find the QR factorization of A=[1 -1 1 ; 2 0 -1 ; 0 1 1]

To find the QR factorization of matrix \( A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 0 & -1 \\ 0 & 1 & 1 \end{bmatrix} \), we need to decompose \( A \) into two matrices: \( Q \) (an orthogonal matrix) and \( R \) (an upper triangular matrix) such that \( A = QR \).

Here’s the step-by-step process to calculate the QR factorization using the **Gram-Schmidt process**:

### Step 1: Define the columns of \( A \)
Let the columns of matrix \( A \) be denoted as:
\[
A_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \quad A_2 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \quad A_3 = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}
\]

### Step 2: Apply Gram-Schmidt process to obtain orthonormal vectors

#### Find \( q_1 \)
To find \( q_1 \), normalize \( A_1 \):
\[
q_1 = \frac{A_1}{\|A_1\|}
\]
The norm of \( A_1 \) is:
\[
\|A_1\| = \sqrt{1^2 + 2^2 + 0^2} = \sqrt{5}
\]
Thus:
\[
q_1 = \frac{1}{\sqrt{5}} \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \\ 0 \end{bmatrix}
\]

#### Find \( q_2 \)
Next, we need to orthogonalize \( A_2 \) against \( q_1 \):
\[
v_2 = A_2 - \text{proj}_{q_1}(A_2)
\]
The projection of \( A_2 \) onto \( q_1 \) is:
\[
\text{proj}_{q_1}(A_2) = \frac{A_2^T q_1}{q_1^T q_1} q_1 = \left( \frac{-1 \cdot \frac{1}{\sqrt{5}} + 0 \cdot \frac{2}{\sqrt{5}} + 1 \cdot 0}{1} \right) q_1 = \left( \frac{-1}{\sqrt{5}} \right) q_1
\]
Thus:
\[
\text{proj}_{q_1}(A_2) = \frac{-1}{\sqrt{5}} \begin{bmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \\ 0 \end{bmatrix} = \begin{bmatrix} -\frac{1}{5} \\ -\frac{2}{5} \\ 0 \end{bmatrix}
\]
Now, subtract this projection from \( A_2 \):
\[
v_2 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} - \begin{bmatrix} -\frac{1}{5} \\ -\frac{2}{5} \\ 0 \end{bmatrix} = \begin{bmatrix} -\frac{4}{5} \\ \frac{2}{5} \\ 1 \end{bmatrix}
\]
Normalize \( v_2 \) to get \( q_2 \):
\[
\|v_2\| = \sqrt{\left( -\frac{4}{5} \right)^2 + \left( \frac{2}{5} \right)^2 + 1^2} = \sqrt{\frac{16}{25} + \frac{4}{25} + 1} = \sqrt{\frac{45}{25}} = \frac{\sqrt{45}}{5} = \frac{3}{\sqrt{5}}
\]
Thus:
\[
q_2 = \frac{v_2}{\|v_2\|} = \frac{1}{\frac{3}{\sqrt{5}}} \begin{bmatrix} -\frac{4}{5} \\ \frac{2}{5} \\ 1 \end{bmatrix} = \frac{\sqrt{5}}{3} \begin{bmatrix} -\frac{4}{5} \\ \frac{2}{5} \\ 1 \end{bmatrix} = \begin{bmatrix} -\frac{4}{3} \\ \frac{2}{3} \\ \frac{\sqrt{5}}

Find the QR factorization of A = [1 -1 1; 2 0 -1; 0 1 1]

Discover how to perform the QR factorization of matrix A = [1 -1 1; 2 0 -1; 0 1 1] using the Gram-Schmidt process. This solution covers linear algebra concepts, including orthogonal matrices, upper triangular matrices, and the Gram-Schmidt orthogonalization technique, suitable for undergraduate students.

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