Spectral Theorem - Linear Algebra

1. If a real matrix $A$ is orthogonally diagonalizable, which of the following statements must be false?
   Answer: $A$ has a complex eigenvalue. When you encounter questions about orthogonally diagonalizable matrices, you're dealing with a very special class of matrices that have strong structural properties. A matrix is orthogonally diagonalizable if there exists an orthogonal matrix $Q$ such that $Q^T A Q = D$, where $D$ is diagonal. The key insight is that orthogonally diagonalizable matrices are precisely the symmetric matrices (for real matrices). This connection leads to a crucial property: symmetric matrices can only have real eigenvalues. Here's why: if $Av = \lambda v$ for some eigenvector $v$, then taking the conjugate transpose gives us $v^* A^T = \bar{\lambda} v^*$. For symmetric matrices where $A = A^T$, we get $v^* A v = \lambda v^* v = \bar{\lambda} v^* v$. Since $v^* v > 0$, we must have $\lambda = \bar{\lambda}$, meaning $\lambda$ is real. Therefore, choice A must be false—orthogonally diagonalizable matrices cannot have complex eigenvalues. The other options can all be true. Choice B is wrong because repeated eigenvalues are perfectly acceptable (consider the identity matrix). Choice C is wrong because singular matrices can be orthogonally diagonalizable—any symmetric matrix with a zero eigenvalue works. Choice D is wrong because while some orthogonally diagonalizable matrices are invertible, others (like the zero matrix) are not. Remember this connection: for real matrices, "orthogonally diagonalizable" essentially means "symmetric," and symmetric matrices always have real eigenvalues. This makes complex eigenvalues impossible.
2. Let $A$ be a symmetric $n \times n$ matrix. According to the Spectral Theorem, $A$ is orthogonally diagonalizable. Which equation correctly represents this decomposition?
   Answer: $A = PDP^T$, where $P$ is an orthogonal matrix whose columns are orthonormal eigenvectors. The spectral decomposition for a symmetric matrix $A$ is $A = PDP^T$. In this form, $D$ is a diagonal matrix of eigenvalues, and $P$ is an orthogonal matrix whose columns are the corresponding orthonormal eigenvectors. Because $P$ is orthogonal, its inverse is its transpose, so $P^{-1} = P^T$. (A) represents standard diagonalization, which applies to a broader class of matrices than just symmetric ones. (C) has the matrices in the incorrect order. (D) represents the Singular Value Decomposition (SVD), which is a different, though related, matrix factorization.
3. Consider the matrix $A$ given by $A = \begin{pmatrix} 3 & 4 \\ -1 & -1 \end{pmatrix}$ Why does the Spectral Theorem for real matrices not apply to $A$?
   Answer: The matrix $A$ is not symmetric. The Spectral Theorem is a statement about symmetric matrices. The first step in applying the theorem is to check if the matrix is symmetric. A matrix is symmetric if $A = A^T$. In this case, $a_{12} = 4$ while $a_{21} = -1$, so $A$ is not symmetric. Therefore, the theorem does not apply. (A) is false; the determinant is $(-3) - (-4) = 1 \neq 0$. (B) is false; the characteristic equation is $(\lambda-1)^2=0$, so the only eigenvalue is $\lambda=1$. (D) is a true statement about this matrix, but it is a consequence of its properties, not the reason the theorem's conditions aren't met. The failure to meet the symmetry hypothesis is the primary reason the theorem is inapplicable.
4. Given a $3 \times 3$ symmetric matrix $H$ with spectral decomposition $H = \sum_{i=1}^3 \lambda_i \mathbf{u}_i \mathbf{u}_i^T$ where $\lambda_1 > \lambda_2 > \lambda_3 > 0$, which of the following optimization problems has solution $\mathbf{x} = \mathbf{u}_2$?
