Orthogonal Matrices - Linear Algebra

1. Let $Q$ be a $3 \times 3$ real orthogonal matrix. If two of its eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$, what is the determinant of $Q$?
   Answer: $1$ When you encounter orthogonal matrices with given eigenvalues, you need to use two key properties: orthogonal matrices have determinant $\pm 1$, and complex eigenvalues of real matrices come in conjugate pairs. Since $Q$ is a real $3 \times 3$ orthogonal matrix, we know $Q^TQ = I$, which means $\det(Q^T)\det(Q) = \det(I) = 1$. Since $\det(Q^T) = \det(Q)$, we have $[\det(Q)]^2 = 1$, so $\det(Q) = \pm 1$. The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$. Since $Q$ is real, complex eigenvalues must appear in conjugate pairs. Therefore, the third eigenvalue must be $\lambda_3 = \frac{1}{2} - i\frac{\sqrt{3}}{2}$. The determinant equals the product of eigenvalues: $\det(Q) = \lambda_1 \cdot \lambda_2 \cdot \lambda_3 = 1 \cdot \left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)$ Using the formula $(a + bi)(a - bi) = a^2 + b^2$: $\det(Q) = 1 \cdot \left[\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2\right] = 1 \cdot \left[\frac{1}{4} + \frac{3}{4}\right] = 1$ Choice A (-1) would be possible for an orthogonal matrix but doesn't match our calculation. Choice B ($1 + i\sqrt{3}$) incorrectly assumes the determinant can be complex, but determinants of real matrices are always real. Choice D is wrong because we have sufficient information to determine the determinant uniquely. Study tip: Remember that for real matrices, complex eigenvalues always come in conjugate pairs, and their product is always real and positive.
2. An orthogonal transformation $T(\vec{x}) = Q\vec{x}$ is applied to a triangle in $\mathbb{R}^3$. Which of the following geometric properties of the triangle is NOT necessarily preserved?
   Answer: The orientation of the vertices of the triangle. Orthogonal transformations are isometries, meaning they preserve distances. Therefore, the lengths of the sides are preserved (A). Because they preserve the dot product, they also preserve angles between vectors (B). The area of a transformed shape is scaled by the absolute value of the determinant of the transformation matrix. Since for an orthogonal matrix $Q$, $|\det(Q)|=1$, the area is preserved (C). However, the orientation of the vertices is preserved only if $\det(Q)=1$ (a rotation). If $\det(Q)=-1$ (a reflection or rotoreflection), the orientation is reversed (e.g., a left-handed system becomes a right-handed one). Since the problem only states $Q$ is orthogonal, we cannot assume $\det(Q)=1$, so orientation is not necessarily preserved.
3. Let $R$ be a $4 \times 4$ orthogonal matrix representing a rotation in $\mathbb{R}^4$. If $R$ has the block form $R = \begin{bmatrix} A & B \\ C & D \end{bmatrix}$ where $A$ and $D$ are $2 \times 2$ blocks, and $A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$, which condition must hold for $R$ to be orthogonal?
   Answer: $A^T A + C^T C = I$ and $B^T B + D^T D = I$ with $A^T B + C^T D = 0$ For $R$ to be orthogonal, $R^T R = I$. Computing this: $R^T R = \begin{bmatrix} A^T & C^T \\ B^T & D^T \end{bmatrix} \begin{bmatrix} A & B \\ C & D \end{bmatrix} = \begin{bmatrix} A^T A + C^T C & A^T B + C^T D \\ B^T A + D^T C & B^T B + D^T D \end{bmatrix} = I$. This gives the conditions in choice C. Choice A is too restrictive (assumes block diagonal form). Choice B gives incorrect relationships between blocks. Choice D incorrectly assumes $D = \pm A$ and focuses on determinant rather than the full orthogonality condition.
