Orthogonal Diagonalization - Linear Algebra

1. Consider the symmetric matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$. During the orthogonal diagonalization process, it is found that one eigenvalue is $\lambda = 2$ with geometric multiplicity 1. If $P$ is the orthogonal matrix that diagonalizes $A$, which statement about the structure of $P^TAP$ is necessarily true?
   Answer: $P^TAP$ has exactly one entry equal to 2 on its main diagonal Since $\lambda = 2$ has geometric multiplicity 1, it appears exactly once on the diagonal of $P^TAP = D$, the diagonal matrix of eigenvalues. The other two eigenvalues are distinct from 2. Choice B would require geometric multiplicity 2, which contradicts the given information. Choice C is incorrect because not all eigenvalues are necessarily positive. Choice D would require $\lambda = 2$ to have geometric multiplicity 3, making $A = 2I$, which contradicts the given matrix.
2. A $3 \times 3$ symmetric matrix $A$ has the property that $A\mathbf{v}_1 = 3\mathbf{v}_1$ and $A\mathbf{v}_2 = -\mathbf{v}_2$ where $\mathbf{v}_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$. During the construction of the orthogonal matrix $P$ for diagonalization, what must be true about the third column $\mathbf{p}_3$?
   Answer: $\mathbf{p}_3$ must be orthogonal to both $\mathbf{v}_1$ and $\mathbf{v}_2$ and satisfy $A\mathbf{p}_3 = \lambda_3\mathbf{p}_3$ for some eigenvalue $\lambda_3$ When you encounter a question about diagonalizing symmetric matrices, remember that the key insight is understanding what makes an orthogonal diagonalization work: you need a complete orthonormal basis of eigenvectors. The given information tells us that $\mathbf{v}_1$ and $\mathbf{v}_2$ are eigenvectors with eigenvalues 3 and -1, respectively. Since $A$ is symmetric, it's guaranteed to have three orthogonal eigenvectors. You can verify that $\mathbf{v}_1 \cdot \mathbf{v}_2 = (2)(0) + (1)(1) + (0)(1) = 1 \neq 0$, so these aren't orthogonal yet, but that's fine—we can orthogonalize them later. For the orthogonal matrix $P = [\mathbf{p}_1 \; \mathbf{p}_2 \; \mathbf{p}_3]$ to diagonalize $A$, each column must be an eigenvector. The third column $\mathbf{p}_3$ must satisfy $A\mathbf{p}_3 = \lambda_3\mathbf{p}_3$ for some third eigenvalue $\lambda_3$. Additionally, since we need an orthogonal matrix, $\mathbf{p}_3$ must be orthogonal to the normalized versions of $\mathbf{v}_1$ and $\mathbf{v}_2$. Answer B captures both requirements. Answer A ignores the eigenvalue requirement—orthogonality alone isn't sufficient. Answer C assumes the cross product automatically gives an eigenvector, which isn't guaranteed. Answer D focuses only on determinant considerations while missing the fundamental eigenvalue requirement. Remember: for orthogonal diagonalization, every column of $P$ must be both an eigenvector and part of an orthonormal set. Both conditions are non-negotiable.
3. A symmetric matrix $A$ is orthogonally diagonalized as $A = PDP^T$ with $P = \begin{pmatrix} 1/\sqrt{5} & -2/\sqrt{5} \\ 2/\sqrt{5} & 1/\sqrt{5} \end{pmatrix}$ and $D = \begin{pmatrix} 10 & 0 \\ 0 & 5 \end{pmatrix}$. What is the matrix $A$?
   Answer: $A = \begin{pmatrix} 6 & 2 \\ 2 & 9 \end{pmatrix}$ To find $A$, we compute the product $PDP^T$. First, find $P^T = \begin{pmatrix} 1/\sqrt{5} & 2/\sqrt{5} \\ -2/\sqrt{5} & 1/\sqrt{5} \end{pmatrix}$. The product is $A = \frac{1}{5} \begin{pmatrix} 1 & -2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 10 & 0 \\ 0 & 5 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -2 & 1 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 10 & -10 \\ 20 & 5 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -2 & 1 \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 10(1)+(-10)(-2) & 10(2)+(-10)(1) \\ 20(1)+5(-2) & 20(2)+5(1) \end{pmatrix} = \frac{1}{5} \begin{pmatrix} 30 & 10 \\ 10 & 45 \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 2 & 9 \end{pmatrix}$. Distractor B results from computing $PDP$. Distractor C results from forgetting the $\frac{1}{5}$ factor from multiplying $P$ and $P^T$. Distractor D contains a sign error in the off-diagonal elements, suggesting a calculation mistake.
