Diagonalization - Linear Algebra

1. Let $A$ be a $4 \times 4$ matrix with characteristic polynomial $p(\lambda) = (\lambda - 1)^2(\lambda + 2)^2$. If the nullspace of $(A - I)$ has dimension 1 and the nullspace of $(A + 2I)$ has dimension 2, what can be concluded about $A$?
   Answer: $A$ is not diagonalizable because the geometric multiplicity of eigenvalue 1 is less than its algebraic multiplicity For diagonalizability, geometric multiplicity must equal algebraic multiplicity for each eigenvalue. Here, eigenvalue $\lambda = 1$ has algebraic multiplicity 2 but geometric multiplicity 1 (dimension of nullspace of $A - I$). Even though eigenvalue $\lambda = -2$ satisfies the condition (both multiplicities are 2), the failure for $\lambda = 1$ means $A$ is not diagonalizable.
2. For what value of $k$ is the matrix $A = \begin{pmatrix} 5 & k \\ 0 & 5 \end{pmatrix}$ not diagonalizable?
   Answer: For any non-zero value of $k$. The matrix $A$ is upper triangular, so its eigenvalues are the diagonal entries. The only eigenvalue is $\lambda=5$ with an algebraic multiplicity of 2. For $A$ to be diagonalizable, the geometric multiplicity of $\lambda=5$ must also be 2. The geometric multiplicity is the dimension of the null space of $A-5I$. $A - 5I = \begin{pmatrix} 0 & k \\ 0 & 0 \end{pmatrix}$ If $k=0$, this is the zero matrix, which has a null space of dimension 2. Thus, for $k=0$, $A$ is diagonalizable. If $k \neq 0$, the equation $(A-5I)\vec{x} = \vec{0}$ becomes $kx_2=0$, which implies $x_2=0$. The eigenvectors are of the form $\begin{pmatrix} x_1 \\ 0 \end{pmatrix}$, spanning a one-dimensional space. Since the geometric multiplicity (1) is less than the algebraic multiplicity (2), the matrix is not diagonalizable for any $k \neq 0$.
3. Consider the matrix $C = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 8 & -12 & 6 \end{pmatrix}$. If the characteristic polynomial of $C$ is $-\lambda^3 + 6\lambda^2 - 12\lambda + 8 = -(\lambda - 2)^3$, which approach would definitively determine whether $C$ is diagonalizable?
   Answer: Check whether $C$ has three linearly independent eigenvectors by computing the rank of $(C - 2I)$ and using the rank-nullity theorem Since $C$ has only one eigenvalue $\lambda = 2$ with algebraic multiplicity 3, $C$ is diagonalizable if and only if the geometric multiplicity of $\lambda = 2$ is also 3. This occurs when $\dim(\text{null}(C - 2I)) = 3$, which happens if and only if $C - 2I = 0$ (i.e., $\text{rank}(C - 2I) = 0$). Computing the rank of $(C - 2I)$ directly determines diagonalizability.
4. The matrix $A = \begin{pmatrix} a & 1 \\ 0 & a \end{pmatrix}$ represents a linear transformation. For which value(s) of the parameter $a$ is this matrix diagonalizable?
   Answer: $A$ is not diagonalizable for any value of $a$ because the off-diagonal entry creates dependency between eigenspaces When you encounter questions about matrix diagonalizability, the key concept is that a matrix is diagonalizable if and only if it has enough linearly independent eigenvectors to form a complete basis. For an $n \times n$ matrix, you need exactly $n$ linearly independent eigenvectors. Let's analyze this upper triangular matrix. The eigenvalues are indeed visible on the diagonal: both eigenvalues equal $a$ (with algebraic multiplicity 2). To find eigenvectors, we solve $(A - aI)\mathbf{v} = \mathbf{0}$: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ This gives us $y = 0$, so all eigenvectors have the form $\begin{pmatrix} x \\ 0 \end{pmatrix}$. The eigenspace is one-dimensional, spanned by $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Since we need 2 linearly independent eigenvectors but only have 1, the matrix is never diagonalizable regardless of $a$'s value. Answer A incorrectly assumes that being upper triangular with visible eigenvalues guarantees diagonalizability—it doesn't. Answer B wrongly focuses on singularity; diagonalizability depends on eigenspace dimensions, not rank. Answer C misses the point entirely by discussing real versus complex values when the fundamental issue is insufficient eigenvectors. Study tip: Remember that geometric multiplicity (dimension of eigenspace) must equal algebraic multiplicity (multiplicity of eigenvalue) for each eigenvalue. When you see a matrix with repeated eigenvalues, always check if there are enough linearly independent eigenvectors—don't assume diagonalizability from eigenvalues alone.