   Answer: Maximize $\mathbf{x}^T H \mathbf{x}$ subject to $\|\mathbf{x}\| = 1$ and $\mathbf{x} \perp \mathbf{u}_1$. By the Spectral Theorem and variational characterization of eigenvalues, maximizing $x^T$ H x subject to ||x|| = 1 gives x = u₁ (largest eigenvalue). When we add the constraint x ⊥ u₁, we maximize over the orthogonal complement, and the maximum is achieved at u₂ (second-largest eigenvalue). Choice B seeks a minimum with wrong constraint. Choice C involves $H^{-1}$, which has eigenvalues 1/λᵢ in reverse order. Choice D over-constrains the problem to a one-dimensional space.
5. A $3 \times 3$ real symmetric matrix $R$ satisfies $R^3 = 8I$. If all eigenvalues of $R$ are real, what is the possible set of eigenvalues for $R$?
   Answer: The eigenvalues must all equal $2$ since complex eigenvalues are prohibited for real symmetric matrices and $\lambda^3 = 8$ requires $\lambda = 2$. Since R³ = 8I, each eigenvalue λ must satisfy λ³ = 8, so λ³ - 8 = 0. The solutions are λ = 2, λ = -1 + i√3, and λ = -1 - i√3. However, by the Spectral Theorem, real symmetric matrices have only real eigenvalues. Since only λ = 2 is real among the cube roots of 8, all eigenvalues must equal 2. Choice A is correct in conclusion but incomplete in reasoning. Choice B ignores the real eigenvalue requirement. Choice D incorrectly allows complex eigenvalues for real symmetric matrices.
6. A student claims that a certain $3 \times 3$ real matrix is symmetric and has eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 3+4i$. Based on this information, what must be true about the third eigenvalue, $\lambda_3$?
   Answer: The student's claim is impossible because the given conditions are contradictory. A fundamental consequence of the Spectral Theorem is that all eigenvalues of a real symmetric matrix must be real numbers. The eigenvalue $\lambda_2 = 3+4i$ is a complex number, not a real number. Therefore, a real matrix with this eigenvalue cannot be symmetric. The student's claim contains a contradiction. (A) is a common trap; while complex eigenvalues of real matrices come in conjugate pairs, a symmetric matrix cannot have them in the first place. (B) and (C) are incorrect because the premise is flawed.
7. Consider the $3 \times 3$ symmetric matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix}$. Which of the following best describes the geometric interpretation of the quadratic form $\mathbf{x}^T A \mathbf{x}$ based on the Spectral Theorem?
   Answer: An ellipsoid with semi-axes determined by eigenvalues $1, 3, 5$ but with principal axes rotated away from coordinate directions. The eigenvalues of A are found by solving det(A - λI) = 0. The 2×2 upper-left block has eigenvalues solving (2-λ)² - 1 = 0, giving λ = 1,3. The third eigenvalue is 5. Since A is not diagonal, the eigenvectors are not aligned with coordinate axes. The upper-left block requires rotation to diagonalize, so the quadratic form represents an ellipsoid with semi-axes proportional to 1/√1, 1/√3, 1/√5, but rotated from coordinate directions. Choice A incorrectly assumes coordinate alignment. Choice C incorrectly identifies eigenvalue 1 as negative. Choice D incorrectly assumes the block structure preserves coordinate alignment.
8. Let $A$ be a $4 \times 4$ symmetric matrix. The eigenspace for $\lambda_1 = 7$ is spanned by the vectors $\{\vec{u}_1, \vec{u}_2\}$, and the eigenspace for $\lambda_2 = -3$ is spanned by the vector $\{\vec{u}_3\}$. Which of the following dot products is guaranteed to be zero by the Spectral Theorem?
   Answer: $\vec{u}_1 \cdot \vec{u}_3$ The Spectral Theorem states that eigenvectors from different eigenspaces of a symmetric matrix are orthogonal. Since $\vec{u}_1$ is in the eigenspace for $\lambda_1 = 7$ and $\vec{u}_3$ is in the eigenspace for $\lambda_2 = -3$, and these eigenvalues are distinct, $\vec{u}_1$ must be orthogonal to $\vec{u}_3$. Their dot product must be zero. (A) is not guaranteed; $\vec{u}_1$ and $\vec{u}_2$ belong to the same eigenspace, and while an orthogonal basis for this space exists (which could be found using Gram-Schmidt), the original spanning vectors are not guaranteed to be orthogonal. (C) is also not guaranteed to be zero as it depends on $\vec{u}_1 \cdot \vec{u}_2$.