4. A $3 \times 3$ symmetric matrix $A$ has the characteristic polynomial $p(\lambda) = -(\lambda-4)^2(\lambda+2)$. If $A=PDP^T$ is an orthogonal diagonalization of $A$, which of the following could be the matrix $D$?
   Answer: $D = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -2 \end{pmatrix}$ The eigenvalues of a matrix are the roots of its characteristic polynomial. The roots of $p(\lambda) = -(\lambda-4)^2(\lambda+2)=0$ are $\lambda=4$ (with algebraic multiplicity 2) and $\lambda=-2$ (with algebraic multiplicity 1). The diagonal matrix $D$ in an orthogonal diagonalization consists of the eigenvalues of $A$ along its main diagonal. The order of the eigenvalues on the diagonal can vary, but the values themselves must be 4, 4, and -2. Choice A correctly lists these eigenvalues. Choices B and C use incorrect signs for the eigenvalues. Choice D shows a Jordan block, which would imply that the matrix is not diagonalizable. However, the Spectral Theorem guarantees that every real symmetric matrix is diagonalizable, so this form is not possible for $D$.
5. A $3 \times 3$ symmetric matrix $A$ has eigenvalues $\lambda_1 = 4$, $\lambda_2 = 1$, and $\lambda_3 = -2$. If the eigenvector corresponding to $\lambda_1 = 4$ is $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$, and after applying the Gram-Schmidt process to obtain an orthonormal basis, the first column of the orthogonal matrix $P$ is $\begin{pmatrix} 1/\sqrt{6} \\ 2/\sqrt{6} \\ 1/\sqrt{6} \end{pmatrix}$. What is the trace of $A^2$?
   Answer: 21 Since $A$ is orthogonally diagonalizable, $A = PDP^T$ where $D = \text{diag}(4, 1, -2)$. Then $A^2 = PD^2P^T$ where $D^2 = \text{diag}(16, 1, 4)$. The trace of $A^2$ equals the trace of $D^2$, which is $16 + 1 + 4 = 21$. Choice B incorrectly uses $4^2 + 1^2 + (-2)^2 = 16 + 1 + 4 = 21$ but makes an arithmetic error. Choice C uses the trace of $A$ instead of $A^2$. Choice D confuses the calculation by mixing eigenvalues incorrectly.
6. The spectral decomposition of a symmetric matrix $A$ is given by $A = \sum_{i=1}^{n} \lambda_i \mathbf{u}_i \mathbf{u}_i^T$, where $\lambda_i$ are eigenvalues and $\mathbf{u}_i$ are the corresponding orthonormal eigenvectors. If $A = \begin{pmatrix} 6 & 2 \\ 2 & 9 \end{pmatrix}$ has eigenvalues $\lambda_1=10$ and $\lambda_2=5$, what is the projection matrix $\mathbf{u}_1 \mathbf{u}_1^T$ associated with $\lambda_1=10$?
   Answer: $\begin{pmatrix} 0.2 & 0.4 \\ 0.4 & 0.8 \end{pmatrix}$ First, find the eigenvector for $\lambda_1=10$. We solve $(A-10I)\mathbf{v}=0$, which is $\begin{pmatrix} -4 & 2 \\ 2 & -1 \end{pmatrix}\mathbf{v}=0$. This gives the equation $2x_1 - x_2 = 0$. An eigenvector is $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. To find the orthonormal eigenvector $\mathbf{u}_1$, we normalize it: $||\mathbf{v}|| = \sqrt{1^2+2^2}=\sqrt{5}$. So, $\mathbf{u}_1 = \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. The projection matrix is $\mathbf{u}_1 \mathbf{u}_1^T = \left( \frac{1}{\sqrt{5}}\begin{pmatrix} 1 \\ 2 \end{pmatrix} \right) \left( \frac{1}{\sqrt{5}}\begin{pmatrix} 1 & 2 \end{pmatrix} \right) = \frac{1}{5} \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 0.2 & 0.4 \\ 0.4 & 0.8 \end{pmatrix}$. Choice B is the projection matrix for $\lambda_2=5$. Choice C is the unnormalized outer product. Choice D is $\lambda_1$ times the projection matrix.