5. Let $A$ be an $n \times n$ matrix. Which condition guarantees that $A$ is diagonalizable?
   Answer: $A$ is a symmetric matrix. When you encounter questions about matrix diagonalizability, you're being tested on which conditions guarantee that a matrix can be written as $A = PDP^{-1}$ where $D$ is diagonal and $P$ is invertible. A symmetric matrix always guarantees diagonalizability because of the Spectral Theorem. When $A = A^T$, the matrix not only has real eigenvalues, but also has orthogonal eigenvectors. This means you can always find a complete set of linearly independent eigenvectors, which is exactly what you need for diagonalization. The orthogonal eigenvectors form the columns of $P$, making the diagonalization process work perfectly. Let's examine why the other options fail. Option A is incorrect because upper triangular matrices can have repeated eigenvalues on the diagonal without sufficient eigenvectors. For example, $\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ has eigenvalue 2 with multiplicity 2, but only one linearly independent eigenvector. Option B is wrong because invertibility only guarantees that zero isn't an eigenvalue—it says nothing about having enough eigenvectors. A matrix can be invertible but still lack sufficient eigenvectors for diagonalization. Option C is tempting but insufficient because having real eigenvalues doesn't guarantee you have enough linearly independent eigenvectors. The geometric multiplicity might be less than the algebraic multiplicity. Remember this key distinction: diagonalizability depends on having enough eigenvectors, not just on the nature of the eigenvalues themselves. Symmetry is the strongest condition here because it guarantees both real eigenvalues and orthogonal eigenvectors.
6. Let $A$ be an $n \times n$ matrix that is diagonalizable with eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_k$ (not necessarily distinct). If $A = PDP^{-1}$ where $D$ is diagonal, what condition must be satisfied for the matrix $A^2 + 3A - 4I$ to be diagonalizable?
   Answer: No additional conditions are needed beyond $A$ being diagonalizable, since $A^2 + 3A - 4I = P(D^2 + 3D - 4I)P^{-1}$ If $A = PDP^{-1}$, then $A^2 + 3A - 4I = PD^2P^{-1} + 3PDP^{-1} - 4I = P(D^2 + 3D - 4I)P^{-1}$. Since $D^2 + 3D - 4I$ is diagonal (polynomials of diagonal matrices are diagonal), and $P$ is the same invertible matrix, $A^2 + 3A - 4I$ is automatically diagonalizable with the same eigenvectors as $A$. The eigenvalues are $\lambda_i^2 + 3\lambda_i - 4$.
7. The characteristic polynomial of a $3 \times 3$ matrix $A$ is given by $p(\lambda) = (4-\lambda)(\lambda-1)^2$. Which of the following statements about $A$ is not necessarily true?
   Answer: $A$ is diagonalizable. From the characteristic polynomial, the eigenvalues are $\lambda=4$ (algebraic multiplicity 1) and $\lambda=1$ (algebraic multiplicity 2). A matrix is invertible if and only if 0 is not an eigenvalue, which is true here, so $A$ is invertible. The statement about algebraic multiplicity is true by definition. However, for $A$ to be diagonalizable, the geometric multiplicity of $\lambda=1$ must also be 2. Without more information about the matrix $A$, we cannot guarantee that the dimension of the eigenspace for $\lambda=1$ is 2. Therefore, $A$ is not necessarily diagonalizable.
8. Let $A$ be a $4 \times 4$ matrix with minimal polynomial $m(\lambda) = (\lambda - 2)^2(\lambda + 1)$. What can be concluded about the diagonalizability of $A$?
