The Central Idea

Let \(A\) be a square matrix.

A nonzero vector \(\mathbf{v}\) is an eigenvector of \(A\) if

\[ \boxed{ A\mathbf{v} = \lambda\mathbf{v} }. \]

The scalar \(\lambda\) is the corresponding eigenvalue.

The equation says that applying \(A\) has the same effect on \(\mathbf{v}\) as multiplying that vector by one scalar.

Why Eigenvectors Are Special

A general matrix transformation may rotate, shear, scale, or otherwise redirect a vector.

An eigenvector stays on its original line through the origin.

Its magnitude and orientation along that line may change, but the line of direction remains invariant.

Eigenvalue as a Scale Factor

In

\[ A\mathbf{v} = \lambda\mathbf{v}, \]

the eigenvalue \(\lambda\) describes the spatial effect along the eigenvector direction.

Example: A Simple Eigenvector

Let

\[ A= \begin{bmatrix} 2&0\\ 0&3 \end{bmatrix} \]

and

\[ \mathbf{i} = \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]

Then

\[ A\mathbf{i} = \begin{bmatrix} 2\\ 0 \end{bmatrix} = 2\mathbf{i}. \]

Therefore, \(\mathbf{i}\) is an eigenvector with eigenvalue \(2\).

A Second Eigenvector of the Same Matrix

Using

\[ \mathbf{j} = \begin{bmatrix} 0\\ 1 \end{bmatrix}, \]

we obtain

\[ A\mathbf{j} = \begin{bmatrix} 0\\ 3 \end{bmatrix} = 3\mathbf{j}. \]

Thus \(\mathbf{j}\) is an eigenvector with eigenvalue \(3\).

The Geometry of a Diagonal Matrix

The matrix

\[ \begin{bmatrix} 2&0\\ 0&3 \end{bmatrix} \]

scales the \(x\)-axis direction by \(2\) and the \(y\)-axis direction by \(3\).

The coordinate axes are therefore invariant directions.

This is the easiest geometric setting in which to see eigenvectors.

Not Every Vector Is an Eigenvector

For the same matrix,

\[ A= \begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}, \]

consider

\[ \mathbf{v} = \begin{bmatrix} 1\\ 1 \end{bmatrix}. \]

Then

\[ A\mathbf{v} = \begin{bmatrix} 2\\ 3 \end{bmatrix}. \]

This is not a scalar multiple of \(\begin{bmatrix}1\\1\end{bmatrix}\), so \(\mathbf{v}\) is not an eigenvector.

The Zero Vector Is Not an Eigenvector

Every matrix satisfies

\[ A\mathbf{0} = \mathbf{0}. \]

If the zero vector were allowed, it would satisfy

\[ A\mathbf{0} = \lambda\mathbf{0} \]

for every scalar \(\lambda\).

That would provide no meaningful directional information.

Therefore, by definition,

\[ \boxed{ \text{an eigenvector must be nonzero} }. \]

Checking an Eigenvector Candidate

To test whether a nonzero vector \(\mathbf{v}\) is an eigenvector:

  1. Compute \(A\mathbf{v}\).
  2. Compare the result with \(\mathbf{v}\).
  3. Determine whether one scalar \(\lambda\) multiplies every component consistently.

If so,

\[ A\mathbf{v} = \lambda\mathbf{v}. \]

Example: Verify an Eigenvector

Let

\[ A= \begin{bmatrix} 3&1\\ 0&2 \end{bmatrix} \]

and

\[ \mathbf{v} = \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]

Then

\[ A\mathbf{v} = \begin{bmatrix} 3\\ 0 \end{bmatrix} = 3\mathbf{v}. \]

Therefore, \(\mathbf{v}\) is an eigenvector with eigenvalue \(3\).

Example: A Candidate That Fails

Using the same matrix, test

\[ \mathbf{w} = \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]

We obtain

\[ A\mathbf{w} = \begin{bmatrix} 1\\ 2 \end{bmatrix}. \]

This output is not a scalar multiple of

\[ \begin{bmatrix} 0\\ 1 \end{bmatrix}. \]

Therefore, \(\mathbf{w}\) is not an eigenvector.

