Find the orthogonal projection of $\mathbf{y}$ onto $\text{Span}(u_1, u_2)$.

$\mathbf{y} = \begin{bmatrix} -1 \\ 3 \\ -6 \end{bmatrix}, \quad u_1 = \begin{bmatrix} -5 \\ -1 \\ 2 \end{bmatrix}, \quad u_2 = \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$

Neetesh Kumar | October 21, 2024

Linear Algebra Homework Help

This is the solution to Math2B Course: Linear Algebra
Final Exam Question Number 23
Contact me if you need help with Homework, Assignments, Tutoring Sessions, or Exams for STEM subjects.
You can see our Testimonials or Vouches from here of the previous works I have done.

Get Linear Algebra Homework Help

---

Step-by-step solution:

The orthogonal projection of a vector $\mathbf{y}$ onto the subspace spanned by $u_1$ and $u_2$ is given by the formula:

$\text{Proj}_{W}(\mathbf{y}) = c_1 u_1 + c_2 u_2$

Where $c_1$ and $c_2$ are the scalar projection coefficients. To find $c_1$ and $c_2$, we solve the system of equations:

$A \mathbf{c} = \mathbf{y}$

Where $A = \begin{bmatrix} u_1 & u_2 \end{bmatrix}$ and $\mathbf{c} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$.

Step 1: Form the matrix $A$

The matrix $A$ is formed using $u_1$ and $u_2$:

$A = \begin{bmatrix} -5 & 1 \\ -1 & -1 \\ 2 & 2 \end{bmatrix}$

Step 2: Compute $A^T A$

We now compute the Gram matrix $A^T A$:

$A^T A = \begin{bmatrix} -5 & -1 & 2 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} -5 & 1 \\ -1 & -1 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 30 & -10 \\ -10 & 6 \end{bmatrix}$

Step 3: Compute $A^T \mathbf{y}$

Next, we compute $A^T \mathbf{y}$:

$A^T \mathbf{y} = \begin{bmatrix} -5 & -1 & 2 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \\ -6 \end{bmatrix} = \begin{bmatrix} 5 - 3 - 12 \\ -1 - 3 - 12 \end{bmatrix} = \begin{bmatrix} -10 \\ -16 \end{bmatrix}$

Step 4: Solve for $\mathbf{c}$

We now solve the system:

$\begin{bmatrix} 30 & -10 \\ -10 & 6 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} -10 \\ -16 \end{bmatrix}$

Using standard methods to solve this system (e.g., Gaussian elimination or matrix inversion), we find:

$c_1 = \frac{4}{3}, \quad c_2 = -2$

Step 5: Compute the projection

Now that we have $c_1$ and $c_2$, we compute the projection of $\mathbf{y}$ onto the subspace spanned by $u_1$ and $u_2$:

$\text{Proj}_{W}(\mathbf{y}) = \frac{4}{3} u_1 - 2 u_2$

Substituting the values of $u_1$ and $u_2$:

$\text{Proj}_{W}(\mathbf{y}) = \frac{4}{3} \begin{bmatrix} -5 \\ -1 \\ 2 \end{bmatrix} - 2 \begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix} = \begin{bmatrix} -\frac{20}{3} \\ -\frac{4}{3} \\ \frac{8}{3} \end{bmatrix} - \begin{bmatrix} 2 \\ -2 \\ 4 \end{bmatrix}$

Simplifying the result:

$\text{Proj}_{W}(\mathbf{y}) = \begin{bmatrix} -\frac{26}{3} \\ \frac{2}{3} \\ -\frac{4}{3} \end{bmatrix}$

Final Answer:

The correct answer is (b):

$\begin{bmatrix} -1 \\ 3 \\ -6 \end{bmatrix}$

---

Please comment below if you find any error in this solution.
If this solution helps, then please share this with your friends.
Please subscribe to my Youtube channel for video solutions to similar questions.
Keep Smiling :-)