Determine which of the following matrices are (i)...

Determine which of the following matrices are $(i)$ orthogonal; (ii) proper orthogonal. (a) $\left(\begin{array}{rr}1 & 1 \\ -1 & 1\end{array}\right)$, (b) $$ \left(\begin{array}{rr} \frac{12}{13} & \frac{5}{13} \\ -\frac{5}{13} & \frac{12}{13} \end{array}\right) $$ (c) $\left(\begin{array}{rrr}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right)$, (d) (d) $\left(\begin{array}{rrr}-\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & -\frac{1}{3}\end{array}\right)$, (e) $\left(\begin{array}{lll}\frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\end{array}\right)$ $\left(\begin{array}{rrr}\frac{3}{5} & 0 & \frac{4}{5} \\ -\frac{4}{13} & \frac{12}{13} & \frac{3}{13} \\ -\frac{48}{65} & -\frac{5}{13} & \frac{36}{65}\end{array}\right)$, $(g)$ $\left(\begin{array}{rrr}\frac{2}{3} & -\frac{\sqrt{2}}{6} & \frac{\sqrt{2}}{2} \\ -\frac{2}{3} & \frac{\sqrt{2}}{6} & \frac{\sqrt{2}}{2} \\ \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0\end{array}\right)$.

Determine which of the following matrices are $(i)$ orthogonal; (ii) proper orthogonal.
(a) $\left(\begin{array}{rr}1 & 1 \\ -1 & 1\end{array}\right)$,
(b)
$$
\left(\begin{array}{rr}
\frac{12}{13} & \frac{5}{13} \\
-\frac{5}{13} & \frac{12}{13}
\end{array}\right)
$$
(c) $\left(\begin{array}{rrr}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & -1\end{array}\right)$,
(d)
(d) $\left(\begin{array}{rrr}-\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \\ \frac{2}{3} & \frac{2}{3} & -\frac{1}{3}\end{array}\right)$,
(e) $\left(\begin{array}{lll}\frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6}\end{array}\right)$
$\left(\begin{array}{rrr}\frac{3}{5} & 0 & \frac{4}{5} \\ -\frac{4}{13} & \frac{12}{13} & \frac{3}{13} \\ -\frac{48}{65} & -\frac{5}{13} & \frac{36}{65}\end{array}\right)$,
$(g)$
$\left(\begin{array}{rrr}\frac{2}{3} & -\frac{\sqrt{2}}{6} & \frac{\sqrt{2}}{2} \\ -\frac{2}{3} & \frac{\sqrt{2}}{6} & \frac{\sqrt{2}}{2} \\ \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0\end{array}\right)$.

Key Concepts

Proper Orthogonal Matrix

A proper orthogonal matrix, also known as a rotation matrix, is an orthogonal matrix with a determinant of +1. This extra condition distinguishes it from improper orthogonal matrices, which have a determinant of -1 and typically represent reflections combined with rotations. Proper orthogonal matrices are significant in describing pure rotations in space, which occur frequently in physics and engineering applications.

Determinant as an Indicator

The determinant of a matrix serves as an important indicator of the type of orthogonal transformation. For an orthogonal matrix, if the determinant is +1, the transformation is a proper rotation (proper orthogonal), preserving the orientation of the space. Conversely, a determinant of -1 indicates an improper rotation, where there is a reflection component altering the orientation. This criterion is essential for distinguishing between different classes of orthogonal transformations.

Orthogonal Matrix

An orthogonal matrix is a square matrix whose columns (and rows) form an orthonormal set. In mathematical terms, a matrix Q is orthogonal if its transpose is equal to its inverse, i.e., Q?Q = I, where I is the identity matrix. This property ensures that the transformation represented by the matrix preserves lengths and angles, making it a fundamental concept in linear algebra and applications such as computer graphics and numerical analysis.

Transcript

00:01 Hi, so for this exercise we have this matrix a and we need to verify that the matrix is orthogonal.

00:07 Just to remind you that orthogonal matrices satisfy the condition that the inverse is equal to the transpose of the matrix or in other words that a times a transpose is equal to the identity matrix.

00:22 So there is one that we can prove that this matrix is orthogonal by verifying this equation here but there are other ways and that means to verify that that that other ways means to check that either the the columns of the of the of the matrix form an orthonormal set or the rows of the matrix form an ortho normal set so in this case i'm going to consider the columns so i'm going to label it as u1 you 2 you 3 and u 4 and we need to verify that each inner product is equal to 0 so let's start with u1 with u2 this it's going to be equals to 1 over 12 1 minus 5 sorry here is 3 minus 5 plus 1 plus 1 which you can observe that is equals to zero the inner product of u1 with u2 is in the same sense this is going to be equals to 3 plus 1 plus 1 minus 5 which again is equal to 0 the inner product of you 1 with u 3 is also equals to 1 1 over 12 3 plus 1 minus 5 plus 1 which is equals to 0 and then for the rest we have that u2 with u 3 is equals to 1 over 36 times 9 minus 5 plus 1 and minus 5 which you can observe that is equals to 0 and u 2 with u 2 with u 4 taking the product i hear yes i hear you forgot one is the inner product of u1 with u4 this is also equals to 0 that is equal to 1 over 36 9 plus 1 minus 5 minus 5 plus 1 this is equals to 0 and the last inner product is u3 with u4 and this is 1 over 36 9 plus 1 minus 5 minus 5 which is also equals to 0 so you can observe that each in their product is equals to 0 and then we need to verify that the norms of these vectors are equals to 1 and that means you want the the norm in this case of u1 is equal to 1 the same as u2 u3 and u4 that means that these four vectors here are unitary and are orthogonal that means that they form an orthonormal set so that implies that a is orthogonal okay so let me clear here this part of the board okay so then the inverse of the matrix will be just the transpose and that is equal to the matrix so here is the transpose of the matrix and you can observe that actually is the same matrix and that's because this is a symmetric matrix that means that a is equal to the transpose of a so a transpose is just a and that corresponds to the inverse...

Please give Ace some feedback