# method1 Schimidt

When orthoganize three $3$ 3-tuple vectors, we have

$\beta _1=\alpha _1$

$\beta _2=\alpha _2-\frac{\left(\beta _1,\alpha _2\right)}{\left(\beta _1,\beta _1\right)}\beta _1$

$\beta _2=\alpha _3-\frac{\left(\beta _2,\alpha _3\right)}{\left(\beta _2,\beta _2\right)}\beta _2-\frac{\left(\beta _1,\alpha _3\right)}{\left(\beta_1,\beta _1\right)}\beta _1$

# method2

In the case of $3$, we can do it use cross product, that is

Choose two vectors, solve the normal vector, and set as $\beta _1$, choose another vector and span a plane, then solve for the normal vector, that is an orthogonal base of $R^3$

$\beta _1=\alpha _1$

$\beta _2=\text{Det}\left[\begin{array}{c} \{i,j,k\} \\ \beta _1 \\ \alpha _2 \\\end{array}\right]$

$\beta _3=\text{Det}\left[\begin{array}{c} \{i,j,k\} \\ \beta _1 \\ \beta _2 \\\end{array}\right]$

My question is what's the name of this method(Is it a valid method?), I've tried a little, I'm not sure the validity.

If valide, the relation between method one and two?

And futhur, how to use this method in general cases, consider we are doing determinant this way

Background is solving for orghogonal bases or diagonalizing a real and symmetric matrix.

The two methods both work correctly in $\mathbb R^3$. But they don't give the same results. To see why, consider the calculation of $\beta_2$ in the two methods.
In method #1, we are subtracting the component of $\alpha_2$ that's in the direction of $\beta_1$. In other words, we are projecting $\alpha_2$ onto the plane normal to $\beta_1$ to get $\beta_2$.
In method #2, we get $\beta_2$ by taking the cross product of $\beta_1$ and $\alpha_2$. So, $\beta_2$ will again lie in the plane normal to $\beta_1$, but it will be perpendicular to $\alpha_2$, rather than being a projection of it.
In $n$-dimensional space, method #1 still works, but (as far as I know) method #2 does not. The problem is that there is no $n$-dimensional version of the cross product, in general.