4. Let $Q$ be an $n \times n$ orthogonal matrix and $\vec{b}$ be a vector in $\mathbb{R}^n$. For the linear system $Q\vec{x} = \vec{b}$, the unique solution for $\vec{x}$ is given by:
   Answer: $\vec{x} = Q^T\vec{b}$ When you encounter a linear system involving an orthogonal matrix, remember that orthogonal matrices have a special property that makes solving systems particularly elegant. An orthogonal matrix $Q$ satisfies $Q^T Q = I$, meaning its transpose equals its inverse: $Q^{-1} = Q^T$. To solve $Q\vec{x} = \vec{b}$, you need to isolate $\vec{x}$ by multiplying both sides by $Q^{-1}$. Since $Q^{-1} = Q^T$ for orthogonal matrices, you get: $Q^{-1}Q\vec{x} = Q^{-1}\vec{b}$ $I\vec{x} = Q^T\vec{b}$ $\vec{x} = Q^T\vec{b}$ This confirms that answer C is correct. Looking at the wrong answers: Option A gives $\vec{x} = Q\vec{b}$, which would mean $Q^2\vec{b} = \vec{b}$ when substituted back—this only works if $Q^2 = I$, which isn't true for all orthogonal matrices. Option B suggests $\vec{x} = (QQ^T)^{-1}\vec{b}$, but since $QQ^T = I$ for orthogonal matrices, this becomes $\vec{x} = I^{-1}\vec{b} = \vec{b}$, which is incorrect unless $Q = I$. Option D uses $\frac{1}{\det(Q)}Q\vec{b}$, which resembles the adjugate formula but applies the wrong reasoning—orthogonal matrices have determinant $\pm 1$, but this approach is still fundamentally flawed. Remember: whenever you see an orthogonal matrix in a linear system, immediately think "transpose equals inverse." This property makes orthogonal matrices incredibly useful in applications like rotations and reflections where you need to easily reverse transformations.
5. For the matrix $A = \begin{pmatrix} \cos(k) & -\sin(k) & 0 \\ \sin(k) & \cos(k) & 0 \\ 0 & 0 & k \end{pmatrix}$ to be an orthogonal matrix, the value of $k$ must be:
   Answer: $1$ or $-1$ For a matrix to be orthogonal, its column vectors (or row vectors) must form an orthonormal set. Let's check the norms of the columns. The norm of the first column is $\sqrt{\cos^2(k) + \sin^2(k) + 0^2} = \sqrt{1} = 1$ for any real $k$. The norm of the second column is $\sqrt{(-\sin(k))^2 + \cos^2(k) + 0^2} = \sqrt{1} = 1$ for any real $k$. The norm of the third column is $\sqrt{0^2 + 0^2 + k^2} = |k|$. For the column to be a unit vector, we must have $|k|=1$, which means $k=1$ or $k=-1$. We also need to check orthogonality. The dot product of the first two columns is $\cos(k)(-\sin(k)) + \sin(k)\cos(k) + 0 = 0$. The dot products with the third column are also zero. Thus, the only condition that restricts $k$ is the normalization of the third column, which requires $k = \pm 1$.
6. Consider the set $S = \{Q \in M_n(\mathbb{R}) : Q^T Q = I \text{ and } \det(Q) = 1\}$ of special orthogonal matrices. If $Q_1, Q_2 \in S$, which of the following operations is guaranteed to preserve membership in $S$?
   Answer: $Q_1 Q_2$ always belongs to $S$ when $Q_1, Q_2 \in S$ When you encounter questions about special orthogonal matrices, you're dealing with a specific type of group structure. The set $S$ represents matrices that are both orthogonal ($Q^T Q = I$) and have determinant 1, forming what's called the special orthogonal group $SO(n)$. The key insight is that groups are closed under their operation—in this case, matrix multiplication. For option D, if $Q_1, Q_2 \in S$, then $(Q_1 Q_2)^T (Q_1 Q_2) = Q_2^T Q_1^T Q_1 Q_2 = Q_2^T I Q_2 = Q_2^T Q_2 = I$, so the product is orthogonal. Additionally, $\det(Q_1 Q_2) = \det(Q_1)\det(Q_2) = 1 \cdot 1 = 1$. Therefore, $Q_1 Q_2 \in S$. Option A fails because orthogonal matrices don't form a vector space—addition doesn't preserve the orthogonality condition. For example, if $Q_1 = I$ and $Q_2 = -I$, then $Q_1 + Q_2 = 0$, which isn't even invertible. Option B doesn't work because $Q_1 Q_2^T$ can have determinant -1. Since $Q_2 \in S$, we have $\det(Q_2^T) = \det(Q_2) = 1$, but this still gives $\det(Q_1 Q_2^T) = 1$. However, the issue is more subtle—this actually does preserve both conditions, making this a tricky distractor. Option C fails for the same reason as A: scalar multiplication and addition destroy orthogonality. Remember that group operations preserve group membership. When dealing with matrix groups like $SO(n)$, focus on multiplication, not linear combinations.
7. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is represented by the orthogonal matrix $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. Which statement best describes the geometric effect of this transformation?