   Answer: $A$ is not diagonalizable because the minimal polynomial contains $(\lambda - 2)^2$ as a factor with exponent greater than 1 When you encounter questions about diagonalizability and minimal polynomials, remember that the minimal polynomial provides crucial information about whether a matrix can be diagonalized. The key theorem is: a matrix is diagonalizable if and only if its minimal polynomial has no repeated roots (i.e., all roots have multiplicity 1). Looking at the given minimal polynomial $m(\lambda) = (\lambda - 2)^2(\lambda + 1)$, we can see that $\lambda = 2$ appears with multiplicity 2, while $\lambda = -1$ appears with multiplicity 1. Since the root $\lambda = 2$ has multiplicity greater than 1 in the minimal polynomial, matrix $A$ cannot be diagonalizable. Let's examine why the other options are incorrect. Option A misunderstands the relationship between minimal polynomial degree and matrix size – the degree being less than the matrix size doesn't guarantee diagonalizability. Option B incorrectly focuses on whether eigenvalues are real versus complex, but diagonalizability depends on multiplicities in the minimal polynomial, not the nature of the roots. Option C makes a similar error, confusing the concept of a polynomial "splitting completely" (having all linear factors) with the condition for diagonalizability. Option D correctly identifies that the factor $(\lambda - 2)^2$ with exponent greater than 1 prevents diagonalization. Study tip: Always check the exponents in the minimal polynomial's factorization. If any factor has an exponent greater than 1, the matrix is not diagonalizable. This is the most reliable test for diagonalizability when given the minimal polynomial.
9. Which of the following $2 \times 2$ matrices is not diagonalizable over the real numbers?
   Answer: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ When you encounter a matrix diagonalizability question, you need to check whether the matrix has enough linearly independent eigenvectors to form a complete basis. A matrix is diagonalizable if and only if it has a full set of linearly independent eigenvectors. For 2×2 matrices, there are two main cases where diagonalization fails: when there are complex eigenvalues (over the reals), or when there's a repeated eigenvalue with insufficient eigenvectors. Let's examine option D: $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. The characteristic polynomial is $\det(A - \lambda I) = \lambda^2 + 1 = 0$, giving eigenvalues $\lambda = \pm i$. Since these are complex numbers, this matrix cannot be diagonalized over the real numbers. Now for the incorrect options: Option A: $\begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$ is symmetric, and all symmetric matrices are diagonalizable over the reals since they have real eigenvalues and orthogonal eigenvectors. Option B: $\begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}$ has characteristic polynomial $(\lambda - 1)^2 = 0$, so $\lambda = 1$ is a repeated eigenvalue. However, checking the eigenspace dimension shows it has two linearly independent eigenvectors, making it diagonalizable. Option C: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ has eigenvalues 0 and 2, which are real and distinct, guaranteeing diagonalizability. Study tip: For 2×2 matrices, quickly check if the eigenvalues are complex by computing the discriminant of the characteristic polynomial. Complex eigenvalues immediately rule out real diagonalizability.
10. A $4 \times 4$ matrix $A$ is known to have eigenvalues $\lambda_1 = 3$, $\lambda_2 = 3$, $\lambda_3 = -2$, and $\lambda_4 = 1$. Which additional piece of information is sufficient to guarantee that $A$ is diagonalizable?
    Answer: The rank of the matrix $A - 3I$ is 2. For a matrix to be diagonalizable, the geometric multiplicity of each eigenvalue must equal its algebraic multiplicity. The eigenvalue $\lambda = 3$ has an algebraic multiplicity of 2. The geometric multiplicity of $\lambda = 3$ is the dimension of its eigenspace, which is given by $\mathrm{dim}(\mathrm{Nul}(A - 3I))$. By the Rank-Nullity Theorem, for a $4 \times 4$ matrix, $\mathrm{rank}(A - 3I) + \mathrm{dim}(\mathrm{Nul}(A - 3I)) = 4$. If the rank is 2, then the nullity (geometric multiplicity) is $4 - 2 = 2$. Since this matches the algebraic multiplicity, and all other eigenvalues are distinct (guaranteeing their geometric multiplicity is 1), the matrix is diagonalizable.
