Linear Algebra Quiz: Diagonalization

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DiagonalizationQuestion 1 of 20

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?

The determinant of

A

-18

The eigenspace corresponding to

\lambda = -2

is one-dimensional.

The rank of the matrix

A - 3I

is 2.

The trace of

A

is 5.

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Linear Algebra Quiz

Linear Algebra Quiz: Diagonalization

Practice Diagonalization in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Diagonalization, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The determinant of $A$ is $-18$ .
The eigenspace corresponding to $\lambda = -2$ is one-dimensional.
The rank of the matrix $A - 3I$ is 2. (correct answer)
The trace of $A$ is 5.

Explanation: 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.

Question 2

Let $A$ be a diagonalizable matrix such that $A = PDP^{-1}$ . Which expression is equivalent to $A^n$ for any positive integer $n$ ?

$P^n D^n (P^{-1})^n$
$P D P^{-n}$
$P D^n P^{-1}$ (correct answer)
$n(PDP^{-1})$

Explanation: 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}

Question 3

Let $A$ be an $n \times n$ matrix. Which condition guarantees that $A$ is diagonalizable?

$A$ is an upper triangular matrix.
$A$ is an invertible matrix.
$A$ has only real eigenvalues.
$A$ is a symmetric matrix. (correct answer)

Explanation: 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}

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.

Question 4

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 $T$ ?

A rotation about the origin.
A projection onto a line. (correct answer)
A reflection across a line.
A shear transformation.

Explanation: 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

Question 5

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 $-1$ ?

$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
$\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 3 \\ -1 \end{pmatrix}$
$\begin{pmatrix} 5 \\ -1 \end{pmatrix}$

Explanation: 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

Question 6

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?

Check whether $C$ has three linearly independent eigenvectors by computing the rank of $(C - 2I)$ and using the rank-nullity theorem (correct answer)
Verify that $C$ is symmetric, since all symmetric matrices are guaranteed to be diagonalizable over the real numbers
Compute $C^3$ and check if it equals $8I$ , since diagonalizable matrices satisfy their characteristic polynomials
Find the minimal polynomial of $C$ and check whether it has the same degree as the characteristic polynomial

Explanation: 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.

Question 7

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?

The matrix $A$ must have distinct eigenvalues, ensuring that $A^2 + 3A - 4I$ inherits the same eigenvectors with transformed eigenvalues
The polynomial $p(\lambda) = \lambda^2 + 3\lambda - 4$ must not map any eigenvalue of $A$ to zero, preventing singularity issues
No additional conditions are needed beyond $A$ being diagonalizable, since $A^2 + 3A - 4I = P(D^2 + 3D - 4I)P^{-1}$ (correct answer)
The eigenvalues of $A$ must satisfy $\lambda_i^2 + 3\lambda_i - 4 \neq \lambda_j^2 + 3\lambda_j - 4$ for $i \neq j$ to ensure linear independence

Explanation: 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

Question 8

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?

$A$ is diagonalizable for all real values of $a$ since it is upper triangular with eigenvalues clearly visible on the diagonal
$A$ is diagonalizable for all values of $a$ except $a = 0$ , where the matrix becomes singular and loses full rank
$A$ is diagonalizable for all complex values of $a$ , but not necessarily for real values depending on the discriminant
$A$ is not diagonalizable for any value of $a$ because the off-diagonal entry creates dependency between eigenspaces (correct answer)

Explanation: 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.

Question 9

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?

The conclusion is always correct because commuting matrices share the same eigenvectors, and $A$ provides a complete basis of eigenvectors
The conclusion is correct if and only if $A$ has three distinct eigenvalues, ensuring that $B$ is forced into diagonal form (correct answer)
The conclusion is correct if and only if $\det(A) \neq 0$ and $\det(B) \neq 0$ , ensuring both matrices are invertible
The conclusion is never guaranteed to be correct based solely on the given information about commutativity and $A$ 's diagonalizability

Explanation: 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

Question 10

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 $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ ?

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{pmatrix} $$ because the transformation acts as scalar multiplication on each basis vector$ (correct answer)
$\begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} $$ because this represents the change of basis from standard coordinates to eigenvector coordinates$
$\begin{pmatrix} 3 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ because the largest eigenvalue appears first when eigenvectors are ordered by their corresponding eigenvalues$
The matrix cannot be determined without additional information about the original matrix $A$ in standard coordinates

Explanation: 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.

Question 11

A $3 \times 3$ matrix $M$ has eigenvalues $\lambda_1 = 2$ , $\lambda_2 = 2$ , and $\lambda_3 = 5$ . The eigenspace corresponding to $\lambda = 2$ is spanned by $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$ , and the eigenspace corresponding to $\lambda = 5$ is spanned by $\mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ . What must be true about $M$ ?

