Theorem : Gram–Schmidt Let $W$ be a $p$ -dimensional subspace of $R^n$ , and let ${w_1, w_2, \dots , w_p}$ be any basis for $W$ . Then the set of vectors ${u_1, u_2, . . . , u_p}$ is an orthogonal basis for $W$ , where

$u_1 = w_1$

$u_2 = w_2 - \frac{u_1^Tw_2}{u_1^Tu_1}u_1$

$u_3 = w_3 - \frac{u_1^Tw_3}{u_1^Tu_1}u_1 - \frac{u_2^Tw_3}{u_2^Tu_2}u_2$

and where in general

$u_i = w_i - \sum_{k=1}^{k = i-1}\frac{u_k^Tw_i}{u_k^Tu_k}u_k, 2 \le i \le p$

By rewriting the above equations for $w_i$ ,

$w_i = u_i - \sum_{k=1}^{k = i-1}\frac{u_k^Tw_i}{u_k^Tu_k}u_k, 2 \le i \le p$

In matrix form

$\begin{bmatrix} w_1^T \\ \vdots \\ w_n^T\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ t_{21} & 1 & 0 & \cdots & 0 \\ t_{31} & t_{32} & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ t_{n1} & t_{n2} & t_{n3} & \cdots & 1 \end{bmatrix} \begin{bmatrix} u_1^T \\ \vdots \\ u_n^T\end{bmatrix}$

$t_{ij} = u_j^T w_{i-1}/(u_j^T u_j), \forall (i > j); t_{ij} = 1,\forall (i = j);t_{ij} = 0, \forall (i < j)$

Given below is an implementation of Gram-Schmidt orthogonalization in Julia

```
using LinearAlgebra
N = 10;
A = rand(N,N); #row vectors to be orthogonalized
U = zeros(N,N);
U[1,:] = A[1,:];
V = zeros(N,N); # for the for loop
V[1,:] = A[1,:];
U[2,:] = A[2,:] - dot(U[1,:],A[2,:])*U[1,:]/dot(U[1,:],U[1,:])
U[3,:] = A[3,:] - dot(U[1,:],A[3,:])*U[1,:]/dot(U[1,:],U[1,:]) - dot(U[2,:],A[3,:])*U[2,:]/dot(U[2,:],U[2,:])
println(dot(U[1,:],U[2,:]))
println(dot(U[1,:],U[3,:]))
println(dot(U[2,:],U[3,:]))
#for loop for a matrix of size N
for i in 2:N
V[i,:] = A[i,:]
for j in 1:(i-1)
V[i,:] -= dot(V[j,:],A[i,:])*V[j,:]/dot(V[j,:],V[j,:])
end
end
println(dot(V[1,:],V[2,:]))
println(dot(V[1,:],V[3,:]))
println(dot(V[2,:],V[3,:]))
println(dot(V[N,:],V[N-1,:]))
```

Using row operations

```
using LinearAlgebra
function test()
N = 4;A = rand(N,N);B = Matrix(1.0I,N,N);U = zeros(N,N);U[1,:] = A[1,:]; V = zeros(N,N);V[1,:] = A[1,:];
row_operations = []
B[2,1] = -dot(U[1,:],A[2,:])/dot(U[1,:],U[1,:])
push!(row_operations,B)
X = B*A
B = Matrix(1.0I,N,N)
B[3,1] = -dot(X[1,:],A[3,:])/dot(X[1,:],X[1,:])
B[3,2] = -dot(X[2,:],A[3,:])/dot(X[2,:],X[2,:])
push!(row_operations,B)
X = B*X
println(dot(X[1,:],X[2,:]))
println(dot(X[1,:],X[3,:]))
println(dot(X[2,:],X[3,:]))
X = zeros(N,N);X = copy(A);#X[1,:] = A[1,:];
row_operations = []
for i in 2:N
C = Matrix(1.0I,N,N)
for j in 1:(i-1)
C[i,j] = -dot(X[j,:],A[i,:])/dot(X[j,:],X[j,:])
end
push!(row_operations,C)
X = C*X
end
println(dot(X[1,:],X[2,:]))
println(dot(X[1,:],X[3,:]))
println(dot(X[2,:],X[3,:]))
println(dot(X[N,:],X[N-1,:]))
end
test()
```

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