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Calculation Results
Eigenvectors Calculator With Steps
Find eigenvectors and eigenvalues with step-by-step explanations.
1. Definition & Formula
An eigenvector of a square matrix $A$ is a nonzero vector $\mathbf{v}$
such that
$$A\mathbf{v}=\lambda\mathbf{v}$$
Here $\lambda$ is the eigenvalue associated with $\mathbf{v}$. To find
eigenvalues solve the characteristic equation:
$$\det(A-\lambda I)=0$$
For each eigenvalue $\lambda$, find eigenvectors by solving:
$$(A-\lambda I)\mathbf{v}=\mathbf{0}$$
The set of all eigenvectors (plus the zero vector) for a given $\lambda$ is
the eigenspace. Eigenvectors can be scaled by any nonzero scalar — they represent
directions.
Since there is only one independent eigenvector for a double root, $A$ is
defective (not diagonalizable).
3. Common Mistakes
Using the zero vector: $\mathbf{0}$ is not an eigenvector.
Stopping after eigenvalues: Finding $\lambda$ is only half the job
— you must solve $(A-\lambda I)\mathbf{v}=0$.
Arithmetic mistakes on determinants: 3×3 determinants and sign
errors are common.
Confusing multiplicities: Check geometric multiplicity (number of
independent eigenvectors) when eigenvalues repeat.
Not simplifying eigenspaces: Present eigenvectors in simplest
integer or rational form when possible.
4. Practice Problems
Try these problems. Click to reveal model eigenvectors.
Exercise 1
$$\begin{bmatrix}2 & 1\\[4pt]0 & 3\end{bmatrix}$$
Show Answer
Eigenvalues $2,3$. Eigenvectors: for $2$, $[1,0]^T$; for $3$,
$[1,0]^T$ (note: check rows). Actually compute: for $\lambda=2$, $(A-2I)=[0\ 1;0\
1]$ gives $y=0\Rightarrow v=[1,0]^T$. For $\lambda=3$, $(A-3I)=[-1\ 1;0\ 0]$ gives
$-x+y=0\Rightarrow v=[1,1]^T$.
Exercise 2
$$\begin{bmatrix}4 & 0\\[4pt]0 & 1\end{bmatrix}$$
Show Answer
Eigenvalues 4 and 1. Eigenvectors: $\lambda=4\Rightarrow [1,0]^T$;
$\lambda=1\Rightarrow [0,1]^T$.
Exercise 3
$$\begin{bmatrix}1 & 2\\[4pt]2 & 1\end{bmatrix}$$
Show Answer
Characteristic polynomial: $(1-\lambda)^2-4 =
\lambda^2-2\lambda-3=(\lambda-3)(\lambda+1)$. Eigenvalues 3 and -1. Eigenvectors:
for 3, $[1,1]^T$; for -1, $[1,-1]^T$.
Exercise 4
$$\begin{bmatrix}5 & 4\\[4pt]2 & 1\end{bmatrix}$$
Show Answer
Characteristic polynomial:
$(5-\lambda)(1-\lambda)-8=\lambda^2-6\lambda-3$. Eigenvalues
$3\pm\sqrt{12}=3\pm2\sqrt3$. For $\lambda=3+2\sqrt3$, solve to get an eigenvector;
for $\lambda=3-2\sqrt3$ similarly. Numerical eigenvector approximations are
acceptable on calculators.
Frequently Asked Questions (FAQ)
Q: What are eigenvectors?
Eigenvectors are vectors that only scale when multiplied by a matrix.
Q: Does this calculator show steps?
Yes. It explains how eigenvalues and eigenvectors are computed.