Every Nonzero Multiple of an Eigenvector Is Also an Eigenvector

Suppose

\[ A\mathbf{v} = \lambda\mathbf{v}. \]

For any scalar \(c\neq0\),

\[ A(c\mathbf{v}) = cA\mathbf{v}. \]

Therefore,

\[ A(c\mathbf{v}) = c\lambda\mathbf{v} = \lambda(c\mathbf{v}). \]

So every nonzero vector on the same eigenvector line shares the same eigenvalue.

Eigendirections

Because all nonzero scalar multiples of an eigenvector are also eigenvectors, it is often more useful geometrically to think about an eigendirection.

An eigendirection is a line through the origin that the transformation maps back onto itself.

Eigenspaces

For a fixed eigenvalue \(\lambda\), the set of all corresponding eigenvectors together with the zero vector forms the eigenspace.

It is the solution set of

\[ \boxed{ (A-\lambda I)\mathbf{v} = \mathbf{0} }. \]

In the plane, an eigenspace is commonly a line through the origin.

From the Eigenvalue Equation to a Homogeneous System

Start from

\[ A\mathbf{v} = \lambda\mathbf{v}. \]

Since

\[ \lambda\mathbf{v} = \lambda I\mathbf{v}, \]

move everything to one side:

\[ A\mathbf{v}-\lambda I\mathbf{v} = \mathbf{0}. \]

Factor:

\[ \boxed{ (A-\lambda I)\mathbf{v} = \mathbf{0} }. \]

Why \(A-\lambda I\) Must Be Singular

An eigenvector must be nonzero.

Therefore,

\[ (A-\lambda I)\mathbf{v} = \mathbf{0} \]

must have a nonzero solution.

An invertible matrix would have only the zero solution.

Therefore,

\[ \boxed{ A-\lambda I \text{ must be noninvertible} }. \]

The Characteristic Equation

A square matrix is noninvertible when its determinant is zero.

Therefore, eigenvalues satisfy

\[ \boxed{ \det(A-\lambda I)=0 }. \]

This is called the characteristic equation.

The Characteristic Polynomial

The expression

\[ \det(A-\lambda I) \]

is a polynomial in \(\lambda\).

It is called the characteristic polynomial.

Its roots are the eigenvalues of the matrix.

Characteristic Equation for a \(2\times2\) Matrix

Let

\[ A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}. \]

Then

\[ A-\lambda I = \begin{bmatrix} a-\lambda&b\\ c&d-\lambda \end{bmatrix}. \]

Therefore,

\[ \det(A-\lambda I) = (a-\lambda)(d-\lambda)-bc. \]

Expanded \(2\times2\) Characteristic Polynomial

Expanding gives

\[ (a-\lambda)(d-\lambda)-bc = ad-a\lambda-d\lambda+\lambda^2-bc. \]

Rearranging,

\[ \boxed{ \lambda^2-(a+d)\lambda+(ad-bc) }. \]

Thus the characteristic equation is

\[ \boxed{ \lambda^2-(a+d)\lambda+(ad-bc)=0 }. \]

Trace and Determinant Appear Naturally

For

\[ A= \begin{bmatrix} a&b\\ c&d \end{bmatrix}, \]

the trace is

\[ \operatorname{tr}(A)=a+d \]

and the determinant is

\[ \det(A)=ad-bc. \]

Therefore,

\[ \boxed{ \lambda^2-\operatorname{tr}(A)\lambda+\det(A)=0 }. \]

Eigenvalue Sum and Product in Two Dimensions

If the eigenvalues are \(\lambda_1\) and \(\lambda_2\), then

\[ \boxed{ \lambda_1+\lambda_2 = \operatorname{tr}(A) } \]

and

\[ \boxed{ \lambda_1\lambda_2 = \det(A) }. \]

These relationships provide useful checks on calculations.