7. A symmetric matrix $A$ is known to satisfy $A^2 = 5A - 6I$. If $A$ is orthogonally diagonalized as $A = PDP^T$, where $D$ is diagonal, which of the following describes the possible diagonal entries of $D$?
   Answer: The diagonal entries are 2 and 3, each appearing with some multiplicity When you encounter a symmetric matrix satisfying a polynomial equation like $A^2 = 5A - 6I$, you're dealing with eigenvalue problems. Since $A$ is symmetric, it can be orthogonally diagonalized, and its eigenvalues appear as the diagonal entries of $D$. The key insight is that if $A$ satisfies the equation $A^2 = 5A - 6I$, then its eigenvalues must satisfy the same polynomial equation. Here's why: if $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, then $Av = \lambda v$. Applying the matrix equation to this eigenvector: $A^2v = 5Av - 6Iv$ $A(\lambda v) = 5(\lambda v) - 6v$ $\lambda^2 v = 5\lambda v - 6v$ Since $v \neq 0$, we can divide by $v$ to get $\lambda^2 = 5\lambda - 6$, or $\lambda^2 - 5\lambda + 6 = 0$. Factoring this quadratic: $(\lambda - 2)(\lambda - 3) = 0$, so $\lambda = 2$ or $\lambda = 3$. Therefore, the diagonal entries of $D$ can only be 2 and 3, making D correct. Looking at the wrong answers: A gives eigenvalues 0 and 5, but these don't satisfy $\lambda^2 - 5\lambda + 6 = 0$. B suggests 5 and -6, which also fail the eigenvalue equation. C proposes 1 and 6, but again, neither satisfies our required polynomial. Study tip: When a matrix satisfies a polynomial equation, its eigenvalues must satisfy the same polynomial. Always substitute potential eigenvalues back into the characteristic equation to verify your answer.
8. Let $A$ be a $4 \times 4$ symmetric matrix that satisfies $(A - 2I)(A + I) = O$. If $\text{rank}(A - 2I) = 3$, what is the correct form of the diagonal matrix $D$ in the orthogonal diagonalization $A = PDP^T$?
   Answer: $D = \text{diag}(2, -1, -1, -1)$ or some permutation of these diagonal entries When you encounter a symmetric matrix satisfying a polynomial equation, you're dealing with eigenvalue constraints. The equation $(A - 2I)(A + I) = O$ tells you that every eigenvalue $\lambda$ of $A$ must satisfy $(\lambda - 2)(\lambda + 1) = 0$, so the only possible eigenvalues are $\lambda = 2$ and $\lambda = -1$. The key insight comes from using the rank information. Since $\text{rank}(A - 2I) = 3$, the nullity of $(A - 2I)$ is $4 - 3 = 1$. This means the eigenspace for $\lambda = 2$ has dimension 1, so $\lambda = 2$ appears exactly once. Since $A$ is $4 \times 4$, the remaining three eigenvalues must all be $\lambda = -1$. Therefore, $D = \text{diag}(2, -1, -1, -1)$ or some permutation, making choice B correct. Choice A is wrong because it assumes $\lambda = 2$ has multiplicity 3, which would mean $\text{rank}(A - 2I) = 1$, not 3. Choice C incorrectly introduces zero eigenvalues, but zero doesn't satisfy $(\lambda - 2)(\lambda + 1) = 0$. Choice D claims all eigenvalues equal 1, but 1 also doesn't satisfy the polynomial constraint. Remember: when a matrix satisfies a polynomial equation, its eigenvalues are roots of that polynomial. Always use rank information to determine the geometric multiplicity of each eigenvalue—this tells you how many times each eigenvalue appears on the diagonal.