9. A real $3 \times 3$ matrix $A$ can be expressed as $A = PDP^T$, where $D$ is a diagonal matrix and $P$ is an orthogonal matrix. Which property must matrix $A$ possess?
   Answer: $A$ is symmetric. The decomposition $A = PDP^T$ with an orthogonal matrix $P$ is the definition of an orthogonally diagonalizable matrix. A key part of the Spectral Theorem is that a real matrix is orthogonally diagonalizable if and only if it is symmetric. We can prove this by taking the transpose: $A^T = (PDP^T)^T = (P^T)^T D^T P^T = PDP^T = A$. Since $A^T = A$, the matrix $A$ must be symmetric. (A) is incorrect because if $D$ contains a zero on its diagonal, $A$ will be singular. (C) is incorrect because the diagonal entries of $D$ (the eigenvalues) can be repeated. (D) is incorrect because rotation matrices are generally not symmetric.
10. Consider a $3 \times 3$ symmetric matrix $B$ with eigenvalues $2, 2, 5$. If $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ is an eigenvector corresponding to eigenvalue $5$, which of the following statements must be true about the eigenspace corresponding to eigenvalue $2$?
    Answer: The eigenspace corresponding to eigenvalue $2$ has dimension $2$ and must contain vectors orthogonal to $\mathbf{v}_1$. By the Spectral Theorem, symmetric matrices are orthogonally diagonalizable, meaning eigenvectors corresponding to distinct eigenvalues are orthogonal. Since 2 has algebraic multiplicity 2 and B is symmetric, the geometric multiplicity equals the algebraic multiplicity, so the eigenspace for eigenvalue 2 has dimension 2. All vectors in this eigenspace must be orthogonal to v₁ since v₁ corresponds to the distinct eigenvalue 5. Choice A has wrong dimension. Choice C incorrectly suggests non-orthogonal eigenvectors for distinct eigenvalues. Choice D incorrectly limits geometric multiplicity for symmetric matrices.
11. Let $F$ be a $4 \times 4$ symmetric matrix with eigenvalues $1, 2, 2, 5$. If $P = \mathbf{v}_1 \mathbf{v}_1^T + \mathbf{v}_4 \mathbf{v}_4^T$ where $\mathbf{v}_1$ and $\mathbf{v}_4$ are orthonormal eigenvectors for eigenvalues $1$ and $5$ respectively, what can be concluded about $PFP$?
    Answer: $PFP$ has eigenvalues $\{1, 0, 0, 5\}$ since $PFP = \mathbf{v}_1 \mathbf{v}_1^T + 5\mathbf{v}_4 \mathbf{v}_4^T$ when computed using spectral decomposition. Using the spectral decomposition F = 1v₁v₁^T + 2v₂v₂^T + 2v₃v₃^T + 5v₄v₄^T, and P = v₁v₁^T + v₄v₄^T, we compute PFP. Since P projects onto span{v₁,v₄}, and using the orthogonality of eigenvectors: PFP = P($1v₁v₁^T$ + $2v₂v₂^T$ + $2v₃v₃^T$ + $5v₄v₄^T$)P = 1Pv₁v₁^TP + 5Pv₄v₄^TP = 1v₁v₁^T + 5v₄v₄^T, since Pv₁ = v₁, Pv₄ = v₄, and Pv₂ = Pv₃ = 0. This gives eigenvalues {1,0,0,5}. Choice A has the right eigenvalues but wrong reasoning. Choice B incorrectly assumes preservation of all eigenvalues. Choice C incorrectly computes the eigenvalues.
12. For a symmetric matrix $A$, the spectral decomposition is given by $A = PDP^T$. Which statement provides the most complete and accurate description of the matrix $P$?