Example: Find Eigenvalues

Let

\[ A= \begin{bmatrix} 3&1\\ 0&2 \end{bmatrix}. \]

Then

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

Its determinant is

\[ (3-\lambda)(2-\lambda). \]

Complete the Eigenvalue Calculation

Set the determinant equal to zero:

\[ (3-\lambda)(2-\lambda)=0. \]

Therefore,

\[ \boxed{ \lambda=3 } \]

or

\[ \boxed{ \lambda=2 }. \]

Finding an Eigenvector after Finding an Eigenvalue

Once an eigenvalue \(\lambda\) is known, solve

\[ \boxed{ (A-\lambda I)\mathbf{v} = \mathbf{0} }. \]

Any nonzero solution is an eigenvector associated with \(\lambda\).

Eigenvector for \(\lambda=3\)

For

\[ A= \begin{bmatrix} 3&1\\ 0&2 \end{bmatrix}, \]

compute

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

Solve

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

Complete the \(\lambda=3\) Eigenspace

The equations require

\[ y=0. \]

The value of \(x\) is free.

Therefore, all vectors of the form

\[ \begin{bmatrix} x\\ 0 \end{bmatrix} \]

belong to the eigenspace.

A convenient eigenvector is

\[ \boxed{ \begin{bmatrix} 1\\ 0 \end{bmatrix} }. \]

Eigenvector for \(\lambda=2\)

Compute

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

The equation

\[ x+y=0 \]

gives

\[ y=-x. \]

A convenient eigenvector is

\[ \boxed{ \begin{bmatrix} 1\\ -1 \end{bmatrix} }. \]

Verify Both Eigenvectors

For

\[ \mathbf{v}_1= \begin{bmatrix} 1\\ 0 \end{bmatrix}, \]

we have

\[ A\mathbf{v}_1 = 3\mathbf{v}_1. \]

For

\[ \mathbf{v}_2= \begin{bmatrix} 1\\ -1 \end{bmatrix}, \]

we have

\[ A\mathbf{v}_2 = 2\mathbf{v}_2. \]

Both eigendirections are confirmed.

Eigenvalue \(1\): Fixed Directions

If

\[ A\mathbf{v} = \mathbf{v}, \]

then

\[ \lambda=1. \]

The vector is unchanged by the transformation.

Such vectors identify fixed directions and, in some transformations, fixed points along lines through the origin.

Example: Reflection and Eigenvalue \(1\)

Reflection across the \(x\)-axis has matrix

\[ F= \begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix}. \]

Every vector along the \(x\)-axis satisfies

\[ F \begin{bmatrix} x\\ 0 \end{bmatrix} = \begin{bmatrix} x\\ 0 \end{bmatrix}. \]

Therefore, the \(x\)-axis is the eigenspace for \(\lambda=1\).

Eigenvalue \(-1\): Reversal without Length Change

If

\[ A\mathbf{v} = -\mathbf{v}, \]

then the eigenvalue is

\[ \lambda=-1. \]

The vector keeps its magnitude but reverses direction.

Reflection and Eigenvalue \(-1\)

For reflection across the \(x\)-axis,

\[ F \begin{bmatrix} 0\\ y \end{bmatrix} = \begin{bmatrix} 0\\ -y \end{bmatrix}. \]

Thus the \(y\)-axis is the eigenspace for

\[ \boxed{\lambda=-1}. \]

Reflection preserves one direction and reverses the perpendicular one.

Eigenvalue \(0\): Collapse

If

\[ A\mathbf{v} = \mathbf{0} \]

for some nonzero vector \(\mathbf{v}\), then

\[ \boxed{\lambda=0} \]

is an eigenvalue.

The transformation collapses that eigendirection to the origin.

Zero Eigenvalues and Noninvertibility

If \(0\) is an eigenvalue, then there is a nonzero vector satisfying

\[ A\mathbf{v} = \mathbf{0}. \]

Therefore, \(A\) loses information and cannot be invertible.

Thus,

\[ \boxed{ 0\text{ is an eigenvalue} \iff A\text{ is singular} }. \]

Negative Eigenvalues

Suppose

\[ \lambda=-3. \]

Then an eigenvector satisfies

\[ A\mathbf{v} = -3\mathbf{v}. \]

The transformation triples the magnitude and reverses the direction along that eigenline.

