I'm given the following matrix:
$$
A = \begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\
0 & 1 & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}
\end{pmatrix}
$$
We're asked to determine if this matrix is orthogonal. I did this successfully using the rule that a matrix $A$ is orthogonal iff $A^T = A^{-1}$. The second part of the question however, I find to be difficult to answer since I have no idea how to start. The question is as follows: How would you determine a solution for the system of linear equations $A\vec{x} = \vec{b}$ using the orthogonal matrix above.
I have determined that $A\vec{x} = \vec{b}$ translates to:
$$
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\
0 & 1 & 0 \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}}
\end{pmatrix}
\begin{pmatrix}
x \\y \\z
\end{pmatrix}
=
\begin{pmatrix}
b_1 \\ b_2 \\ b_3
\end{pmatrix}
$$ yet I have no idea how the above matrix could help me solve this system easily. Anyone have any idea on how you would answer this type of question?
Best Answer
The point of showing your matrix was orthogonal was that you then know the inverse. Once you know that the columns/row are perpendicular (dot product zero) for all columns that are not themselves and each column/row are unit vectors, then you know the matrix is orthogonal. $$ \begin{aligned} v_1&= \begin{pmatrix} \dfrac{1}{\sqrt{2}} \\ 0 \\ \dfrac{1}{\sqrt{2}} \end{pmatrix} \\ v_2&= \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \\ v_3&= \begin{pmatrix} -\dfrac{1}{\sqrt{2}} \\ 0 \\ \dfrac{1}{\sqrt{2}} \end{pmatrix} \\ v_1 \cdot v_1&= \dfrac{1}{2} + 0 + \dfrac{1}{2}= 1 \\ v_2 \cdot v_2&= 0 + 1 + 0 = 1 \\ v_3 \cdot v_3&= \dfrac{1}{2} + 0 + \dfrac{1}{2} \\ v_1 \cdot v_2&= 0 \\ v_1 \cdot v_3&= 0 \\ v_2 \cdot v_3&= 0 \end{aligned} $$ This shows that $U$ is orthogonal. Then if $U$ is a orthogonal matrix, we know $U^{-1}= U^T$. Then knowing the inverse, we can solve the system $Ux=b$ via $$ \begin{aligned} Ux&= b \\ U^{-1}Ux&= U^{-1}b \\ U^TUx&= U^Tb \\ x&= U^Tb \end{aligned} $$
NOTE. I originally used the word unitary. A unitary matrix is just the complex number version of an orthogonal matrix. So if the matrix is real then it is orthogonal if and only if it is unitary. I do prefer the word unitary, because it reminds what you need to check - the columns (or rows) are perpendicular to each other and each row/column has unit length (length 1).