find the orthogonal basis for the column space of the matrix [
6 1 −5
2 1 1
−2
6
−2
8
5
7
]
try typing your matrix in a more readable manner, such as
(6 1 -5)
(2 1 1)
...
so it looks kinda like a matrix!
To find the orthogonal basis for the column space of a matrix, we need to use the Gram-Schmidt process. Here are the step-by-step instructions:
Step 1: Write the given matrix as a combination of column vectors.
The given matrix [6 1 -5; 2 1 1; -2; 6; -2; 8; 5; 7] can be written in terms of vectors as:
v1 = [6; 2; -2; 6; -2; 8; 5; 7]
v2 = [1; 1; 0; 0; 0; 0; 0; 0]
v3 = [-5; 1; 0; 0; 0; 0; 0; 0]
Step 2: Normalize the first vector.
Let u1 = v1 / ||v1||, where ||v1|| denotes the Euclidean norm (length) of v1.
Calculating the Euclidean norm of v1:
||v1|| = √(6^2 + 2^2 + (-2)^2 + 6^2 + (-2)^2 + 8^2 + 5^2 + 7^2) = √(36 + 4 + 4 + 36 + 4 + 64 + 25 + 49) = √(222) ≈ 14.89
Then, u1 = v1 / ||v1|| = [6/14.89; 2/14.89; -2/14.89; 6/14.89; -2/14.89; 8/14.89; 5/14.89; 7/14.89]
Step 3: Project v2 onto u1 and subtract the projection.
Let proj_u1(v2) denote the projection of v2 onto u1.
proj_u1(v2) = (v2 · u1) * u1, where · denotes the dot product between vectors.
Calculating the dot product between v2 and u1:
(v2 · u1) = (1*6/14.89) + (1*2/14.89) + (0*(-2)/14.89) + (0*6/14.89) + (0*(-2)/14.89) + (0*8/14.89) + (0*5/14.89) + (0*7/14.89) = 8/14.89
Then, proj_u1(v2) = (8/14.89) * [6/14.89; 2/14.89; -2/14.89; 6/14.89; -2/14.89; 8/14.89; 5/14.89; 7/14.89] = [8*6/14.89^2; 8*2/14.89^2; 8*(-2)/14.89^2; 8*6/14.89^2; 8*(-2)/14.89^2; 8*8/14.89^2; 8*5/14.89^2; 8*7/14.89^2]
= [2.56/14.89; 0.854/14.89; -0.854/14.89; 2.56/14.89; -0.854/14.89; 3.413/14.89; 2.133/14.89; 2.989/14.89]
Subtracting the projection from v2:
w2 = v2 - proj_u1(v2) = [1 - 2.56/14.89; 1 - 0.854/14.89; 0 + 0.854/14.89; 0 - 2.56/14.89; 0 + 0.854/14.89; 0 - 3.413/14.89; 0 - 2.133/14.89; 0 - 2.989/14.89]
= [11.33/14.89; 13.04/14.89; 0.854/14.89; 2.56/14.89; -0.854/14.89; 4.32/14.89; 2.857/14.89; 3.964/14.89]
Step 4: Normalize w2.
Let u2 = w2 / ||w2||, where ||w2|| denotes the Euclidean norm of w2.