11. Let $A$ be a $3 \times 3$ matrix with eigenvalues $\lambda=0, 2, 3$. Which of the following matrices is guaranteed to be similar to $A$?
    Answer: $\begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ Since the $3 \times 3$ matrix $A$ has three distinct eigenvalues (0, 2, and 3), it is guaranteed to be diagonalizable. This means $A$ is similar to a diagonal matrix $D$ whose diagonal entries are the eigenvalues of $A$. Two matrices are similar if one can be obtained from the other by a similarity transformation ($B = P^{-1}AP$). Therefore, $A$ must be similar to the diagonal matrix formed by its eigenvalues. The matrix in choice B is the correct diagonal matrix. The order of the eigenvalues on the diagonal can vary, but any such diagonal matrix is similar to any other permutation.
12. A student claims that the matrix $B = \begin{pmatrix} 3 & -1 \\ 4 & -1 \end{pmatrix}$ is diagonalizable and provides the factorization $B = PDP^{-1}$ where $P = \begin{pmatrix} 1 & 1 \\ 1 & 4 \end{pmatrix}$ and $D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. What is wrong with this claim?
    Answer: The factorization is incorrect because $PDP^{-1} \neq B$ when computed with the given matrices The eigenvalues of $B$ are found from $\det(B - \lambda I) = (3-\lambda)(-1-\lambda) - (-1)(4) = \lambda^2 - 2\lambda + 1 = (\lambda-1)^2 = 0$, giving $\lambda = 1$ with algebraic multiplicity 2. However, $\dim(\text{null}(B-I)) = 1$ since $B-I = \begin{pmatrix} 2 & -1 \\ 4 & -2 \end{pmatrix}$ has rank 1. While the student's $D$ has the correct eigenvalue, the matrix $B$ is actually not diagonalizable, so any claimed diagonalization must fail when verified by multiplication.
13. Consider the matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}$. Which statement about the diagonalizability of $A$ is correct?
    Answer: $A$ is not diagonalizable because the eigenvalue $\lambda = 2$ has algebraic multiplicity 2 but geometric multiplicity 1 The characteristic polynomial is $(\lambda - 2)^2(\lambda - 3) = 0$, giving eigenvalues $\lambda = 2$ (algebraic multiplicity 2) and $\lambda = 3$ (algebraic multiplicity 1). For $\lambda = 2$, we solve $(A - 2I)\mathbf{v} = \mathbf{0}$, which gives the eigenspace spanned by $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ only, so geometric multiplicity is 1. Since algebraic multiplicity (2) ≠ geometric multiplicity (1), $A$ is not diagonalizable.
14. A linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ has matrix representation $A$ with respect to the standard basis. If $A$ has eigenvalues $\lambda_1 = 1$, $\lambda_2 = 1$, $\lambda_3 = 3$ and corresponding eigenvectors $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$, $\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$, $\mathbf{v}_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$, what is the matrix representation of $T$ with respect to the basis $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$?
    Answer: $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ because the transformation acts as scalar multiplication on each basis vector When a linear transformation is represented with respect to a basis of eigenvectors, the matrix representation is diagonal with eigenvalues on the diagonal. Since $T(\mathbf{v}_1) = 1 \cdot \mathbf{v}_1$, $T(\mathbf{v}_2) = 1 \cdot \mathbf{v}_2$, and $T(\mathbf{v}_3) = 3 \cdot \mathbf{v}_3$, the matrix representation is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix}$. The fact that we have three linearly independent eigenvectors confirms that $A$ is diagonalizable.
15. Let $A = \begin{pmatrix} 1 & 0 & -2 \\ 0 & 5 & 0 \\ -2 & 0 & 4 \end{pmatrix}$. Given that $A$ is diagonalizable, what must be true about the matrix $P$ in the diagonalization $A = PDP^{-1}$?
    Answer: The columns of $P$ are mutually orthogonal. The matrix $A$ is a real symmetric matrix. A key property of symmetric matrices (part of the Spectral Theorem) is that eigenvectors corresponding to distinct eigenvalues are orthogonal. Since this matrix is symmetric, it has an orthogonal basis of eigenvectors. Therefore, the columns of the matrix $P$, which are the eigenvectors, can be chosen to be mutually orthogonal. If they are also normalized to unit length, $P$ becomes an orthogonal matrix.