9. Consider a $4 \times 4$ symmetric matrix $A$ with the property that $\text{rank}(A - 2I) = 2$ and $\text{rank}(A + I) = 3$. If $A$ is orthogonally diagonalizable, what is the geometric multiplicity of the eigenvalue $\lambda = 2$?
   Answer: 2 When you encounter questions about eigenvalue multiplicities with rank conditions, you're working with the fundamental relationship between eigenvalues, eigenvectors, and matrix rank. The key insight is that $\text{rank}(A - \lambda I)$ directly tells you about the geometric multiplicity of eigenvalue $\lambda$. For any $n \times n$ matrix, the geometric multiplicity of eigenvalue $\lambda$ equals $n - \text{rank}(A - \lambda I)$. This is because the geometric multiplicity is the dimension of the eigenspace, which is the nullspace of $(A - \lambda I)$. Since $A$ is $4 \times 4$ and $\text{rank}(A - 2I) = 2$, the geometric multiplicity of $\lambda = 2$ is $4 - 2 = 2$. The condition $\text{rank}(A + I) = 3$ tells us that $\lambda = -1$ has geometric multiplicity $4 - 3 = 1$. Looking at the wrong answers: Choice A) gives geometric multiplicity 1, which would require $\text{rank}(A - 2I) = 3$, not 2. Choice D) suggests geometric multiplicity 3, which would need $\text{rank}(A - 2I) = 1$. Choice B) claims geometric multiplicity 4, which would only occur if $A = 2I$, making $\text{rank}(A - 2I) = 0$. Since we know two eigenvalues account for multiplicities 2 and 1, the remaining eigenvalue(s) must account for multiplicity 1, confirming our answer is consistent. Study tip: Remember the formula $\text{geometric multiplicity} = n - \text{rank}(A - \lambda I)$ for $n \times n$ matrices. This relationship appears frequently in linear algebra problems involving eigenspaces.
10. Let $A$ be a $3 \times 3$ symmetric matrix with distinct eigenvalues $\lambda_1, \lambda_2, \lambda_3$ and corresponding eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$. Which of the following statements must be true?
    Answer: The dot product $\mathbf{v}_i \cdot \mathbf{v}_j = 0$ for all $i \neq j$. A key theorem regarding symmetric matrices states that eigenvectors corresponding to distinct eigenvalues are orthogonal. Therefore, the dot product of any two eigenvectors from different eigenspaces must be zero. Choice A is false; the sum of two eigenvectors is generally not an eigenvector unless they belong to the same eigenspace. Choice B is not necessarily true; the set is orthogonal, but the eigenvectors are not guaranteed to be of unit length unless they have been normalized. Choice D is false; a set of non-zero orthogonal vectors is always linearly independent.
11. Let $A$ and $B$ be two $n \times n$ symmetric matrices. Which of the following statements is not always true?
    Answer: $AB$ is a symmetric matrix. When you encounter questions about symmetric matrices, you need to recall which operations preserve symmetry and which don't. A matrix is symmetric if $A = A^T$, meaning it equals its transpose. Let's examine why option B is not always true. For the product $AB$ to be symmetric, we need $(AB)^T = AB$. Using the transpose property $(AB)^T = B^T A^T$, and since both matrices are symmetric (so $A^T = A$ and $B^T = B$), we get $(AB)^T = BA$. For $AB$ to be symmetric, we need $AB = BA$, meaning the matrices must commute. However, symmetric matrices don't always commute, so $AB$ isn't always symmetric. Now let's verify why the other options are always true. Option A works because $(A + B)^T = A^T + B^T = A + B$, so the sum of symmetric matrices is always symmetric. Option C holds since $(cA)^T = cA^T = cA$ for any scalar $c$. Option D is true by the Spectral Theorem, which guarantees that every symmetric matrix is orthogonally diagonalizable. The key insight is that while addition and scalar multiplication preserve symmetry, matrix multiplication requires the additional condition of commutativity. When studying symmetric matrices, remember that their fundamental properties (like orthogonal diagonalizability) are preserved, but products require careful consideration of whether the matrices commute.
12. A $3 \times 3$ symmetric matrix $A$ has eigenvalues $\lambda_1 = 4$ with multiplicity 2 and $\lambda_2 = -1$. Let $E_4$ and $E_{-1}$ be the corresponding eigenspaces. Which statement is necessarily true about the orthogonal diagonalization of $A$?
