Questions: Find the orthogonal projection of v = [ 15, -14, 8, -20 ] onto the subspace V of R^4 spanned by x1 = [ -3, -2, 1, -1 ] and x2 = [ 2, 5, 16, 0 ]

Solution Steps

To find the orthogonal projection of a vector \(\mathbf{v}\) onto a subspace \(V\) spanned by vectors \(\mathbf{x}_1\) and \(\mathbf{x}_2\), we can use the following steps:

1. Form the matrix \(A\) whose columns are the basis vectors \(\mathbf{x}_1\) and \(\mathbf{x}_2\).
2. Compute the projection matrix \(P = A(A^T A)^{-1} A^T\).
3. Multiply the projection matrix \(P\) by the vector \(\mathbf{v}\) to get the orthogonal projection.

Given the basis vectors \(\mathbf{x}_1\) and \(\mathbf{x}_2\): \[ \mathbf{x}_1 = \begin{bmatrix} -3 \\ -2 \\ 1 \\ -1 \end{bmatrix}, \quad \mathbf{x}_2 = \begin{bmatrix} 2 \\ 5 \\ 16 \\ 0 \end{bmatrix} \] We form the matrix \(A\) by using these vectors as columns: \[ A = \begin{bmatrix} -3 & 2 \\ -2 & 5 \\ 1 & 16 \\ -1 & 0 \end{bmatrix} \]

The projection matrix \(P\) is given by: \[ P = A (A^T A)^{-1} A^T \] First, compute \(A^T A\): \[ A^T A = \begin{bmatrix} -3 & -2 & 1 & -1 \\ 2 & 5 & 16 & 0 \end{bmatrix} \begin{bmatrix} -3 & 2 \\ -2 & 5 \\ 1 & 16 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 15 & -1 \\ -1 & 285 \end{bmatrix} \] Next, compute \((A^T A)^{-1}\): \[ (A^T A)^{-1} = \begin{bmatrix} 15 & -1 \\ -1 & 285 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{285}{4284} & \frac{1}{4284} \\ \frac{1}{4284} & \frac{15}{4284} \end{bmatrix} \] Now, compute \(P\): \[ P = A \begin{bmatrix} \frac{285}{4284} & \frac{1}{4284} \\ \frac{1}{4284} & \frac{15}{4284} \end{bmatrix} A^T = \begin{bmatrix} 0.6140 & 0.4351 & -0.0877 & 0.2 \\ 0.4351 & 0.3544 & 0.1474 & 0.1333 \\ -0.0877 & 0.1474 & 0.9649 & -0.0667 \\ 0.2 & 0.1333 & -0.0667 & 0.0667 \end{bmatrix} \]

Given the vector \(\mathbf{v}\): \[ \mathbf{v} = \begin{bmatrix} 15 \\ -14 \\ 8 \\ -20 \end{bmatrix} \] The orthogonal projection of \(\mathbf{v}\) onto the subspace \(V\) is: \[ \text{projection} = P \mathbf{v} = \begin{bmatrix} 0.6140 & 0.4351 & -0.0877 & 0.2 \\ 0.4351 & 0.3544 & 0.1474 & 0.1333 \\ -0.0877 & 0.1474 & 0.9649 & -0.0667 \\ 0.2 & 0.1333 & -0.0667 & 0.0667 \end{bmatrix} \begin{bmatrix} 15 \\ -14 \\ 8 \\ -20 \end{bmatrix} = \begin{bmatrix} -1.5825 \\ 0.0772 \\ 5.6737 \\ -0.7333 \end{bmatrix} \]

Final Answer