16. A matrix $A$ is diagonalized by $A = PDP^{-1}$, where $P = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}$ and $D = \begin{pmatrix} 5 & 0 \\ 0 & -1 \end{pmatrix}$. Which of the following is an eigenvector of $A$ with a corresponding eigenvalue of $-1$?
    Answer: $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ In the diagonalization $A = PDP^{-1}$, the columns of the matrix $P$ are the eigenvectors of $A$, and the diagonal entries of $D$ are the corresponding eigenvalues. The order must be consistent. The first column of $P$, $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$, is the eigenvector corresponding to the first diagonal entry of $D$, which is 5. The second column of $P$, $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$, is the eigenvector corresponding to the second diagonal entry of $D$, which is $-1$.
17. Suppose a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is represented by a matrix $A$ that is diagonalizable. If the eigenvalues of $A$ are $\lambda_1 = 1$ and $\lambda_2 = 0$, what is the geometric interpretation of the transformation $T$?
    Answer: A projection onto a line. Since $A$ is diagonalizable, there is a basis of eigenvectors. Let $\vec{v}_1$ be the eigenvector for $\lambda_1=1$ and $\vec{v}_2$ be the eigenvector for $\lambda_2=0$. For any vector $\vec{x}$ in the direction of $\vec{v}_1$, $T(\vec{x}) = 1 \cdot \vec{x} = \vec{x}$. For any vector $\vec{y}$ in the direction of $\vec{v}_2$, $T(\vec{y}) = 0 \cdot \vec{y} = \vec{0}$. Any vector in $\mathbb{R}^2$ can be written as a linear combination of $\vec{v}_1$ and $\vec{v}_2$. The transformation leaves the component along $\vec{v}_1$ unchanged and annihilates the component along $\vec{v}_2$. This is the definition of a projection onto the line spanned by $\vec{v}_1$.
18. If a matrix $A$ is diagonalized as $A=PDP^{-1}$, what is the relationship between the determinant of $A$ and the trace of $A$?
    Answer: $\det(A)$ is the product of the diagonal entries of $D$, and $\mathrm{tr}(A)$ is the sum of the diagonal entries of $D$. The determinant and trace are similarity invariants. That is, if $A$ and $D$ are similar matrices ($A=PDP^{-1}$), then $\det(A) = \det(D)$ and $\mathrm{tr}(A) = \mathrm{tr}(D)$. The matrix $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $A$. The determinant of a diagonal matrix is the product of its diagonal entries, and its trace is the sum of its diagonal entries. Therefore, $\det(A)$ is the product of the eigenvalues, and $\mathrm{tr}(A)$ is the sum of the eigenvalues.
19. Consider two $3 \times 3$ matrices $A$ and $B$ where $A$ is diagonalizable and $AB = BA$. A student concludes that $B$ must also be diagonalizable. Under what condition is this conclusion correct?
    Answer: The conclusion is correct if and only if $A$ has three distinct eigenvalues, ensuring that $B$ is forced into diagonal form If $A$ has three distinct eigenvalues, then $A$ has three linearly independent eigenvectors that form a basis for $\mathbb{R}^3$. Since $AB = BA$, matrix $B$ must map each eigenspace of $A$ to itself. When $A$ has distinct eigenvalues, each eigenspace is one-dimensional, so $B$ acts as scalar multiplication on each eigenspace. Therefore, $B$ shares the same eigenvectors as $A$ and is diagonalizable. However, if $A$ has repeated eigenvalues, $B$ might not be diagonalizable even though it commutes with $A$.
20. Let $A$ be a diagonalizable matrix such that $A = PDP^{-1}$. Which expression is equivalent to $A^n$ for any positive integer $n$?
    Answer: $P D^n P^{-1}$ To find $A^n$, we can write out the product: $A^n = (PDP^{-1})(PDP^{-1})\cdots(PDP^{-1})$. The adjacent $P^{-1}P$ terms cancel out to the identity matrix $I$. This results in $A^n = P D (I) D (I) \cdots D P^{-1} = P D^n P^{-1}$.

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