Eigenvalues Practice

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Difficulty

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1.1Check an Eigenvector

Exam I | Problem 1.1 |

Let

A = \begin{bmatrix} 2 & 0 \\ 0 & 5 \end{bmatrix}

and

v = \begin{bmatrix} 3 \\ 0 \end{bmatrix}.

Is $v$ an eigenvector of $A$? If so, what is the eigenvalue?

Solution

Compute

Av = \begin{bmatrix} 2 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 6 \\ 0 \end{bmatrix}.

Since

\begin{bmatrix} 6 \\ 0 \end{bmatrix} = 2 \begin{bmatrix} 3 \\ 0 \end{bmatrix},

$v$ is an eigenvector and the eigenvalue is $2$.

1.2Read Eigenvalues from a Triangular Matrix

Exam I | Problem 1.2 |

Find the eigenvalues of

A = \begin{bmatrix} 4 & 1 & 0 \\ 0 & -2 & 7 \\ 0 & 0 & 5 \end{bmatrix}.

Solution

For a triangular matrix, the eigenvalues are the diagonal entries.

So the eigenvalues are

4,\quad -2,\quad 5.

1.3Solve a Characteristic Equation

Exam I | Problem 1.3 |

Find the eigenvalues of

A = \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix}.

Solution

Form the characteristic equation:

\det(A - \lambda I) = \begin{vmatrix} 3-\lambda & 1 \\ 2 & 2-\lambda \end{vmatrix}

=(3-\lambda)(2-\lambda) - 2

= \lambda^2 - 5\lambda + 4.

Set this equal to $0$:

\lambda^2 - 5\lambda + 4 = 0

Factor:

(\lambda - 1)(\lambda - 4) = 0.

So the eigenvalues are

\lambda = 1,\quad 4.

1.4Find an Eigenspace

Exam I | Problem 1.4 |

For

A = \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix},

find a basis for the eigenspace corresponding to $\lambda = 4$.

Solution

Compute

A - 4I = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}.

Solve

\begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.

This gives

$$ -x + y = 0, $$

so $y = x$.

A basis for the eigenspace is

\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}.

1.5Use Trace to Find the Missing Eigenvalue

Exam I | Problem 1.5 |

A $2 \times 2$ matrix has eigenvalues $6$ and $k$. Its trace is $11$.

What is $k$?

Solution

The trace equals the sum of the eigenvalues:

$$ 6 + k = 11. $$

$$ k = 5. $$

1.6Decide Whether a Matrix Is Invertible

Exam I | Problem 1.6 |

A matrix has eigenvalues $3$, $0$, and $-2$.

Is the matrix invertible?

Solution

A matrix is invertible if and only if $0$ is not an eigenvalue.

Since $0$ is one of the eigenvalues here, the matrix is not invertible.

1.7Identify Algebraic Multiplicity

Exam I | Problem 1.7 |

The characteristic polynomial of a matrix is

(\lambda - 2)^3(\lambda + 1).

What is the algebraic multiplicity of $\lambda = 2$?

Solution

The algebraic multiplicity is the multiplicity of the root in the characteristic polynomial.

Since $(\lambda - 2)$ appears three times, the algebraic multiplicity of $2$ is $3$.

1.8Use a Projection Matrix

Exam I | Problem 1.8 |

What are the only possible eigenvalues of a projection matrix $P$ satisfying

$$ P^2 = P? $$

Solution

If $Pv = \lambda v$ for an eigenvector $v \ne 0$, then

P^2v = \lambda^2 v.

But $P^2 = P$, so also

P^2v = Pv = \lambda v.

Thus

\lambda^2 = \lambda,

\lambda(\lambda - 1) = 0.

Therefore the only possible eigenvalues are

0 \quad \text{and} \quad 1.

1.9Find the Eigenvalues of a Rotation Matrix

Exam I | Problem 1.9 |

Find the eigenvalues of

A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.

Solution

Compute the characteristic equation:

\det(A - \lambda I) = \begin{vmatrix} -\lambda & -1 \\ 1 & -\lambda \end{vmatrix}

= \lambda^2 + 1.

\lambda^2 + 1 = 0.

Therefore the eigenvalues are

\lambda = i,\quad -i.

