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Let \( W \) be a finite-dimensional subspace of \(\mathbb{R}^2\) spanned by the vector \((3,4)\).

Find the orthogonal projection \( E \) of \(\mathbb{R}^2\) onto \( W \).

Answer :

So, the basis for W is (0.6, 0.8), the projection matrix E is [[0.36, 0.48], [0.48, 0.64]], and the orthogonal projection W^1 is [[0.64, -0.48], [-0.48, 0.36]].

is the subspace of R² orthogonal to W. Let's break down the solution step-by-step.

1) First, we need to find the basis for subspace W. This is done by normalizing the given vector (3,4). The normalization is done by dividing each element of the vector with its length(norm). The length of vector will be √(3² + 4²) = 5, therefore the normalized vector, which is our basis for W, becomes (3/5, 4/5) = (0.6, 0.8).

3) The next step is calculating the projection matrix (E). It can be calculated using the formula E = vv^T, where v represents the basis vector and v^T indicates the transpose of the vector. Furthermore, vv^T for our basis vector (0.6, 0.8) is calculated as follows:

E = vv^T = [0.6 * 0.6, 0.6 * 0.8;
0.8 * 0.6, 0.8 * 0.8]

which simplifies to E = [[0.36, 0.48];
[0.48, 0.64]]

4) The last calculation we need to make is to find the orthogonal projection W^1. The orthogonal projection W^1 can be calculated using the formula W^1 = I - E, where I represents the identity matrix in R². The identity matrix in R² is a 2x2 matrix with ones on its diagonal and zeros elsewhere:

I = [[1, 0];
[0, 1]]

By applying the formula, we obtain:
W^1 = I - E = [[1 - 0.36, 0 - 0.48];
[0 - 0.48, 1 - 0.64]]

This results in:
W^1 = [[0.64, -0.48];
[-0.48, 0.36]]

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