Fractional Eigenvalues

Suppose

\[ \lambda=\frac12. \]

Then the transformation contracts the eigenvector to half its original magnitude while preserving its direction.

Repeated application continues to shrink that component.

Repeated Transformations along an Eigenvector

If

\[ A\mathbf{v} = \lambda\mathbf{v}, \]

then applying \(A\) again gives

\[ A^2\mathbf{v} = A(\lambda\mathbf{v}) = \lambda A\mathbf{v}. \]

Therefore,

\[ A^2\mathbf{v} = \lambda^2\mathbf{v}. \]

General Repeated-Application Rule

Continuing the pattern,

\[ \boxed{ A^n\mathbf{v} = \lambda^n\mathbf{v} }. \]

Matrix powers become extremely simple along eigendirections.

Long-Term Behavior When \(|\lambda|>1\)

If

\[ |\lambda|>1, \]

then

\[ |\lambda|^n \]

grows as \(n\) increases.

Repeated transformations therefore magnify the component along that eigendirection.

Long-Term Behavior When \(0<|\lambda|<1\)

If

\[ 0<|\lambda|<1, \]

then

\[ |\lambda|^n \to0. \]

Repeated transformations shrink the component along that eigendirection toward zero.

Long-Term Behavior When \(\lambda=-1\)

If

\[ \lambda=-1, \]

then

\[ \lambda^n \]

alternates between \(1\) and \(-1\).

Repeated transformation alternates the vector between two opposite orientations of equal magnitude.

Long-Term Behavior When \(\lambda=1\)

If

\[ \lambda=1, \]

then

\[ A^n\mathbf{v} = \mathbf{v} \]

for every positive integer \(n\).

The eigenvector remains fixed under repeated application.

Different Eigenvalues Create Different Spatial Rates

Suppose a matrix has two independent eigenvectors

\[ \mathbf{v}_1,\qquad\mathbf{v}_2 \]

with eigenvalues

\[ \lambda_1,\qquad\lambda_2. \]

A general vector may be written

\[ \mathbf{x} = c_1\mathbf{v}_1+c_2\mathbf{v}_2. \]

Then

\[ A\mathbf{x} = c_1\lambda_1\mathbf{v}_1 + c_2\lambda_2\mathbf{v}_2. \]

Repeated Effects on a General Vector

If

\[ \mathbf{x} = c_1\mathbf{v}_1+c_2\mathbf{v}_2, \]

then

\[ \boxed{ A^n\mathbf{x} = c_1\lambda_1^n\mathbf{v}_1 + c_2\lambda_2^n\mathbf{v}_2 }. \]

This reveals how different directional components grow, shrink, or reverse at different rates.

Dominant Eigenvalues

When one eigenvalue has a larger absolute value than the others, its eigendirection can dominate repeated transformations.

For large \(n\), the component associated with the largest

\[ |\lambda| \]

often becomes comparatively more important.

This idea appears in dynamical systems, population models, ranking algorithms, and numerical methods.

Eigenvectors and Change of Basis

Suppose a matrix has enough independent eigenvectors to form a basis.

If we choose those eigenvectors as basis vectors, the transformation becomes especially simple.

Each basis direction is merely scaled by its eigenvalue.

Eigenbasis

A basis made entirely of eigenvectors of a matrix is called an eigenbasis.

In an eigenbasis,

\[ A\mathbf{v}_i = \lambda_i\mathbf{v}_i \]

for every basis vector.

The matrix representation becomes diagonal.

Diagonal Form in an Eigenbasis

Suppose

\[ \mathbf{v}_1,\mathbf{v}_2 \]

form an eigenbasis with eigenvalues

\[ \lambda_1,\lambda_2. \]

Then the transformation matrix in that basis is

\[ \boxed{ D= \begin{bmatrix} \lambda_1&0\\ 0&\lambda_2 \end{bmatrix} }. \]

Why Diagonal Matrices Are So Convenient

A diagonal matrix acts independently on each coordinate direction.

