwise x =F(-a/b) which contradicts the assumption that x != Fy. Hence Fa = 0 which implies that a=0 since the columns of F are linearly independent. Thus the columns of [F x] are also linearly independent.

• We have A = A_[m#n] I_[n#n] = I_[m#m]A_[m#n] . Since rank(A) is the minimum inner dimension of such a decomposition, we must have rank(A) <= m and rank(A) <= n.

• If A=FG is a full-rank decomposition, then we can decompose AB as F^.GB. Hence rank(AB) is <= the inner dimension of this decomposition which is rank(A). A similar argument shows that rank(AB) <= rank(B).

• If the columns of A are linearly independent and A=F_[m#r]G_[r#n] is a full-rank decomposition, then from 1.1, we have rank(A) = r >= n, but from 1.3, rank(A)<=n. Hence rank(A)=n.
• If A's columns are not linearly independent then from 1.2 we can find a linearly independent subset F with A=F_[m#r]G_[r#n]. This implies that rank(A) <= r < n since the rank is the minimum inner dimension of such a decomposition.

• The columns of I are linearly independent since Ix=0 implies x=0. Hence from 1.5 its rank is n.

• We have A=IA so from 1.1 we can find D and B such that I_[m#m]=[D A]_[m#m]B. From 1.6, 1.4 and 1.3 we have m = rank(I) <= rank([D A]) <= m. Hence rank([D A])=m and by 1.5 its columns are linearly independent.

We prove this by induction on n. If n=1, we set Q=a/||a||, R=||a||  and S=||a||^-1 where a=A. Note that ||a|| cannot equal 0 since this would violate the assumption that the columns of A are linearly independent.

Now assume the theorem to be true for A_[m#n] with Q, R and S defined as in the theorem. Given an m#(n+1) matrix A⁺ = [A x] with linearly independent columns, we define y = x - QQ^Hx and note that y cannot equal 0 since we would then have x = QQ^Hx = ASQ^Hx contradicting the assumption that the columns of A⁺ are linearly independent.

We set Q⁺ = [Q  y/||y|| ], R⁺ = [R  Q^Hx; 0^T  ||y|| ] and S⁺ = [S  -SQ^Hx/||y||; 0^T  ||y||^-1 ]. It is straightforward to show that A⁺, Q⁺, R⁺ and S⁺ satisfy the theorem by performing the appropriate multiplications. Note that if A is real then all the above calculations give real results.

This recursive procedure is called Gram-Schmitt orthogonalization but is not recommended because of its poor numerical properties.

• From 1.2 we decompose A=FG with G upper triangular and F_[m#r] linearly independent.
• From 1.7 we find D such that [F D]_[m#m] has linearly independent columns.
• From 1.8 we decompose [F d] = QU where Q^HQ = I.

We now have A = FG = [F D] [I; 0] G = Q U [I; 0] G = QR where R = U [I; 0] G is the product of three upper triangular matrices and is therefore upper triangular itself. If A is real then all of thee operations preserve realness.

If m>=n, the lower m-n rows of R are all zero and so the product A=QR is unaffected if we delete the last n-r columns of Q and the last n-r rows of R.

• If A is skew symmetric then by definition a(i,j)=-a(j,i). By setting j=i, we get a(i,i)=-a(i,i) from which a(i,i)=0.

• We prove this by induction starting with a 1#1 skew-symmetric matrix which must equal 0 and hence has even rank.
• Given a skew-symmetric A_n+1#n+1, we can, by [1.10] partition it as A = [C b ; -b^T 0] where C_n#n is skew-symmetric and rank(C) is even, by the induction hypothesis.
• If  b = Cx for some x, then A = [I x]^T C [I x] and also C = [I 0] A [I 0]^T  and so, by [1.4], rank(A) = rank(C) which is even.
• If, on the other hand, b is linearly independent of the columns of C then, -b^T is also linearly independent of the rows of C = -C^T. Hence rank(A) = rank([C b ; -b^T 0]) = rank([C b]) + 1 = rank(C) + 2 which is even.

• (AD)_i,j = a_i,jd_j,j = (DA)_i,j = d_i,ia_i,j  ==> a_i,j (d_i,i - d_j,j) = 0  ==> a_i,j = 0 or d_i,i = d_j,j

• From [1.12] a_i,j=0 except when d_i,i = d_j,j. However since the c_k are distinct, d_i,i = d_j,j only within one of the c_kI_k blocks. It follows that a_i,j=0 except when i and j lie in the same block. The result follows.

• Define the unitary matrices P=U^HL and Q=V^HM
• UD²U^H = AA^H = LD²L^H = UPD²P^HU^H ==> D²P = PD²
• From [1.13], it follows that P (and also Q, by a similar argument based on A^HA) is block diagonal with orthogonal blocks whose sizes match the multiplicity of the singular values in D. It follows that PD = DP.
• We have U^HAV = D = U^HLDM^H V= PDQ^H ==> DQ = PD = DP
• Hence D(P - Q) = 0 and P must equal Q except for rows in which d_ii = 0. The result follows.

• (XDX^T)_pq = sum_i(x_pid_ix_qi) = sum_i(d_i × x_pix_qi) = sum_i(d_i × (x_ix_i^T)_pq) = (sum_i(d_i × x_ix_i^T))_pq
• The proof for XDX^H is identical.

• tr(XDX^T) = tr(sum_i(d_i × x_ix_i^T)) [1.15] = tr(sum_i(d_i × x_i^Tx_i)) = sum_i(d_i × x_i^Tx_i)
• The proof for tr(XDX^H) is identical.

• (AB)_i,j = sumk(a_ikb_kj)
• tr(AB) = sumi(AB)_ii = sumi,k(a_ikb_ki) = sum_k,i(b_kia_ik) = sum_k(BA)_kk = tr(BA)

• Note that conformality implies that the matrix dimensions are A_[n#m], B_[n#k], C_[n#m] and D_[n#k]
• tr([A B]^T [C D]) = tr([A^TC A^TD ; B^TC B^TD]) = tr(A^TC) + tr(B^TD) since these diagonal blocks are square.

• Assume that both X⁺ and Y satisfy the pseudo inverse equations. We will show that Y = X⁺.
  1. (a) XX⁺X=X  and (b)  XYX=X
  2. (a) X⁺XX⁺=X⁺ and (b)  YXY=Y
  3. (a) (XX⁺)^H=XX⁺  and (b)  (XY)^H=XY
  4. (a) (X⁺X)^H=X⁺X  and (b)  (YX)^H=YX
• Y = YXY [2b] = YY^HX^H [3b] = YY^HX^H(X⁺)^HX^H [1a] = YXY(X⁺)^HX^H [3b] = YXYXX⁺ [3a] = YXX⁺ [2b] = X^HY^HX⁺ [4b] = X^H(X⁺)^HX^HY^HX⁺ [1a] = X⁺XX^HY^HX⁺ [4a] = X⁺XYXX⁺ [4b] = X⁺XX⁺ [1b] = X⁺ [2a]