Calculating the Euclidean norm of w2:
||w2|| = √((11.33/14.89)^2 + (13.04/14.89)^2 + (0.854/14.89)^2 + (2.56/14.89)^2 + (-0.854/14.89)^2 + (4.32/14.89)^2 + (2.857/14.89)^2 + (3.964/14.89)^2)
= √(0.469 + 0.901 + 0.000454 + 0.0639 + 0.000454 + 0.102 + 0.0431 + 0.0852) ≈ 1.32
Then, u2 = w2 / ||w2|| = [11.33/14.89 / 1.32; 13.04/14.89 / 1.32; 0.854/14.89 / 1.32; 2.56/14.89 / 1.32; -0.854/14.89 / 1.32; 4.32/14.89 / 1.32; 2.857/14.89 / 1.32; 3.964/14.89 / 1.32]
= [8.56/14.89; 9.89/14.89; 0.646/14.89; 1.94/14.89; -0.646/14.89; 3.27/14.89; 2.18/14.89; 3.00/14.89]
Since we have obtained two vectors u1 and u2 that are orthogonal, they form an orthogonal basis for the column space of the given matrix. Therefore, the orthogonal basis is:
[u1, u2] = [[6/14.89; 2/14.89; -2/14.89; 6/14.89; -2/14.89; 8/14.89; 5/14.89; 7/14.89], [8.56/14.89; 9.89/14.89; 0.646/14.89; 1.94/14.89; -0.646/14.89; 3.27/14.89; 2.18/14.89; 3.00/14.89]]
To find the orthogonal basis for the column space of a matrix, we can use the Gram-Schmidt process.
The column space of a matrix is simply the span of its column vectors. So, we need to find a set of column vectors that span the same space but are mutually orthogonal (i.e., perpendicular to each other).
Let's start by considering the first column vector [6, 2, -2, 6, -2, 5, 7].
To make it orthogonal, we subtract its projection onto any previously determined vectors from itself. Since it is the first vector, we can leave it as it is.
Next, let's consider the second column vector [1, 1, 0, 0, 0, 0, 0].
To make it orthogonal, we subtract its projection onto the previous vector. The projection of the second vector onto the first vector can be calculated as follows:
proj_v1_v2 = (v2 dot v1) / (v1 dot v1) * v1
where v1 and v2 are the first and second vectors, respectively.
So, proj_v1_v2 = ([1, 1, 0, 0, 0, 0, 0] dot [6, 2, -2, 6, -2, 5, 7]) / ([6, 2, -2, 6, -2, 5, 7] dot [6, 2, -2, 6, -2, 5, 7]) * [6, 2, -2, 6, -2, 5, 7]
Performing the calculations, we get:
proj_v1_v2 = (6 + 2 + 0 + 0 + 0 + 0 + 0) / (6^2 + 2^2 + (-2)^2 + 6^2 + (-2)^2 + 5^2 + 7^2) * [6, 2, -2, 6, -2, 5, 7]
Simplifying, we have:
proj_v1_v2 = 8/104 * [6, 2, -2, 6, -2, 5, 7]
Therefore, the orthogonal vector to the second vector is given by:
orthogonal_v2 = [1, 1, 0, 0, 0, 0, 0] - proj_v1_v2
Now, we can repeat this process for the remaining columns:
For the third column vector [−5, 1, 0, 0, 0, 0, 0], the orthogonal vector is: orthogonal_v3 = [−5, 1, 0, 0, 0, 0, 0] - proj_v1_v3 - proj_v2_v3
For the fourth column vector [0, 0, 6, -2, 8, 0, 0], the orthogonal vector is: orthogonal_v4 = [0, 0, 6, -2, 8, 0, 0] - proj_v1_v4 - proj_v2_v4 - proj_v3_v4
For the fifth column vector [0, 0, 0, 0, 0, 1, 0], the orthogonal vector is: orthogonal_v5 = [0, 0, 0, 0, 0, 1, 0] - proj_v1_v5 - proj_v2_v5 - proj_v3_v5 - proj_v4_v5
For the sixth column vector [0, 0, 0, 0, 0, 0, 1], the orthogonal vector is: orthogonal_v6 = [0, 0, 0, 0, 0, 0, 1] - proj_v1_v6 - proj_v2_v6 - proj_v3_v6 - proj_v4_v6 - proj_v5_v6
Finally, we have the orthogonal basis for the column space of the matrix as the set of orthogonal vectors: [orthogonal_v1, orthogonal_v2, orthogonal_v3, orthogonal_v4, orthogonal_v5, orthogonal_v6].