    Answer: Any vector in $E_4$ is orthogonal to any vector in $E_{-1}$. For any symmetric matrix, eigenvectors corresponding to distinct eigenvalues are orthogonal. Therefore, every vector in the eigenspace $E_4$ is orthogonal to every vector in the eigenspace $E_{-1}$. Choice A is false; a standard basis for an eigenspace may not be orthogonal. The Gram-Schmidt process is required to produce an orthogonal basis for $E_4$. Choice B is false; for a symmetric matrix, the geometric multiplicity of an eigenvalue equals its algebraic multiplicity, so the dimension of $E_4$ must be 2. Choice D is false; the Spectral Theorem guarantees that any real symmetric matrix is orthogonally diagonalizable, regardless of repeated eigenvalues.
13. A student claims to have found an orthogonal diagonalization for the matrix $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$. Why must this claim be incorrect?
    Answer: The matrix $A$ is not symmetric. When you encounter questions about orthogonal diagonalization, remember that this is a very special process that only works for certain types of matrices. Orthogonal diagonalization means finding an orthogonal matrix $P$ such that $P^TAP = D$ where $D$ is diagonal. The fundamental theorem here is that a matrix can be orthogonally diagonalized if and only if it is symmetric (meaning $A = A^T$). Let's check if matrix $A$ is symmetric by computing its transpose: $A^T = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$. Since $A \neq A^T$, the matrix is not symmetric, which immediately tells us orthogonal diagonalization is impossible. Looking at the incorrect options: Choice A is wrong because $A$ is invertible—its determinant is $1 \cdot 3 - 2 \cdot 0 = 3 \neq 0$. Choice B is incorrect since the eigenvalues (found by solving $\det(A - \lambda I) = 0$) are $\lambda = 1$ and $\lambda = 3$, both real numbers. Choice C is also wrong because having two distinct eigenvalues guarantees linearly independent eigenvectors, so $A$ is diagonalizable—just not orthogonally diagonalizable. The correct answer is D: the matrix is not symmetric. Study tip: For orthogonal diagonalization questions, always check symmetry first. If $A \neq A^T$, you can immediately rule out orthogonal diagonalization without computing eigenvalues or eigenvectors. Only symmetric matrices have the special property that their eigenvectors can be chosen to be orthogonal.
14. Given a symmetric matrix $A$, a student performs the following steps to find an orthogonal matrix $P$ for diagonalization: \\ 1. Find the distinct eigenvalues $\lambda_i$. \\ 2. For each $\lambda_i$, find a basis for the eigenspace $E_{\lambda_i}$. \\ 3. Form $P$ by placing the basis vectors found in step 2 as columns. \\ The student finds their matrix $P$ is not orthogonal. Which of the following describes the critical omission in their procedure?
    Answer: They did not convert the collected basis vectors into an orthonormal set before forming the matrix $P$. When diagonalizing a symmetric matrix, you're looking to find an orthogonal matrix $P$ such that $P^TAP = D$, where $D$ is diagonal. The key insight is that while symmetric matrices guarantee orthogonal eigenvectors between different eigenspaces, they don't guarantee that basis vectors within the same eigenspace are orthogonal to each other. The student's procedure correctly finds eigenvalues and eigenspace bases, but misses a crucial step. When an eigenvalue has multiplicity greater than 1 (meaning its eigenspace has dimension > 1), any basis for that eigenspace will work algebraically, but those basis vectors might not be orthogonal to each other. Since $P$ must be orthogonal (meaning $P^TP = I$), all columns of $P$ must form an orthonormal set. Answer B identifies the critical omission: the student must apply the Gram-Schmidt process (or another orthonormalization procedure) to convert their collection of basis vectors into an orthonormal set before forming $P$. Answer A is wrong because eigenvalues of symmetric matrices are automatically real - this isn't something that needs verification. Answer C is incorrect because the order of eigenvalues in $D$ doesn't affect whether $P$ is orthogonal. Answer D is false because eigenvectors must be columns of $P$ in the standard diagonalization formula $P^TAP = D$. Study tip: Remember that "orthogonal matrix" means orthonormal columns. When diagonalizing symmetric matrices, always orthonormalize your eigenvectors, especially when dealing with repeated eigenvalues where you have freedom in choosing the eigenspace basis.