$M$ is not diagonalizable, and its Jordan canonical form contains exactly one $2 \times 2$ Jordan block
$M$ is diagonalizable if and only if there exists another eigenvector for $\lambda = 2$ that is orthogonal to $\mathbf{v}_1$
$M$ cannot be diagonalizable because the geometric multiplicity of $\lambda = 2$ is less than its algebraic multiplicity (correct answer)
$M$ is diagonalizable only if it is symmetric, since repeated eigenvalues require special orthogonality conditions

Explanation: The eigenvalue

\lambda = 2

has algebraic multiplicity 2 but geometric multiplicity 1 (since its eigenspace is one-dimensional, spanned only by

\mathbf{v}_1

). For diagonalizability, geometric multiplicity must equal algebraic multiplicity for every eigenvalue. Since this condition fails for

\lambda = 2

, matrix

M

cannot be diagonalizable.

Question 12

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 $A$ ?

$A$ is diagonalizable because the sum of geometric multiplicities equals 4, which matches the size of the matrix
$A$ is not diagonalizable because the geometric multiplicity of eigenvalue 1 is less than its algebraic multiplicity (correct answer)
$A$ is diagonalizable only if there exists an invertible matrix $P$ such that $A = PDP^{-1}$ where $D$ has repeated diagonal entries
$A$ is not diagonalizable because the characteristic polynomial has degree 4 but contains only 2 distinct roots

Explanation: 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.

Question 13

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}

The matrix $P$ is not invertible because its columns are linearly dependent, making the diagonalization impossible
The diagonal matrix $D$ has incorrect entries; the eigenvalues of $B$ are actually $\lambda = 2$ and $\lambda = 0$
The factorization is incorrect because $PDP^{-1} \neq B$ when computed with the given matrices (correct answer)
The matrix $B$ is not diagonalizable because it has complex eigenvalues, not real eigenvalues as assumed

Explanation: 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.

Question 14

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 $A = PDP^{-1}$ ?

The columns of $P$ are mutually orthogonal. (correct answer)
The columns of $P$ sum to the zero vector.
$P$ is an upper triangular matrix.
The determinant of $P$ is zero.

Explanation: 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.

Question 15

For what value of $k$ is the matrix $A = \begin{pmatrix} 5 & k \\ 0 & 5 \end{pmatrix}$ not diagonalizable?

Only for $k=0$ .
Only for $k=5$ .
For any non-zero value of $k$ . (correct answer)
For no value of $k$ .

Explanation: 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}

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

Question 16

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 $A$ ?

$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 3 \end{pmatrix}$
$\begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ (correct answer)
$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$
$\begin{pmatrix} 0 & 2 & 3 \\ 2 & 0 & 3 \\ 3 & 2 & 0 \end{pmatrix}$

Explanation: 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.

Question 17

If a matrix $A$ is diagonalized as $A=PDP^{-1}$ , what is the relationship between the determinant of $A$ and the trace of $A$ ?

$\det(A)$ is the product of the diagonal entries of $D$ , and $\mathrm{tr}(A)$ is the sum of the diagonal entries of $D$ . (correct answer)
$\det(A)$ is the sum of the diagonal entries of $D$ , and $\mathrm{tr}(A)$ is the product of the diagonal entries of $D$ .
$\det(A)$ is the product of the diagonal entries of $P$ , and $\mathrm{tr}(A)$ is the sum of the diagonal entries of $P$ .
$\det(A)$ and $\mathrm{tr}(A)$ are equal to $\det(D)$ and $\mathrm{tr}(D)$ respectively, but only if $P$ is an orthogonal matrix.

Explanation: 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.

Question 18

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?

$A$ has an eigenvalue of 4.
$A$ is invertible.
The algebraic multiplicity of $\lambda=1$ is 2.
$A$ is diagonalizable. (correct answer)

Explanation: 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.

Question 19

Which of the following $2 \times 2$ matrices is not diagonalizable over the real numbers?

$\begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$
$\begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}$
$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$
$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ (correct answer)

Explanation: 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}

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.

Question 20

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 $A$ ?

$A$ is diagonalizable because the minimal polynomial has degree 3, which is less than the matrix size 4
$A$ is diagonalizable because all roots of the minimal polynomial are real numbers with no complex eigenvalues
$A$ is diagonalizable because the minimal polynomial splits completely over the real numbers into linear and quadratic factors
$A$ is not diagonalizable because the minimal polynomial contains $(\lambda - 2)^2$ as a factor with exponent greater than 1 (correct answer)

Explanation: 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.