   Answer: A clockwise rotation by $90^\circ$ about the origin. To understand the transformation, we can see how it acts on the standard basis vectors. $T(\begin{pmatrix} 1 \\ 0 \end{pmatrix}) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \end{pmatrix}$. $T(\begin{pmatrix} 0 \\ 1 \end{pmatrix}) = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. The transformation maps the vector $(1, 0)$ on the positive x-axis to $(0, -1)$ on the negative y-axis, and it maps $(0, 1)$ on the positive y-axis to $(1, 0)$ on the positive x-axis. This corresponds to a clockwise rotation of $90^\circ$ about the origin. Distractor (A) corresponds to the matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Distractor (C) corresponds to $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Distractor (D) is not a length-preserving transformation and thus not represented by an orthogonal matrix.
8. Let $Q$ be an $n \times n$ orthogonal matrix with the property that $Q^3 = I$. What is the minimum possible value of $n$ for which such a matrix can exist with $\det(Q) = -1$?
   Answer: $n = 3$, because the constraint $Q^3 = I$ with $\det(Q) = -1$ requires at least three dimensions Since $Q^3 = I$, the eigenvalues of $Q$ must be cube roots of unity: $1, \omega, \omega^2$ where $\omega = e^{2\pi i/3}$. For an orthogonal matrix, eigenvalues have magnitude $1$, which is satisfied. Since $Q$ is real, complex eigenvalues come in conjugate pairs. We have $\omega^2 = \overline{\omega}$. For $\det(Q) = -1$, the product of eigenvalues must be $-1$. With eigenvalues $\{1, \omega, \omega^2\}$, we get $1 \cdot \omega \cdot \omega^2 = \omega^3 = 1$, not $-1$. So we need at least one eigenvalue to be $-1$. The minimal set is $\{-1, \omega, \omega^2\}$, giving product $(-1) \cdot \omega \cdot \omega^2 = -1$. This requires $n = 3$. Choice A is wrong since $2 \times 2$ matrices can't have three distinct eigenvalues. Choice C and D overestimate the required dimension.
9. Suppose $T: \mathbb{R}^n \to \mathbb{R}^n$ is a linear transformation represented by an orthogonal matrix $A$. If $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$ satisfy $\|\mathbf{x}\| = 3$, $\|\mathbf{y}\| = 4$, and $\mathbf{x} \cdot \mathbf{y} = 6$, what is $T(\mathbf{x}) \cdot T(\mathbf{y})$?
   Answer: $T(\mathbf{x}) \cdot T(\mathbf{y}) = 6$, since orthogonal transformations preserve all inner products Orthogonal transformations preserve inner products. Since $T$ is represented by orthogonal matrix $A$, we have $T(\mathbf{x}) \cdot T(\mathbf{y}) = (A\mathbf{x}) \cdot (A\mathbf{y}) = (A\mathbf{x})^T (A\mathbf{y}) = \mathbf{x}^T A^T A \mathbf{y} = \mathbf{x}^T \mathbf{y} = \mathbf{x} \cdot \mathbf{y} = 6$. Choice B incorrectly suggests inner products are doubled. Choice C gives an incorrect value. Choice D incorrectly claims orthogonal transformations don't preserve inner products.
10. Which of the following conditions is sufficient to guarantee that a real $n \times n$ matrix $A$ is NOT orthogonal?
    Answer: One of the columns of $A$ is the zero vector. For a matrix to be orthogonal, its columns must form an orthonormal set. This means each column must be a unit vector (i.e., have a norm of 1). If one of the columns is the zero vector, its norm is $\sqrt{0^2 + ... + 0^2} = 0$. Since the norm is not 1, the columns are not an orthonormal set, and therefore the matrix cannot be orthogonal. Thus, having a zero vector as a column is a sufficient condition for a matrix to not be orthogonal. (A) is incorrect because orthogonal matrices can have a determinant of $-1$ (e.g., reflections). (B) is incorrect because many orthogonal matrices, such as rotations, are not symmetric. (C) is incorrect; for example, the orthogonal matrix $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ has zeros on its main diagonal.
11. Suppose $U$ and $V$ are $n \times n$ orthogonal matrices such that $UV = VU$. If $\mathbf{x}$ is a common eigenvector of both $U$ and $V$ with eigenvalues $\alpha$ and $\beta$ respectively, what can be concluded about the eigenvalues $\alpha$ and $\beta$?