1.10Compute Eigenvalues of a 2x2 Matrix

Exam I | Problem 1.10 |

Find the eigenvalues of

A = \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix}.

Solution

For a $2 \times 2$ matrix, use the characteristic polynomial:

\det(A - \lambda I) = \begin{vmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{vmatrix}

=(4-\lambda)(3-\lambda) - 2

= \lambda^2 - 7\lambda + 10.

Factor:

(\lambda - 5)(\lambda - 2) = 0.

So the eigenvalues are

5,\quad 2.

Difficulty

Integrated

2.1Find Eigenvectors for Both Eigenvalues

Exam II | Problem 2.1 |

For

A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix},

find a basis for each eigenspace.

Solution

From the characteristic equation, the eigenvalues are $5$ and $2$.

For $\lambda = 5$,

A - 5I = \begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}.

This gives $y = x$, so a basis is

\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\}.

For $\lambda = 2$,

A - 2I = \begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}.

This gives $2x + y = 0$, so a basis is

\left\{ \begin{bmatrix} 1 \\ -2 \end{bmatrix} \right\}.

2.2Compare Multiplicities

Exam II | Problem 2.2 |

For

A = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix},

find the algebraic multiplicity and geometric multiplicity of the eigenvalue $3$.

Solution

The characteristic polynomial is

\det(A - \lambda I) = (3-\lambda)^2.

So the algebraic multiplicity of $3$ is $2$.

Now compute the eigenspace:

A - 3I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.

Solving

\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

gives $y = 0$.

So the eigenspace is

\operatorname{span} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right\},

which has dimension $1$.

Therefore the geometric multiplicity is $1$.

2.3Decide Whether a Matrix Is Diagonalizable

Exam II | Problem 2.3 |

A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{bmatrix}

diagonalizable?

Solution

The matrix is upper triangular, so its eigenvalues are the diagonal entries:

2,\quad 2,\quad 5.

The eigenvalue $2$ has algebraic multiplicity $2$.

Now compute

A - 2I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \end{bmatrix}.

From $(A-2I)v=0$, we get $y=0$ and $z=0$, with $x$ free. So the eigenspace for $\lambda=2$ has dimension $1$.

That is not enough independent eigenvectors. The matrix is not diagonalizable.

2.4Build P and D from Eigenpairs

Exam II | Problem 2.4 |

Suppose a matrix has the eigenpairs

\lambda_1 = 5,\quad v_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},

\lambda_2 = 1,\quad v_2 = \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix},

and

\lambda_3 = -2,\quad v_3 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}.

Write the matrices $P$ and $D$ for the diagonalization $A = P D P^{-1}$.

Solution

Put the eigenvectors into the columns of $P$ in the same order as the eigenvalues in $D$:

P = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{bmatrix}.

The diagonal matrix is

D = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix}.

2.5Use Trace and Determinant to Check Your Work

Exam II | Problem 2.5 |

You found that the eigenvalues of

A = \begin{bmatrix} 4 & 1 \\ 1 & 2 \end{bmatrix}

are $5$ and $1$.

Use the trace and determinant to check whether this is correct.

Solution

The trace of $A$ is

$$ 4 + 2 = 6. $$

The sum of the proposed eigenvalues is

$$ 5 + 1 = 6. $$

The determinant of $A$ is

4 \cdot 2 - 1 \cdot 1 = 7.

The product of the proposed eigenvalues is

5 \cdot 1 = 5.

Since the product does not match the determinant, the proposed eigenvalues are not correct.

2.6Use a Symmetric Matrix

Exam II | Problem 2.6 |

For

A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix},

find the eigenvalues and one eigenvector for each. Then check whether the two eigenvectors are orthogonal.

Solution

Compute the characteristic polynomial:

\det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix}

=(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3.

Factor:

(\lambda - 1)(\lambda - 3) = 0.

So the eigenvalues are $1$ and $3$.

For $\lambda = 3$,

A - 3I = \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix},

so $y = x$ and one eigenvector is

\begin{bmatrix} 1 \\ 1 \end{bmatrix}.

For $\lambda = 1$,

A - I = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix},

so $x + y = 0$ and one eigenvector is

\begin{bmatrix} 1 \\ -1 \end{bmatrix}.