For

\[ D= \begin{bmatrix} \lambda_1&0\\ 0&\lambda_2 \end{bmatrix}, \]

we have

\[ D \begin{bmatrix} c_1\\ c_2 \end{bmatrix} = \begin{bmatrix} \lambda_1c_1\\ \lambda_2c_2 \end{bmatrix}. \]

No coordinate mixing occurs.

Diagonalization

Let the columns of \(P\) be independent eigenvectors of \(A\).

Then

\[ \boxed{ P^{-1}AP=D }. \]

Equivalently,

\[ \boxed{ A=PDP^{-1} }. \]

This process is called diagonalization.

Connection to the Previous Lesson

The change-of-basis formula

\[ [T]_B=P_B^{-1}AP_B \]

becomes

\[ P^{-1}AP=D \]

when the chosen basis is an eigenbasis.

Diagonalization is therefore a strategically chosen coordinate change.

Matrix Powers through Diagonalization

If

\[ A=PDP^{-1}, \]

then

\[ A^2 = PDP^{-1}PDP^{-1}. \]

Since

\[ P^{-1}P=I, \]

this simplifies to

\[ A^2=PD^2P^{-1}. \]

General Matrix-Power Formula

Continuing the cancellation pattern,

\[ \boxed{ A^n = PD^nP^{-1} }. \]

Since powers of a diagonal matrix are easy,

\[ D^n = \begin{bmatrix} \lambda_1^n&0\\ 0&\lambda_2^n \end{bmatrix}. \]

Eigenvalues therefore simplify repeated transformations dramatically.

Not Every Matrix Has an Eigenbasis

A matrix may fail to have enough independent eigenvectors to form a basis.

In that case, the matrix is not diagonalizable over the coordinate system being used.

Having eigenvalues is not by itself enough; enough independent eigenvectors are required.

Repeated Eigenvalues

A characteristic polynomial may have the same root more than once.

Such an eigenvalue is called a repeated eigenvalue.

A repeated eigenvalue may have many independent eigenvectors or too few for diagonalization.

The eigenvectors must be examined directly.

Example: Every Direction as an Eigendirection

Consider uniform scaling:

\[ A= \begin{bmatrix} 3&0\\ 0&3 \end{bmatrix} = 3I. \]

For every vector \(\mathbf{v}\),

\[ A\mathbf{v} = 3\mathbf{v}. \]

Therefore, every nonzero vector is an eigenvector with eigenvalue \(3\).

Shears and Eigenvectors

Consider the horizontal shear

\[ H= \begin{bmatrix} 1&k\\ 0&1 \end{bmatrix}. \]

The horizontal direction

\[ \begin{bmatrix} 1\\ 0 \end{bmatrix} \]

satisfies

\[ H \begin{bmatrix} 1\\ 0 \end{bmatrix} = \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]

Thus the shear has eigenvalue \(1\) along the horizontal direction.

Why a Shear Is Not Usually Diagonalizable in the Plane

For a nonzero horizontal shear factor \(k\), the matrix

\[ \begin{bmatrix} 1&k\\ 0&1 \end{bmatrix} \]

has only one independent eigendirection.

It therefore does not have an eigenbasis of two independent plane vectors.

Rotations and Real Eigenvectors

Consider a genuine plane rotation through an angle other than \(0^\circ\) or \(180^\circ\).

Every nonzero real vector changes to a different real line of direction.

Therefore, such a rotation has no nonzero real eigenvectors.

Example: \(90^\circ\) Rotation

The matrix

\[ R= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} \]

rotates every nonzero real vector by \(90^\circ\).

No nonzero real vector remains on its original line.

Thus \(R\) has no real eigenvectors.

Complex Eigenvalues: A Gentle Preview

A matrix can have eigenvalues that are not real numbers.

For the \(90^\circ\) rotation matrix, the characteristic equation is

\[ \lambda^2+1=0. \]

Its solutions are

\[ \lambda=\pm i. \]

Complex eigenvalues encode rotational behavior that cannot be represented by real invariant lines.

Spatial Meaning before Computation

Eigenvalue calculations are more useful when connected to geometry.

Before solving equations, ask:

These clues often predict eigenvectors and eigenvalues.

