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Determine whether or not the following three vectors in $\mathbb{R}^4$ form an orthogonal set: $$\vec{u}_1=\begin{bmatrix} 1 \\ 2 \\ 1 \\ 0 \end{bmatrix}, \vec{u}_2=\begin{bmatrix} 2 \\ -1 \\ 0 \\ 1 \end{bmatrix}, \vec{u}_3=\begin{bmatrix} -1 \\ 1 \\ -1 \\ 3 \end{bmatrix}$$ Do they form a linearly independent set? Let $W=\mathbf{Span}\{\vec{u}_1,\vec{u}_2, \vec{u}_3\}$ and let $W^{\perp}$ be the orthogonal complement of $W$. Find a vector $\vec{u}_4 \in W^{\perp}$. Do the vectors $\{\vec{u}_1,\vec{u}_2, \vec{u}_3,\vec{u}_4\}$ form a basis for $\mathbb{R}^4$?

Can anyone help me out? Thank you in advance!

Kapitolina K
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2 Answers2

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Take the matrix $A=[u_1,u_2,u_3].$ Then anything in the kernel of $A^T$ (the transpose) will be orthogonal to $W.$ You can see this because $A^T\vec{x}=0,$ precisely when $x\cdot u_i=0,$ (dot product) for $i=1,2,3,$ as this is part of the definition of matrix multiplication.

Chickenmancer
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For each two of the vectors $u_1,u_2,u_3$ check if their inner product($<x,y>= x_1y_1+x_2y_2+x_3y_3+x_4y_4$ in $\mathbb{R}^4$)$<u_i,u_j>=0$ If so they are an orthogonal set.

Secondly i think you can prove, if these vectors are linearly independent or not.Also you can use the fact that, if they are orthogonal then they are linearly independent.

Now $W^{\bot}=\{y \in \mathbb{R}^4|<y,x>=0, \forall x \in W\}$.Take an element $u_4 \in W^{\bot},u_4=(a_1,a_2,a_3,a_4)$ and compute $<u_4,uj>=0$ for $j=1,2,3$ .you will get a $3\times4$ system.Solve it in terms of $a_i$.

In this way you will find the form of an element in $W^{\bot}$ in terms of $a_i$ (if you have free variables after solving the system).

Take an arbitrary element $u_4$ you found in $W^{\bot}$ (if this element has free variables then plug any value you want on these variables) and check if the set $\{u_1,u_2,u_3,u_4\}$ is linearly independent.

Marios Gretsas
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