Calculating the basis and ortogonalizing it using the Gram-Schmidt method

Question

I have a subspace of a space in $\mathbb{R}^4$: $$U=lin\{(1, 2, -2, 1), (1, 1, 0, 2), (1, 8, 1, 0)\}$$ Could someone explain to me, step by step, how to get the basis of the subspace U and then how to ortogonalize the basis using the Gram-Schmidt method? I do not understand it very well and would appreciate an explaination on how to solve this.

Answer 1

If a subspace is given in this form, namely $U$ is the linear hull/set of linear combinations of three vectors, then we already know that these three vectors span $U$. That means any vector in $U$ can be expressed as a linear combination of the three vectors $v_1 = (1,2,-2,1), v_2 = (1,1,0,2), v_3 = (1,8,1,0)$. To find a basis we need to make sure that all vectors are linearly independent (because a basis is a system of vectors that spans $U$ and is linearly independent). To check for linear independence we can for example calculate the rank of the matrix with the three vectors as rows (or columns - as you like), namely $$A= \begin{pmatrix}1&2&-2&1\\1&1&0&2\\1&8&1&0 \end{pmatrix}$$ If the rank is $3$, the three vectors are linearly independent and form a basis of $U$. If the rank is $2$, one of the three vectors can be written as a linear combination of the other two and is therefore "unnecessary" and can be removed and it would still span the whole set.

To calculate the rank we use "Gauss-Elimination-like" row manipulations to bring it in row echelon form:

$$A \sim \begin{pmatrix}1&2&-2&1\\0&1&-2&-2\\0&6&3&-1 \end{pmatrix} \sim \begin{pmatrix}1&2&-2&1\\0&1&-2&-2\\0&0&-15&-11 \end{pmatrix}$$ we see that we have no "zero-row" and therefore $A$ has rank $3$ and a basis of $U$ is just given by $\{v_1, v_2, v_3\}$.

To orthogonalize them we will just keep the first vector $w_1=v_1 = (1,2,-2,1)$ as it is and will set (with $\langle \cdot , \cdot \rangle$ I mean the inner product - I am assuming that you want the standard inner product here) $$w_2 = v_2 - \frac{\langle v_2, w_1 \rangle}{\langle w_1, w_1 \rangle}w_1=v_2 - \frac 12 w_1= (\frac 12, 0,1, \frac 32)$$

and then $$w_3 = v_3 - \frac{\langle v_3, w_1 \rangle}{\langle w_1, w_1 \rangle}w_1-\frac{\langle v_3, w_2 \rangle}{\langle w_2, w_2 \rangle}w_2=v_3 -\frac 32 w_1 -\frac 37 w_2= (-\frac{5}{7}, 5,\frac{25}{7},-\frac{15}{7})$$

Now $\{w_1, w_2, w_3\}$ is an orthogonal basis of $U$.