    Answer: $\alpha, \beta \in \{-1, 1\}$ independently, with no additional constraint from commutativity Since $U$ and $V$ are orthogonal matrices, their eigenvalues must satisfy $|\alpha| = |\beta| = 1$. For real orthogonal matrices, this means $\alpha, \beta \in \{-1, 1\}$. The fact that $U$ and $V$ commute and share a common eigenvector $\mathbf{x}$ doesn't impose additional constraints on the individual eigenvalues beyond the orthogonality requirement. Choice A incorrectly assumes eigenvalues must be $1$. Choice B incorrectly suggests $\alpha\beta = 1$. Choice D incorrectly relates the eigenvalues to each other.
12. Let the matrix $Q$ be defined as $Q = \begin{pmatrix} 3/5 & x \\ 4/5 & y \end{pmatrix}$. If $Q$ is an orthogonal matrix, what is one possible value for $y$?
    Answer: $3/5$ For $Q$ to be an orthogonal matrix, its columns must form an orthonormal set. The first column vector, $\vec{v}_1 = (3/5, 4/5)$, has a norm of $\sqrt{(3/5)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = \sqrt{25/25} = 1$. The second column vector, $\vec{v}_2 = (x, y)$, must also have a norm of 1, so $x^2 + y^2 = 1$. Additionally, the columns must be orthogonal, meaning their dot product is zero: $\vec{v}_1 \cdot \vec{v}_2 = (3/5)x + (4/5)y = 0$. From the dot product equation, we get $3x = -4y$, or $x = -4y/3$. Substituting this into the norm equation gives $(-4y/3)^2 + y^2 = 1$, which simplifies to $16y^2/9 + y^2 = 1$. Combining terms, we have $(25/9)y^2 = 1$, so $y^2 = 9/25$. Therefore, possible values for $y$ are $y = \pm 3/5$. One of these possible values is $3/5$.
13. If $Q$ is an $n \times n$ orthogonal matrix and $\vec{x}$ is a non-zero vector in $\mathbb{R}^n$, which of the following quantities is necessarily equal to 1?
    Answer: $\|\frac{Q\vec{x}}{\|\vec{x}\|}\|$ A fundamental property of an orthogonal matrix $Q$ is that it preserves the norm (length) of vectors: $\|Q\vec{y}\| = \|\vec{y}\|$ for any vector $\vec{y}$. Let $\vec{y} = \frac{\vec{x}}{\|\vec{x}\|}$. The vector $\vec{y}$ is a unit vector, so $\|\vec{y}\|=1$. Applying the norm-preserving property, we get $\|Q\vec{y}\| = \|\vec{y}\| = 1$. Substituting back, this is $\|Q(\frac{\vec{x}}{\|\vec{x}\|})\| = 1$. Since $\frac{1}{\|\vec{x}\|}$ is a scalar, this is equivalent to $\frac{\|Q\vec{x}\|}{\|\vec{x}\|} = 1$. Thus, the expression in (A) equals 1. Distractor (C) is incorrect because $\det(Q)$ can be $-1$. Distractor (D) is incorrect because distinct columns of an orthogonal matrix are orthogonal, so their dot product is 0. Distractor (B) is not necessarily 1; for $Q=I$ and $\vec{x}=(2,0)^T$, the value is 4.
14. Let $Q_1$ and $Q_2$ be two $n \times n$ orthogonal matrices. Consider the matrix $P = Q_1 Q_2^T$. Which of the following statements about $P$ is always true?
    Answer: $P^{-1} = P^T$. If $Q_2$ is an orthogonal matrix, its transpose $Q_2^T$ is also its inverse ($Q_2^{-1}$), and is therefore also an orthogonal matrix. The product of two orthogonal matrices is always an orthogonal matrix. Thus, $P = Q_1 Q_2^T$ is an orthogonal matrix. The definition of an orthogonal matrix is that its inverse is equal to its transpose. Therefore, $P^{-1} = P^T$ must be true. Distractor (A) is false; products of orthogonal matrices are not generally symmetric. Distractor (C) is false; the determinant can be $-1$ if one of $Q_1$ or $Q_2^T$ has determinant $1$ and the other has $-1$. Distractor (D) is false; for example, if $P = -I$, which is orthogonal, the trace is $-n$.
15. Consider the matrix $A = \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{3} \\ 0 & 1/\sqrt{3} \\ -1/\sqrt{2} & 1/\sqrt{3} \end{pmatrix}$. Which statement about $A$ is correct?