    Answer: The columns of $P$ form an orthonormal basis for $\mathbb{R}^n$ consisting of eigenvectors of $A$. For the decomposition $A = PDP^T$ to hold with $P^T$ as the inverse of $P$, the matrix $P$ must be orthogonal. An orthogonal matrix is a square matrix whose columns are not just orthogonal, but also have a norm (length) of 1. Thus, the columns form an orthonormal set. Since these columns are eigenvectors that must span the entire space, they form an orthonormal basis for $\mathbb{R}^n$. (A) is too weak; it describes any diagonalizable matrix. (B) is also too weak as it misses the unit vector (normalization) requirement. (C) is incorrect because the eigenvectors form a basis for the entire domain $\mathbb{R}^n$, not just the column space (unless the matrix is invertible).
13. Let $A$ be a symmetric matrix. Suppose $\vec{v}_1$ is an eigenvector corresponding to eigenvalue $\lambda_1 = 5$, and $\vec{v}_2$ is an eigenvector corresponding to eigenvalue $\lambda_2 = -2$. What is the value of the dot product $\vec{v}_1 \cdot \vec{v}_2$?
    Answer: 0 The Spectral Theorem guarantees that eigenvectors corresponding to distinct eigenvalues of a symmetric matrix are orthogonal. Since $\lambda_1 = 5$ and $\lambda_2 = -2$ are distinct eigenvalues, their corresponding eigenvectors $\vec{v}_1$ and $\vec{v}_2$ must be orthogonal. The dot product of two orthogonal vectors is 0. (C) is a distractor based on multiplying the eigenvalues. (D) is incorrect because this property holds for all symmetric matrices.
14. Let $A$ be a non-zero, real symmetric matrix such that $A^3 = A$. Which of the following sets contains all possible real eigenvalues of $A$?
    Answer: $\{0, 1, -1\}$ When you encounter a matrix equation like $A^3 = A$ combined with eigenvalue questions, think about what this condition tells you about the possible eigenvalues. Since eigenvalues must satisfy the same polynomial relationships as the matrix itself, any eigenvalue $\lambda$ of $A$ must satisfy $\lambda^3 = \lambda$. To find which values work, solve $\lambda^3 - \lambda = 0$. Factoring gives $\lambda(\lambda^2 - 1) = 0$, which factors further as $\lambda(\lambda - 1)(\lambda + 1) = 0$. Therefore, the only possible eigenvalues are $\lambda = 0, 1,$ or $-1$. Since $A$ is a real symmetric matrix, all its eigenvalues are real, and since $A$ is non-zero, it has at least one non-zero eigenvalue. But the constraint $A^3 = A$ limits us to exactly the three values we found. All three values are indeed possible: you can construct examples of real symmetric matrices with various combinations of these eigenvalues that satisfy $A^3 = A$. Choice B $\{1, -1\}$ is incorrect because it excludes 0, but matrices satisfying $A^3 = A$ can have 0 as an eigenvalue (think of a matrix that projects onto a subspace). Choice C $\{0, 1\}$ is wrong because $-1$ is also a valid eigenvalue. Choice D is incorrect because the polynomial constraint $\lambda^3 = \lambda$ severely restricts the possible values—most non-negative numbers don't satisfy this equation. Study tip: When you see matrix polynomial equations, always check what they imply about eigenvalues by substituting $\lambda$ for the matrix and solving the resulting polynomial equation.
15. The spectral decomposition of a symmetric matrix $A$ can be written as $A = \sum_{i=1}^n \lambda_i \vec{u}_i \vec{u}_i^T$, where $\{\vec{u}_1, \dots, \vec{u}_n\}$ is an orthonormal basis of eigenvectors and $\lambda_i$ are the corresponding real eigenvalues. What is the geometric interpretation of the matrix $P_i = \vec{u}_i \vec{u}_i^T$?
    Answer: $P_i$ is the matrix for the orthogonal projection onto the subspace spanned by the eigenvector $\vec{u}_i$. The matrix $P_i = \vec{u}_i \vec{u}_i^T$ is a rank-1 matrix that acts on any vector $\vec{x}$ as $P_i\vec{x} = (\vec{u}_i \vec{u}_i^T)\vec{x} = \vec{u}_i (\vec{u}_i^T \vec{x}) = (\vec{x} \cdot \vec{u}_i) \vec{u}_i$. This is the standard formula for the orthogonal projection of the vector $\vec{x}$ onto the line (1-dimensional subspace) spanned by the unit vector $\vec{u}_i$. The full decomposition shows how $A$ acts as a sum of scaled projections onto its orthogonal eigenspaces. (B) is imprecise; projection is the correct term. (A) and (D) describe different types of transformations with different matrix forms.