Their dot product is

$$ (1)(1) + (1)(-1) = 0, $$

so the eigenvectors are orthogonal.

2.7Use a Nilpotent Matrix

Exam II | Problem 2.7 |

Suppose a matrix satisfies

$$ A^3 = 0. $$

What can you conclude about its eigenvalues and determinant?

Solution

If $Av = \lambda v$ for an eigenvector $v \ne 0$, then

A^3v = \lambda^3 v.

But $A^3 = 0$, so $A^3v = 0$. Since $v \ne 0$, this forces

\lambda^3 = 0,

so $\lambda = 0$.

Therefore every eigenvalue is $0$.

The determinant is the product of the eigenvalues, so

\det(A) = 0.

2.8Find Complex Eigenvalues from a Real Matrix

Exam II | Problem 2.8 |

Find the eigenvalues of

A = \begin{bmatrix} 2 & -5 \\ 1 & 2 \end{bmatrix}.

Solution

Compute the characteristic polynomial:

\det(A - \lambda I) = \begin{vmatrix} 2-\lambda & -5 \\ 1 & 2-\lambda \end{vmatrix}

=(2-\lambda)^2 + 5 = \lambda^2 - 4\lambda + 9.

Set this equal to $0$:

\lambda^2 - 4\lambda + 9 = 0.

Use the quadratic formula:

\lambda = \frac{4 \pm \sqrt{16 - 36}}{2} = \frac{4 \pm \sqrt{-20}}{2} = 2 \pm i\sqrt{5}.

So the eigenvalues are

2 + i\sqrt{5}

and

2 - i\sqrt{5}.

Difficulty

Applied

3.1Predict Long-Term Behavior from Eigenvalues

Final | Problem 3.1 |

A discrete system is defined by

x_{k+1} = Ax_k.

Suppose the eigenvalues of $A$ are

\lambda_1 = \frac{1}{4} \quad \text{and} \quad \lambda_2 = \frac{3}{2}.

What happens to the two eigenmodes as $k$ gets large?

Solution

Each eigenmode is scaled by its eigenvalue at every step.

The mode with eigenvalue $\frac{1}{4}$ decays toward $0$ because

\left(\frac{1}{4}\right)^k \to 0.

The mode with eigenvalue $\frac{3}{2}$ grows without bound because

\left(\frac{3}{2}\right)^k \to \infty.

So the component in the $\frac{1}{4}$-direction dies out, while the component in the $\frac{3}{2}$-direction dominates.

3.2Analyze a Differential Equation

Final | Problem 3.2 |

Consider the system

$$ x'(t) = Ax(t) $$

where $A$ has eigenvalues $-2$ and $0$.

What do these eigenvalues say about the two modes of the solution?

Solution

For a linear system $x'(t) = Ax(t)$, an eigenvalue $\lambda$ produces a mode that behaves like $e^{\lambda t}$.

The mode for $\lambda = -2$ is

e^{-2t},

so it decays to $0$ as $t$ increases.

The mode for $\lambda = 0$ is

e^{0t} = 1,

so it stays constant in size.

Thus one mode decays and the other remains neutral.

3.3Find a Steady-State Eigenvector

Final | Problem 3.3 |

Let

P = \begin{bmatrix} 0.8 & 0.1 \\ 0.2 & 0.9 \end{bmatrix}.

Find a nonzero vector $v$ such that

$$ Pv = v. $$

Solution

The equation $Pv = v$ means $v$ is an eigenvector for eigenvalue $1$.

So solve

$$ (P - I)v = 0. $$

That gives

\begin{bmatrix} -0.2 & 0.1 \\ 0.2 & -0.1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.

From the first row,

$$ -0.2x + 0.1y = 0, $$

$$ y = 2x. $$

One nonzero fixed vector is

\begin{bmatrix} 1 \\ 2 \end{bmatrix}.

3.4Use Diagonalization to Track Repeated Action

Final | Problem 3.4 |

Suppose $v_1$ and $v_2$ are eigenvectors of $A$ with eigenvalues $3$ and $\frac{1}{3}$, respectively. Let

$$ x = 2v_1 - v_2. $$

Find a formula for $A^k x$.

Solution

Use the eigenvector rules:

$$ A^k v_1 = 3^k v_1 $$

and

A^k v_2 = \left(\frac{1}{3}\right)^k v_2.

