# Is there a more concise way to find the linear combination of a vector with a set of orthonormal vectors?

So I have a set of vectors that I verified are orthonormal. $$\begin{equation*} A = \begin{pmatrix} \frac{1}{3\sqrt{2}} & \frac{2}{3} & \frac{1}{\sqrt{2}} \\ \frac{1}{3\sqrt{2}} & \frac{2}{3} & -\frac{1}{\sqrt{2}} \\ -\frac{4}{3\sqrt{2}} & \frac{1}{3} & 0 \end{pmatrix} \end{equation*}$$ The column vectors of $$A$$ are orthonormal since I multiplied $$A^TA$$ to get $$I_3$$. I want to get the vector $$x= \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}$$ written as a linear combination of the column vectors of $$A$$ above.

I know in order to get a linear combination of a vector in terms of another set you would append the additional column vector to matrix $$A$$, get the matrix into reduced row echelon form, and multiply each column(vector) by the corresponding components in the last column then sum them. But this matrix is an orthonormal set. Is there a more concise way to do this, knowing we have a orthonormal set of vectors, and use that to get $$||x||^2$$?

Yes, there is a concise way of doing it, your intuition is correct. Let $$u_1 = (\frac{1}{3 \sqrt{2}}, \frac{1}{3 \sqrt{2}}, \frac{-4}{3 \sqrt{2}})$$, $$u_2 = (\frac{2}{3}, \frac{2}{3}, \frac{1}{3})$$, and $$u_3 = (\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}, 0).$$
We know from general theory of orthonormal bases that any vector $$v \in \mathbb R^3$$ can be uniquely expressed as $$$$\begin{split} v = \langle v, u_1 \rangle u_1 + \langle v, u_2 \rangle u_2 + \langle v, u_3 \rangle u_3. \end{split}$$$$
• Thanks I know how to find the squared magnitude of $x$, it would be 30. But I did that by dotting the vector with itself and the square root and squared cancel each other out. But how do I use the above information to help with finding $||x||^2$ Apr 21, 2022 at 19:22
• I'm confused about what you're asking. If you have calculated $||x||^2$ easily as you have done, then why do you need to use 'the above information' to do it a different way? If you're asking whether there is a more concise way to do it, the answer is no, it's very straightforward to dot the vector (1,2,5) with itself. Apr 22, 2022 at 0:40