\[ \begin{array}{l} A=\left[\begin{array}{rrr} 2 & -3 & 2 \\ 2 & 3 & -4 \\ -2 & -6 & 7 \end{array}\right] \\ W=\operatorname{Col}(A) \\ \vec{V}=\left[\begin{array}{l} 4 \\ 5 \\ 0 \end{array}\right] \end{array} \] 4) Determine an orthogonal basis for W. Describe your steps.
Solution
Step 1 ：First, we need to find the column space of matrix A, which is denoted as \(W = Col(A)\). The column space of a matrix is the set of all possible linear combinations of its column vectors.
Step 2 ：The column vectors of matrix A are \(\begin{bmatrix} 2 \ 2 \ -2 \end{bmatrix}\), \(\begin{bmatrix} -3 \ 3 \ -6 \end{bmatrix}\), and \(\begin{bmatrix} 2 \ -4 \ 7 \end{bmatrix}\).
Step 3 ：We can see that the second column vector is a multiple of the first column vector, and the third column vector is a linear combination of the first and second column vectors. Therefore, the column space of A is spanned by the first column vector only.
Step 4 ：So, the basis for the column space of A is \(\begin{bmatrix} 2 \ 2 \ -2 \end{bmatrix}\).
Step 5 ：Next, we need to find an orthogonal basis for the column space of A. An orthogonal basis is a basis where all the vectors are orthogonal (perpendicular) to each other.
Step 6 ：Since the column space of A is one-dimensional, any non-zero vector in the column space can be an orthogonal basis. Therefore, the orthogonal basis for the column space of A is \(\begin{bmatrix} 2 \ 2 \ -2 \end{bmatrix}\).
Step 7 ：Finally, we need to check if our result is correct. A basis is orthogonal if the dot product of any two different vectors in the basis is zero. Since our basis only contains one vector, it is indeed orthogonal.