Trace as Total Eigenvalue Sum

In two dimensions, if the eigenvalues are

\[ \lambda_1,\lambda_2, \]

then

\[ \boxed{ \operatorname{tr}(A) = \lambda_1+\lambda_2 }. \]

The trace therefore summarizes the eigenvalue sum.

Determinant as Eigenvalue Product

In two dimensions,

\[ \boxed{ \det(A) = \lambda_1\lambda_2 }. \]

This links eigenvalue scaling to the total signed area-scaling effect of the transformation.

Eigenvalues and Area Scaling

If a plane transformation has two independent eigenvectors with eigenvalues \(\lambda_1\) and \(\lambda_2\), then an eigenvector-aligned parallelogram has its side directions scaled by those factors.

Its signed area is therefore scaled by

\[ \lambda_1\lambda_2. \]

The ordinary area factor is

\[ \boxed{ |\lambda_1\lambda_2| }. \]

Eigenvalues and Orientation

In two dimensions, a negative determinant means the transformation reverses orientation.

When real eigenvalues describe the transformation, a negative product

\[ \lambda_1\lambda_2<0 \]

means one eigendirection reverses relative orientation while the other does not.

Eigenvectors in Applications

Eigenvectors and eigenvalues appear whenever repeated linear behavior or preferred directions matter.

Applications include:

Common Error: Allowing the Zero Vector as an Eigenvector

The zero vector satisfies

\[ A\mathbf{0} = \lambda\mathbf{0} \]

for every \(\lambda\).

For that reason, eigenvectors are required to be nonzero.

Common Error: Assuming the Output Must Equal the Input

An eigenvector does not need to be fixed.

It only needs to satisfy

\[ A\mathbf{v} = \lambda\mathbf{v}. \]

Only the special case

\[ \lambda=1 \]

leaves the vector unchanged.

Common Error: Ignoring Direction Reversal

A negative eigenvalue still corresponds to an eigenvector.

The output lies on the same line but points in the opposite direction.

Same eigendirection does not mean same orientation.

Common Error: Solving \(\det(A)=0\) Instead of \(\det(A-\lambda I)=0\)

To find eigenvalues, the required equation is

\[ \boxed{ \det(A-\lambda I)=0 }. \]

The equation

\[ \det(A)=0 \]

only tests whether \(0\) is an eigenvalue.

Common Error: Finding Eigenvalues but Not Eigenvectors

Eigenvalues tell us the scale factors.

Eigenvectors identify the directions to which those scale factors apply.

A complete geometric interpretation normally requires both.

Common Error: Using the Wrong Eigenvalue in the Eigenspace Equation

For each eigenvalue \(\lambda\), solve

\[ (A-\lambda I)\mathbf{v} = \mathbf{0}. \]

Different eigenvalues generally produce different eigenspaces.

Do not mix equations from one eigenvalue with another.

Common Error: Assuming Distinct Eigenvalues Have the Same Eigenvector

A nonzero vector cannot be an eigenvector for two different eigenvalues of the same matrix.

If

\[ A\mathbf{v} = \lambda_1\mathbf{v} \]

and

\[ A\mathbf{v} = \lambda_2\mathbf{v}, \]

then

\[ (\lambda_1-\lambda_2)\mathbf{v} = \mathbf{0}. \]

Since \(\mathbf{v}\neq\mathbf{0}\), we must have

\[ \lambda_1=\lambda_2. \]

Common Error: Assuming Every Matrix Is Diagonalizable

Diagonalization requires enough independent eigenvectors to form a basis.

Some matrices, such as a nontrivial plane shear, do not have enough independent eigenvectors.

The existence of eigenvalues alone does not guarantee diagonalizability.

Common Error: Confusing a Basis Change with a Different Transformation

In

\[ P^{-1}AP=D, \]

the matrices \(A\) and \(D\) can represent the same underlying transformation in different bases.

Diagonalization changes the coordinate description, not the transformation itself.