16. A $4 \times 4$ symmetric matrix $C$ has the property that $C^2 = 2C + 3I$. Using the Spectral Theorem, what can be concluded about the eigenvalues of $C$?
    Answer: All eigenvalues of $C$ must satisfy $\lambda^2 = 2\lambda + 3$, so the eigenvalues are $3$ and $-1$ with total multiplicity $4$. By the Spectral Theorem, C has an orthogonal diagonalization C = $PDP^T$ where D contains the eigenvalues. The equation C² = 2C + 3I becomes PD²$P^T$ = $2PDP^T$ + 3I = P(2D + 3I)$P^T$, which simplifies to D² = 2D + 3I. This means each eigenvalue λ satisfies λ² = 2λ + 3, giving λ² - 2λ - 3 = 0, so (λ - 3)(λ + 1) = 0. Therefore λ = 3 or λ = -1. Choice B ignores the constraint equation. Choice C incorrectly solves the quadratic. Choice D uses an irrelevant criterion about integer entries.
17. Which of the following conditions on a real $n \times n$ matrix $A$ is sufficient to guarantee that $A$ is orthogonally diagonalizable?
    Answer: $A$ is a symmetric matrix. The Spectral Theorem states that a real matrix is orthogonally diagonalizable if and only if it is symmetric. Therefore, the condition that $A$ is symmetric is sufficient. (A) is a condition that guarantees diagonalizability, but not necessarily orthogonal diagonalizability, as the eigenvectors for a non-symmetric matrix are not guaranteed to be orthogonal. (B) is not sufficient; many invertible, diagonalizable matrices are not symmetric. (D) is the condition for a matrix to be diagonalizable over the real numbers, which is a weaker condition than being orthogonally diagonalizable.
18. A real $n \times n$ matrix $A$ is symmetric. Which of the following is a direct and guaranteed consequence of the Spectral Theorem?
    Answer: There exists an orthonormal basis for $\mathbb{R}^n$ consisting entirely of eigenvectors of $A$. The Spectral Theorem states that a real symmetric matrix is orthogonally diagonalizable. This means that there is an orthonormal basis for the entire vector space $\mathbb{R}^n$ composed of the eigenvectors of the matrix. (A) is incorrect because a symmetric matrix can have repeated eigenvalues (e.g., the identity matrix). (B) is incorrect because the determinant is the product of all eigenvalues, counted with their algebraic multiplicity, not just the distinct ones. (D) is incorrect because eigenvectors corresponding to the same eigenvalue are not necessarily orthogonal, although an orthogonal basis for that eigenspace can be constructed.
19. A $2 \times 2$ symmetric matrix $M$ has eigenvalues $\alpha$ and $\beta$ with $\alpha > \beta > 0$. If $\mathbf{p}$ is a unit vector making angle $\theta$ with the eigenvector corresponding to $\alpha$, which expression gives $\mathbf{p}^T M \mathbf{p}$ in terms of $\alpha, \beta,$ and $\theta$?
    Answer: $\mathbf{p}^T M \mathbf{p} = \alpha \cos^2 \theta + \beta \sin^2 \theta$ since eigenvalues weight the squared projections onto eigenvectors. By the Spectral Theorem, M can be written as M = αu₁u₁^T + βu₂u₂^T where u₁, u₂ are orthonormal eigenvectors. If p makes angle θ with u₁, then p = (cos θ)u₁ + (sin θ)u₂. Therefore $p^T$ M p = $p^T$($αu₁u₁^T$ + $βu₂u₂^T$)p = α($p^T$ u₁)² + β($p^T$ u₂)² = α cos² θ + β sin² θ. Choice B incorrectly takes a square root. Choice C incorrectly factors out eigenvalues. Choice D uses linear rather than quadratic weighting.

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