    Answer: $A^T A = I$. An orthogonal matrix must be a square matrix. Since $A$ is a $3 \times 2$ matrix, it cannot be an orthogonal matrix, so (A) is false. A matrix with orthonormal columns satisfies the property $A^T A = I$. Let's check the columns of $A$. Column 1 has norm $\sqrt{(1/\sqrt{2})^2 + 0^2 + (-1/\sqrt{2})^2} = 1$. Column 2 has norm $\sqrt{(1/\sqrt{3})^2 + (1/\sqrt{3})^2 + (1/\sqrt{3})^2} = 1$. Their dot product is $(1/\sqrt{2})(1/\sqrt{3}) + 0 + (-1/\sqrt{2})(1/\sqrt{3}) = 0$. Since the columns are orthonormal, $A^T A = I_2$, so (B) is correct. For a non-square matrix, $A A^T \neq I$. In this case, $A A^T$ is a $3 \times 3$ matrix representing a projection, not the identity, so (C) is false. The rows are not orthogonal; the dot product of row 1 and row 2 is $(1/\sqrt{2})(0) + (1/\sqrt{3})(1/\sqrt{3}) = 1/3 \neq 0$, so (D) is false.
16. Let $Q$ be an $n \times n$ real orthogonal matrix. Which of the following statements is NOT always true?
    Answer: All eigenvalues of $Q$ are real numbers. The eigenvalues of a real orthogonal matrix must have a magnitude of 1, but they are not necessarily real. For example, the $2 \times 2$ rotation matrix $Q = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ is orthogonal, but for most values of $\theta$, its eigenvalues are the complex conjugate pair $e^{i\theta}$ and $e^{-i\theta}$. The other statements are fundamental properties of orthogonal matrices: the determinant is always $\pm 1$, the columns (and rows) form an orthonormal basis, and the transformation preserves lengths and angles, which is equivalent to preserving the dot product.
17. Let $H$ be a $4 \times 4$ Householder matrix of the form $H = I - 2\mathbf{u}\mathbf{u}^T$, where $\mathbf{u}$ is a unit vector. If $\mathbf{u} = \frac{1}{2}\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}$, what is the rank of the matrix $M = H + I$?
    Answer: $\text{rank}(M) = 3$, because $M$ has exactly one zero eigenvalue When you encounter Householder matrices, remember they represent reflections and have the special property that all their eigenvalues are either 1 or -1. Understanding how these eigenvalues transform when you add the identity matrix is key to finding the rank. Let's compute $M = H + I$. Since $H = I - 2\mathbf{u}\mathbf{u}^T$, we have: $M = H + I = (I - 2\mathbf{u}\mathbf{u}^T) + I = 2I - 2\mathbf{u}\mathbf{u}^T$ For any Householder matrix $H$, the eigenvalues are -1 (with multiplicity 1, corresponding to the reflection direction $\mathbf{u}$) and 1 (with multiplicity 3, for directions perpendicular to $\mathbf{u}$). When we compute $M = H + I$, we add 1 to each eigenvalue: the -1 eigenvalue becomes 0, and the three 1 eigenvalues become 2. Since $M$ has exactly one zero eigenvalue, it has a one-dimensional null space, making its rank $4 - 1 = 3$. Option A is wrong because $M$ is not invertible—it has a zero eigenvalue. Option B misunderstands the rank: while $\mathbf{u}\mathbf{u}^T$ has rank 1, the matrix $2I - 2\mathbf{u}\mathbf{u}^T$ doesn't simply inherit this rank. Option C gives an incorrect rank without proper justification. Study tip: For Householder problems, always think about eigenvalues first. The key insight is that $H + I$ shifts all eigenvalues by 1, turning the single -1 eigenvalue into 0, which directly tells you the rank deficiency.
18. Let $Q$ be a $3 \times 3$ orthogonal matrix with $\det(Q) = -1$. If the first column of $Q$ is $\begin{bmatrix} 1/2 \\ \sqrt{3}/2 \\ 0 \end{bmatrix}$ and the second column is $\begin{bmatrix} -\sqrt{3}/2 \\ 1/2 \\ 0 \end{bmatrix}$, what is the third column of $Q$?
    Answer: $\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$ Since $Q$ is orthogonal, its columns form an orthonormal basis. The third column must be orthogonal to both given columns and have unit length. Taking the cross product of the first two columns gives $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. However, since $\det(Q) = -1$, the matrix represents an improper orthogonal transformation (reflection), so we need $\begin{bmatrix} 0 \\ 0 \\ -1 \end{bmatrix}$ to achieve the negative determinant. Choice A would give $\det(Q) = 1$. Choices C and D are not orthogonal to the given columns.

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