A Reliable Procedure for Testing an Eigenvector

  1. Verify that the candidate vector is nonzero.
  2. Compute \(A\mathbf{v}\).
  3. Determine whether the output is a scalar multiple of \(\mathbf{v}\).
  4. If so, identify that scalar as \(\lambda\).
  5. Check the geometric meaning of the sign and magnitude of \(\lambda\).

A Reliable Procedure for Finding Eigenvalues

  1. Confirm that \(A\) is square.
  2. Form \(A-\lambda I\).
  3. Compute \(\det(A-\lambda I)\).
  4. Set the determinant equal to zero.
  5. Solve the resulting characteristic equation.
  6. Check the eigenvalue sum and product against trace and determinant when convenient.

A Reliable Procedure for Finding Eigenvectors

  1. Choose one eigenvalue \(\lambda\).
  2. Form \(A-\lambda I\).
  3. Solve \((A-\lambda I)\mathbf{v}=\mathbf{0}\).
  4. Describe the nonzero solution vectors.
  5. Select a convenient representative eigenvector.
  6. Verify it by checking \(A\mathbf{v}=\lambda\mathbf{v}\).
  7. Repeat for each eigenvalue.

A Reliable Procedure for Spatial Interpretation

  1. Identify each eigendirection.
  2. Interpret \(|\lambda|\) as the scale factor along that direction.
  3. Use the sign of \(\lambda\) to determine whether direction reverses.
  4. Check for fixed directions when \(\lambda=1\).
  5. Check for collapsed directions when \(\lambda=0\).
  6. For repeated transformations, examine \(\lambda^n\).

A Reliable Procedure for Diagonalization

  1. Find the eigenvalues.
  2. Find corresponding eigenvectors.
  3. Determine whether enough independent eigenvectors exist to form a basis.
  4. Place those eigenvectors into the columns of \(P\).
  5. Place their eigenvalues in matching order on the diagonal of \(D\).
  6. Use \(P^{-1}AP=D\) or \(A=PDP^{-1}\).
  7. Verify the column-eigenvalue pairing carefully.

Example: Full Eigenvalue Workflow

Let

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

Form

\[ A-\lambda I = \begin{bmatrix} 4-\lambda&1\\ 2&3-\lambda \end{bmatrix}. \]

Then

\[ \det(A-\lambda I) = (4-\lambda)(3-\lambda)-2. \]

Complete the Characteristic Equation

Expand:

\[ (4-\lambda)(3-\lambda)-2 = 12-7\lambda+\lambda^2-2. \]

Therefore,

\[ \lambda^2-7\lambda+10=0. \]

Factor:

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

Thus,

\[ \boxed{ \lambda_1=5,\qquad \lambda_2=2 }. \]

Find an Eigenvector for \(\lambda=5\)

Compute

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

The system gives

\[ -x+y=0. \]

Therefore,

\[ y=x. \]

A convenient eigenvector is

\[ \boxed{ \mathbf{v}_1= \begin{bmatrix} 1\\ 1 \end{bmatrix} }. \]

Find an Eigenvector for \(\lambda=2\)

Compute

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

The system gives

\[ 2x+y=0. \]

Therefore,

\[ y=-2x. \]

A convenient eigenvector is

\[ \boxed{ \mathbf{v}_2= \begin{bmatrix} 1\\ -2 \end{bmatrix} }. \]

Interpret the Full Example Geometrically

Along

\[ \begin{bmatrix} 1\\ 1 \end{bmatrix}, \]

the transformation stretches by a factor of \(5\).

Along

\[ \begin{bmatrix} 1\\ -2 \end{bmatrix}, \]

it stretches by a factor of \(2\).

These two independent eigendirections form an eigenbasis.

Diagonalize the Full Example

Put the eigenvectors into

\[ P= \begin{bmatrix} 1&1\\ 1&-2 \end{bmatrix}. \]

Put the corresponding eigenvalues into

\[ D= \begin{bmatrix} 5&0\\ 0&2 \end{bmatrix}. \]

Then

\[ \boxed{ P^{-1}AP=D } \]

and

\[ \boxed{ A=PDP^{-1} }. \]

Practice 1: Verify an Eigenvector

Let

\[ A= \begin{bmatrix} 4&0\\ 0&2 \end{bmatrix} \]

and

\[ \mathbf{v} = \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]

Determine whether \(\mathbf{v}\) is an eigenvector and, if so, find its eigenvalue.

