Eigenvalues Practice

Compute

Since

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

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

So the eigenvalues are

Form the characteristic equation:

Set this equal to $0$:

Factor:

So the eigenvalues are

Compute

Solve

This gives

so $y = x$.

A basis for the eigenspace is

The trace equals the sum of the eigenvalues:

So

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.

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 $3$.

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

But $P^2 = P$, so also

Thus

so

Therefore the only possible eigenvalues are

Compute the characteristic equation:

So

Therefore the eigenvalues are

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

Factor:

So the eigenvalues are

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

For $\lambda = 5$,

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

For $\lambda = 2$,

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

The characteristic polynomial is

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

Now compute the eigenspace:

Solving

gives $y = 0$.

So the eigenspace is

which has dimension $1$.

Therefore the geometric multiplicity is $1$.

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

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

Now compute

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.

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

The diagonal matrix is

The trace of $A$ is

The sum of the proposed eigenvalues is

The determinant of $A$ is

The product of the proposed eigenvalues is

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

Compute the characteristic polynomial:

Factor:

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

For $\lambda = 3$,

so $y = x$ and one eigenvector is

For $\lambda = 1$,

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

Their dot product is

so the eigenvectors are orthogonal.

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

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

so $\lambda = 0$.

Therefore every eigenvalue is $0$.

The determinant is the product of the eigenvalues, so

Compute the characteristic polynomial:

Set this equal to $0$:

Use the quadratic formula:

So the eigenvalues are

and

Each eigenmode is scaled by its eigenvalue at every step.

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

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

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

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

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

The mode for $\lambda = 0$ is

so it stays constant in size.

Thus one mode decays and the other remains neutral.

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

So solve

That gives

From the first row,

so

One nonzero fixed vector is

Use the eigenvector rules:

and

Therefore

Find the eigenvalues:

Factor:

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

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

Because $A$ is triangular, the eigenvalues are

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

If $k = 2$, then

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

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

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

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

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.

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.

So

and

This gives the characteristic equation

Thus the eigenvalues are

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

So here it is

Set this equal to $0$:

Use the quadratic formula:

So the eigenvalues are

and

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