15. A $4 \times 4$ real symmetric matrix $A$ has characteristic polynomial $p(\lambda) = (\lambda - 3)^2(\lambda + 1)^2$. During orthogonal diagonalization, it is determined that the eigenspace for $\lambda = 3$ has dimension 1, while the eigenspace for $\lambda = -1$ has dimension 2. What can be concluded about the diagonalizability of $A$?
    Answer: $A$ is not diagonalizable because the geometric multiplicity does not equal the algebraic multiplicity for $\lambda = 3$ For $\lambda = 3$, the algebraic multiplicity is 2 but the geometric multiplicity is 1. Since geometric multiplicity must equal algebraic multiplicity for diagonalizability, $A$ is not diagonalizable. Choice B incorrectly assumes all symmetric matrices are diagonalizable (true only when geometric = algebraic multiplicities). Choice C is contradictory since non-orthogonal diagonalizability still requires geometric = algebraic multiplicities. Choice D is incorrect because the given information is sufficient to determine non-diagonalizability.
16. Suppose $A = PDP^T$ is an orthogonal diagonalization of a non-zero symmetric matrix $A$. If the diagonal matrix $D$ is replaced by $D^k$ for a positive integer $k$, what matrix does the expression $PD^kP^T$ represent?
    Answer: $A^k$ We can write out the power of $A$: $A^k = (PDP^T)^k = (PDP^T)(PDP^T)...(PDP^T)$. Since $P$ is an orthogonal matrix, $P^TP = I$. The product telescopes: $A^k = P D (P^T P) D (P^T P) ... D P^T = P D I D I ... D P^T = PD^kP^T$. Therefore, the expression represents $A^k$. Choice B confuses matrix exponentiation with scalar multiplication. Choice C is a nonsensical operation. Choice D incorrectly applies the exponent property $(XY)^k=X^kY^k$, which is not generally true for matrices.
17. Let $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$. The matrix $A$ is orthogonally diagonalized by the equation $A=PDP^T$. If the diagonal entries of $D$ are ordered such that $\lambda_1 > \lambda_2$, which of the following is a possible matrix for $P$?
    Answer: $P = \begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$ First, find the eigenvalues of $A$. The characteristic equation is $(2-\lambda)^2 - 1 = 0$, which simplifies to $\lambda^2 - 4\lambda + 3 = 0$, or $(\lambda-3)(\lambda-1) = 0$. The eigenvalues are $\lambda_1 = 3$ and $\lambda_2 = 1$. Since we must have $\lambda_1 > \lambda_2$, the first column of $P$ must be the eigenvector for $\lambda_1=3$ and the second for $\lambda_2=1$. For $\lambda_1=3$, we solve $(A-3I)\mathbf{v}=0$, which gives $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{v}=0$. An eigenvector is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, which normalizes to $\begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}$. For $\lambda_2=1$, we solve $(A-I)\mathbf{v}=0$, which gives $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\mathbf{v}=0$. An eigenvector is $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$ or $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$. Normalizing $\begin{pmatrix} -1 \\ 1 \end{pmatrix}$ gives $\begin{pmatrix} -1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}$. Thus, $P = \begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$. Choice A swaps the second eigenvector. Choice B has incorrect eigenvectors. Choice C has correct eigenvectors, but they are not normalized.
18. A $3 \times 3$ symmetric matrix $A$ has orthogonal diagonalization $A = PDP^T$ where $P = \begin{pmatrix} \mathbf{p}_1 & \mathbf{p}_2 & \mathbf{p}_3 \end{pmatrix}$ and $D = \text{diag}(4, 1, 1)$. If $\mathbf{p}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, what constraint must $\mathbf{p}_2$ and $\mathbf{p}_3$ satisfy beyond being unit vectors?
