from edumath.linear_algebra import eigenvalues_2x2, validate_eigenpair
A = [[3, 0], [0, 1]]
v = [1, 0]
print(eigenvalues_2x2(A))
print(validate_eigenpair(A, v, 3).message)((3+0j), (1+0j))
Correct.
An eigenvector is a nonzero vector whose direction is preserved by a matrix transformation. The vector may stretch, shrink, or flip, but it stays on the same line through the origin.
The defining equation is
\[ A\mathbf{v}=\lambda \mathbf{v}. \]
Here \(\mathbf{v}\) is the eigenvector and \(\lambda\) is the eigenvalue, the scale factor.
By the end of this lesson, you should be able to:
A vector \(\mathbf{v}\) is an eigenvector of \(A\) if:
For example, let
\[ A=\begin{bmatrix}3&0\\0&1\end{bmatrix} \quad \text{and} \quad \mathbf{v}=\begin{bmatrix}1\\0\end{bmatrix}. \]
Then
\[ A\mathbf{v} = \begin{bmatrix}3\\0\end{bmatrix} =3\begin{bmatrix}1\\0\end{bmatrix}. \]
So \(\mathbf{v}\) is an eigenvector and its eigenvalue is \(3\).
The zero vector would satisfy \(A\mathbf{0}=\lambda\mathbf{0}\) for every \(\lambda\), which would make the definition meaningless. Eigenvectors must be nonzero because they represent directions.
To check whether \(\mathbf{v}\) is an eigenvector:
Let
\[ A=\begin{bmatrix}2&0\\0&5\end{bmatrix}. \]
Check \(\mathbf{v}=\begin{bmatrix}1\\0\end{bmatrix}\):
\[ A\mathbf{v}=\begin{bmatrix}2\\0\end{bmatrix}=2\mathbf{v}. \]
So \(\mathbf{v}\) is an eigenvector with eigenvalue 2.
Now check \(\mathbf{w}=\begin{bmatrix}1\\1\end{bmatrix}\):
\[ A\mathbf{w}=\begin{bmatrix}2\\5\end{bmatrix}. \]
This is not a scalar multiple of \(\begin{bmatrix}1\\1\end{bmatrix}\), because a scalar multiple would have equal coordinates. Therefore \(\mathbf{w}\) is not an eigenvector of this matrix.
For a diagonal matrix
\[ \begin{bmatrix} a&0\\0&d \end{bmatrix}, \]
the coordinate axes are eigenvector directions. The eigenvalues are \(a\) and \(d\). The first coordinate direction is scaled by \(a\) and the second coordinate direction is scaled by \(d\).
If \(A\mathbf{v}=\lambda\mathbf{v}\), then repeated multiplication is easy:
\[ A^2\mathbf{v}=\lambda^2\mathbf{v}, \qquad A^3\mathbf{v}=\lambda^3\mathbf{v}. \]
This idea helps explain long-term behavior in dynamical systems, differential equations, Markov chains, vibrations, computer graphics, and data analysis.
Eigenvectors reveal the directions a transformation treats simply. Even if a matrix mixes most vectors, it may have special directions that only scale.
Eigenvalue problems ask one central question: does the matrix keep this vector on the same line through the origin?
Guess first. If \(A\mathbf{v}=\begin{bmatrix}6\\0\end{bmatrix}\) and \(\mathbf{v}=\begin{bmatrix}3\\0\end{bmatrix}\), is \(\mathbf{v}\) an eigenvector?
Guided exercise.
Let
\[ A=\begin{bmatrix}2&0\\0&5\end{bmatrix}, \qquad \mathbf{v}=\begin{bmatrix}3\\0\end{bmatrix}. \]
Compute
\[ A\mathbf{v}=\begin{bmatrix}6\\0\end{bmatrix}. \]
Now compare this with \(\mathbf{v}\):
\[ \begin{bmatrix}6\\0\end{bmatrix}=2\begin{bmatrix}3\\0\end{bmatrix}. \]
So \(\mathbf{v}\) is an eigenvector and the eigenvalue is \(2\).
Checkpoint. Check whether \(\begin{bmatrix}0\\4\end{bmatrix}\) is an eigenvector of \(\begin{bmatrix}2&0\\0&5\end{bmatrix}\). If yes, find the eigenvalue.
Guess first. Is \(\begin{bmatrix}2\\5\end{bmatrix}\) a scalar multiple of \(\begin{bmatrix}1\\1\end{bmatrix}\)?
Guided exercise.
A scalar multiple of \(\begin{bmatrix}1\\1\end{bmatrix}\) must have equal coordinates:
\[ c\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}c\\c\end{bmatrix}. \]
Since \(\begin{bmatrix}2\\5\end{bmatrix}\) has unequal coordinates, it is not a scalar multiple. Therefore if \(A\begin{bmatrix}1\\1\end{bmatrix}= \begin{bmatrix}2\\5\end{bmatrix}\), then \(\begin{bmatrix}1\\1\end{bmatrix}\) is not an eigenvector of \(A\).
Checkpoint. Is \(\begin{bmatrix}-6\\3\end{bmatrix}\) a scalar multiple of \(\begin{bmatrix}-2\\1\end{bmatrix}\)? If yes, what is the scalar?
Guess first. What are the eigenvalues of \(\begin{bmatrix}-3&0\\0&4\end{bmatrix}\)?
Guided exercise.
A diagonal matrix scales coordinate directions separately:
\[ \begin{bmatrix}-3&0\\0&4\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}-3\\0\end{bmatrix} =-3\begin{bmatrix}1\\0\end{bmatrix}, \]
and
\[ \begin{bmatrix}-3&0\\0&4\end{bmatrix} \begin{bmatrix}0\\1\end{bmatrix} = \begin{bmatrix}0\\4\end{bmatrix} =4\begin{bmatrix}0\\1\end{bmatrix}. \]
So the eigenvalues are \(-3\) and \(4\).
Checkpoint. Find the eigenvalues and coordinate-axis eigenvectors of \(\begin{bmatrix}7&0\\0&\frac12\end{bmatrix}\).
Guess first. If \(\lambda=-2\), does the vector shrink, grow, flip, or both grow and flip?
Guided exercise.
If \(A\mathbf{v}=-2\mathbf{v}\), then applying \(A\) makes the vector twice as long and points it in the opposite direction. The negative sign flips direction. The absolute value \(2\) tells the stretch factor.
Checkpoint. Describe what happens along an eigenvector when \(\lambda=0.25\), \(\lambda=-1\), and \(\lambda=3\).
Guess first. If \(A\mathbf{v}=3\mathbf{v}\), what is \(A^4\mathbf{v}\)?
Guided exercise.
Each application of \(A\) multiplies the vector by \(3\):
\[ A\mathbf{v}=3\mathbf{v}, \quad A^2\mathbf{v}=3^2\mathbf{v}, \quad A^3\mathbf{v}=3^3\mathbf{v}, \quad A^4\mathbf{v}=3^4\mathbf{v}=81\mathbf{v}. \]
This is why eigenvectors are useful for understanding long-term behavior.
Checkpoint. If \(A\mathbf{v}=\frac12\mathbf{v}\), what is \(A^5\mathbf{v}\)?
Use this checkpoint to identify eigenvalues, eigenvectors, and repeated behavior.
SymPy is very useful for eigenvalue work because it can return exact symbolic answers. In a PyScript/Pyodide page, the same commands can be used after loading SymPy.
((3+0j), (1+0j))
Correct.