Practice 2: Test a Non-Eigenvector

For the same matrix,

\[ A= \begin{bmatrix} 4&0\\ 0&2 \end{bmatrix}, \]

determine whether

\[ \mathbf{v} = \begin{bmatrix} 1\\ 1 \end{bmatrix} \]

is an eigenvector.

Practice 3: Interpret an Eigenvalue

A vector is an eigenvector with eigenvalue

\[ \lambda=-3. \]

Describe the spatial effect of the transformation on that vector.

Practice 4: Repeated Transformation

If

\[ A\mathbf{v}=2\mathbf{v}, \]

find

\[ A^5\mathbf{v}. \]

Practice 5: Characteristic Equation

Find the eigenvalues of

\[ A= \begin{bmatrix} 5&0\\ 0&-1 \end{bmatrix}. \]

Practice 6: Reflection Eigenvalues

For

\[ F= \begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix}, \]

identify the eigendirections associated with eigenvalues \(1\) and \(-1\).

Practice 7: Zero Eigenvalue

Suppose a matrix has eigenvalue \(0\).

What does this imply about invertibility?

Practice 8: Trace and Determinant Check

A \(2\times2\) matrix has eigenvalues \(3\) and \(7\).

Find its trace and determinant.

Practice 9: Diagonalization Requirement

What additional condition, beyond having eigenvalues, is required for a \(2\times2\) matrix to be diagonalizable over a given field?

Practice 10: Identify the Error

A student says that

\[ \mathbf{0} \]

is an eigenvector of every matrix because

\[ A\mathbf{0} = \mathbf{0}. \]

Explain the error.

Practice Answers

Practice 1

\[ A\mathbf{v} = \begin{bmatrix} 4\\ 0 \end{bmatrix} = 4 \begin{bmatrix} 1\\ 0 \end{bmatrix}. \]

Therefore,

\[ \boxed{ \mathbf{v} \text{ is an eigenvector with }\lambda=4 }. \]

Practice 2

\[ A \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} 4\\ 2 \end{bmatrix}. \]

This is not one scalar multiple of

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

Therefore, the vector is not an eigenvector.

Practice 3

The magnitude is multiplied by \(3\), and the direction is reversed because the eigenvalue is negative.

Practice 4

\[ A^5\mathbf{v} = 2^5\mathbf{v} = \boxed{32\mathbf{v}}. \]

Practice 5

The matrix is diagonal, so its eigenvalues are its diagonal entries:

\[ \boxed{ \lambda=5,\qquad \lambda=-1 }. \]

Practice 6

The \(x\)-axis is the eigendirection for

\[ \boxed{\lambda=1}. \]

The \(y\)-axis is the eigendirection for

\[ \boxed{\lambda=-1}. \]

Practice 7

A zero eigenvalue means some nonzero vector is sent to zero.

Therefore, the matrix is not invertible.

Practice 8

\[ \operatorname{tr}(A) = 3+7 = \boxed{10}. \] \[ \det(A) = 3(7) = \boxed{21}. \]

Practice 9

The matrix must have enough linearly independent eigenvectors to form a basis.

For a \(2\times2\) matrix, that means two independent eigenvectors.

Practice 10

The zero vector is explicitly excluded from the definition of eigenvector.

It satisfies the eigenvalue equation for every scalar and therefore cannot identify a meaningful invariant direction or unique eigenvalue.

What to Remember

Completing Vectors, Matrices, and Spatial Algebra

This lesson completes the current sequence on vectors, matrices, and spatial algebra. The section began with vectors as directed quantities, developed component methods, vector operations, angular relationships, lines and planes, matrix transformations, composition, and basis changes, and now ends with the invariant directions revealed by eigenvectors.

Together, these ideas provide a bridge from elementary spatial reasoning to linear algebra: vectors describe directed quantities, matrices describe linear actions on space, bases describe coordinate frameworks, and eigenvectors expose the directions in which those actions become simplest.