Therefore

$$ A^k x = A^k(2v_1 - v_2) $$

$$ = 2A^k v_1 - A^k v_2 $$

= 2 \cdot 3^k v_1 - \left(\frac{1}{3}\right)^k v_2.

3.5Identify Principal Directions

Final | Problem 3.5 |

For the symmetric matrix

A = \begin{bmatrix} 4 & 1 \\ 1 & 4 \end{bmatrix},

which direction is stretched more, and by how much?

Solution

Find the eigenvalues:

\det(A - \lambda I) = \begin{vmatrix} 4-\lambda & 1 \\ 1 & 4-\lambda \end{vmatrix}

=(4-\lambda)^2 - 1 = \lambda^2 - 8\lambda + 15.

Factor:

(\lambda - 3)(\lambda - 5) = 0.

So the stretch factors are $5$ and $3$.

The larger stretch is in the direction of the eigenvector for $\lambda = 5$, which satisfies $y = x$. So the direction is

\begin{bmatrix} 1 \\ 1 \end{bmatrix}.

The smaller stretch is in the direction of the eigenvector for $\lambda = 3$, which satisfies $y = -x$. So the direction is

\begin{bmatrix} 1 \\ -1 \end{bmatrix}.

Difficulty

Challenge

4.1When Is a Triangular Matrix Diagonalizable?

Final | Problem 4.1 |

For

A = \begin{bmatrix} k & 1 \\ 0 & 2 \end{bmatrix},

for what values of $k$ is $A$ diagonalizable?

Solution

Because $A$ is triangular, the eigenvalues are

k \quad \text{and} \quad 2.

If $k \ne 2$, then the matrix has two distinct eigenvalues, so it is diagonalizable.

If $k = 2$, then

A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}.

This is the repeated-eigenvalue case from the notes. Its eigenspace has dimension $1$, so it is not diagonalizable.

Therefore $A$ is diagonalizable exactly when

k \ne 2.

4.2A Repeated Eigenvalue in Three Dimensions

Final | Problem 4.2 |

Consider

B = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}.

Find the eigenvalues, their algebraic multiplicities, and decide whether $B$ is diagonalizable.

Solution

Since $B$ is triangular, the eigenvalues are the diagonal entries:

1,\quad 1,\quad -1.

So the algebraic multiplicity of $1$ is $2$, and the algebraic multiplicity of $-1$ is $1$.

Now check the eigenspace for $\lambda = 1$:

B - I = \begin{bmatrix} 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -2 \end{bmatrix}.

Solving $(B-I)v = 0$ gives $y = 0$ and $z = 0$, with $x$ free. So the eigenspace for $1$ has dimension $1$.

That is fewer than its algebraic multiplicity, so $B$ does not have three independent eigenvectors.

Therefore $B$ is not diagonalizable.

4.3Use Orthogonality and the Characteristic Equation

Final | Problem 4.3 |

A real $2 \times 2$ orthogonal matrix has determinant $1$ and trace $0$.

What are its eigenvalues?

Solution

Let the eigenvalues be $\lambda_1$ and $\lambda_2$.

For a $2 \times 2$ matrix, the sum of the eigenvalues equals the trace and the product equals the determinant.

\lambda_1 + \lambda_2 = 0

and

\lambda_1\lambda_2 = 1.

This gives the characteristic equation

\lambda^2 + 1 = 0.

Thus the eigenvalues are

i \quad \text{and} \quad -i.

4.4Complex Eigenvalues from Trace and Determinant

Final | Problem 4.4 |

A real $2 \times 2$ matrix has trace $4$ and determinant $13$.

Find its eigenvalues.

Solution

For a $2 \times 2$ matrix, the characteristic polynomial is

\lambda^2 - (\operatorname{tr} A)\lambda + \det(A).

So here it is

\lambda^2 - 4\lambda + 13.

Set this equal to $0$:

\lambda^2 - 4\lambda + 13 = 0.

Use the quadratic formula:

\lambda = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i.

So the eigenvalues are

$$ 2 + 3i $$

and

$$ 2 - 3i. $$

Since neither eigenvalue is real, this matrix has no real eigenvectors associated with these eigenvalues.