    Answer: $\mathbf{p}_2$ and $\mathbf{p}_3$ must be orthogonal to $\mathbf{p}_1$ and to each other, with first components equal to zero When you encounter orthogonal diagonalization problems, focus on the fundamental requirements: the matrix $P$ must have orthonormal columns that are eigenvectors of $A$. Since $A = PDP^T$ with $D = \text{diag}(4, 1, 1)$, we know $\mathbf{p}_1$ corresponds to eigenvalue 4, while $\mathbf{p}_2$ and $\mathbf{p}_3$ correspond to the repeated eigenvalue 1. Given $\mathbf{p}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, the orthogonality requirement in matrix $P$ means $\mathbf{p}_2$ and $\mathbf{p}_3$ must be orthogonal to $\mathbf{p}_1$. Since $\mathbf{p}_1$ has only a first component, both $\mathbf{p}_2$ and $\mathbf{p}_3$ must have zero first components to ensure $\mathbf{p}_1 \cdot \mathbf{p}_2 = \mathbf{p}_1 \cdot \mathbf{p}_3 = 0$. Additionally, for $P$ to be orthogonal, $\mathbf{p}_2$ and $\mathbf{p}_3$ must be orthogonal to each other. Choice A incorrectly suggests $\mathbf{p}_2$ and $\mathbf{p}_3$ don't need mutual orthogonality, but orthogonal diagonalization requires all columns of $P$ to be orthonormal. Choice B ignores the constraint from $\mathbf{p}_1$—if $\mathbf{p}_2$ and $\mathbf{p}_3$ had nonzero first components, they couldn't be orthogonal to $\mathbf{p}_1$. Choice D misses the crucial orthogonality requirements entirely, focusing only on the eigenvector property. The answer is C because orthogonal diagonalization demands that all eigenvectors be mutually orthogonal, which completely determines the structure of $\mathbf{p}_2$ and $\mathbf{p}_3$. Remember: In orthogonal diagonalization, "orthogonal" applies to the entire set of eigenvectors, not just within eigenspaces. This constraint often determines more structure than students initially realize.
19. The quadratic form $Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$ for a symmetric matrix $A = \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix}$ can be simplified by an orthogonal change of variables $\mathbf{x} = P\mathbf{y}$, which eliminates the cross-product term. Which expression represents the quadratic form in the new variables $\mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$?
    Answer: $5y_1^2 + y_2^2$ The change of variables $\mathbf{x}=P\mathbf{y}$ transforms the quadratic form to $\mathbf{x}^T A \mathbf{x} = (P\mathbf{y})^T A (P\mathbf{y}) = \mathbf{y}^T P^T A P \mathbf{y} = \mathbf{y}^T D \mathbf{y}$. The new quadratic form has no cross-product terms, and the coefficients are the eigenvalues of $A$. We find the eigenvalues of $A$: $\det(A-\lambda I) = (3-\lambda)^2 - 4 = \lambda^2 - 6\lambda + 5 = (\lambda-5)(\lambda-1)=0$. The eigenvalues are $\lambda_1=5$ and $\lambda_2=1$. Therefore, the new quadratic form is $\lambda_1 y_1^2 + \lambda_2 y_2^2 = 5y_1^2 + y_2^2$ (or $y_1^2 + 5y_2^2$, depending on the ordering). Choice A is the original form with different variables. Choice C incorrectly uses only the diagonal elements of $A$. Choice D is an arbitrary combination of numbers related to $A$.
20. A real matrix $A$ is orthogonally diagonalizable. Which of the following conditions is necessary, but not sufficient, for $A$ to be orthogonally diagonalizable?
    Answer: $A$ must be diagonalizable and have only real eigenvalues. If a matrix $A$ is orthogonally diagonalizable, it can be written as $A=PDP^T$ where $P$ is orthogonal and $D$ is a real diagonal matrix. This implies $A$ is diagonalizable and its eigenvalues (the entries of $D$) are real. So, the condition is necessary. However, it is not sufficient. For example, the non-symmetric matrix $A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ is diagonalizable with real eigenvalues 1 and 2, but it is not orthogonally diagonalizable because it is not symmetric. Choice A is incorrect because an orthogonally diagonalizable matrix can have an eigenvalue of 0, making it non-invertible. Choice B is incorrect because this is a sufficient condition, but not a necessary one (e.g., the identity matrix is orthogonally diagonalizable but has repeated eigenvalues). Choice D is incorrect because this is both a necessary and sufficient condition for a matrix to be orthogonally diagonalizable.

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