A square matrix can rotate, reflect, shear, stretch, and squash. Amid that motion, some lines through the origin are special. The matrix stretches or flips vectors on those lines, but it does not knock them onto a different line. Those lines are spanned by eigenvectors. The stretch factors are eigenvalues.

If you came here asking what $\lambda$ means in $A\mathbf{v}=\lambda\mathbf{v}$, the short answer is this. Applying $A$ has the same effect as multiplying by the scalar $\lambda$, at least on that one direction. Most directions are not like that. The matrix usually turns them as it stretches them. This page sits in the explainer library beside Jacobian matrices in plain language, where a derivative is a matrix and these stretch factors describe that linear approximation. The course map is the linear algebra first course.

The defining equation

A nonzero vector $\mathbf{v}$ is an eigenvector of a square matrix $A$ with eigenvalue $\lambda$ when

$$A\mathbf{v}=\lambda\mathbf{v}.$$

The zero vector is banned because $A\mathbf{0}=\lambda\mathbf{0}$ is true for every $\lambda$ and therefore says nothing. Rearranged, the equation is $(A-\lambda I)\mathbf{v}=\mathbf{0}$. So $\lambda$ is an eigenvalue exactly when $A-\lambda I$ fails to be invertible, which for a square matrix is the same as $\det(A-\lambda I)=0$. That determinant, as a polynomial in $\lambda$, is the characteristic polynomial.

For a $2\times 2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ the polynomial collapses to a quadratic you can finish by hand:

$$\lambda^2-(\mathrm{tr}\,A)\lambda+\det A=0,\qquad \mathrm{tr}\,A=a+d,\quad \det A=ad-bc.$$

The two roots (counted with multiplicity, and allowed to be complex) are the eigenvalues. Their sum is the trace. Their product is the determinant. Those two identities are the fastest audit of any list you compute.

A picture you can draw on graph paper

Example Stretch the plane by $2$ in the $x$ direction and $3$ in the $y$ direction:

$$A=\begin{pmatrix}2&0\\0&3\end{pmatrix}.$$

Then $\mathbf{e}_1=(1,0)$ satisfies $A\mathbf{e}_1=2\mathbf{e}_1$, and $\mathbf{e}_2=(0,1)$ satisfies $A\mathbf{e}_2=3\mathbf{e}_2$. The axes are the invariant lines. A vector such as $(1,1)$ becomes $(2,3)$. It is longer, and it is no longer on the same line through the origin. So $(1,1)$ is not an eigenvector, even though it is a perfectly ordinary input.

That is the geometric test. Draw $\mathbf{v}$ and $A\mathbf{v}$. If they lie on one line, you have found an eigen direction. If they form a parallelogram that is not flattened onto a line, you have not.

A quieter 2 by 2 that is not diagonal

Example Take $B=\begin{pmatrix}4&1\\2&3\end{pmatrix}$. The trace is $7$ and the determinant is $10$, so the characteristic polynomial is $\lambda^2-7\lambda+10=0$, which factors as $(\lambda-5)(\lambda-2)=0$. The eigenvalues are $5$ and $2$.

For $\lambda=5$, solve $(B-5I)\mathbf{v}=\mathbf{0}$:

$$B-5I=\begin{pmatrix}-1&1\\2&-2\end{pmatrix}.$$

The rows are multiples, so the system is $ -v_1+v_2=0$. One eigenvector is $(1,1)$. Check: $B(1,1)=(5,5)=5(1,1)$. For $\lambda=2$, the same work produces a line spanned by $(1,-2)$. The two lines are not the coordinate axes, yet each is still a direction that $B$ refuses to tip.

Once you have a basis of eigenvectors, $B$ becomes diagonal in those coordinates. Powers of $B$ become easy because you only raise the two scalars to a power. That is why people hunt for these directions. They are a coordinate system adapted to the map.

When the plane has no real invariant line

A rotation by a right angle, $R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, sends $(1,0)$ to $(0,1)$ and $(0,1)$ to $(-1,0)$. No nonzero real vector stays on its own line. The characteristic polynomial is $\lambda^2+1=0$, with roots $\pm i$. Over the reals the map rotates and scales (here the scale is $1$). Over the complex numbers those roots are ordinary eigenvalues. A real matrix is allowed to have a conjugate pair. That is not a bug in the definition.

A repeated real root is a different warning. The Jordan block $\begin{pmatrix}3&1\\0&3\end{pmatrix}$ has a double eigenvalue $3$, but only one independent real eigenvector, along $(1,0)$. Algebraic multiplicity and geometric multiplicity can disagree. “Has a repeated eigenvalue” does not automatically mean “has a basis of eigenvectors.”

Pitfall. Eigenvectors are defined only up to nonzero scaling. If $\mathbf{v}$ works, so does $7\mathbf{v}$ and so does $-\mathbf{v}$. Also, the entries of $A$ are not the eigenvalues except in special cases such as triangular matrices. Do not read $a$ and $d$ as the spectrum unless you know the matrix is triangular.

A negative factor flips the line

Example Let $F=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. This reflection keeps the $x$ axis fixed and reverses the $y$ axis. Then $F(1,0)=(1,0)$, so $\lambda=1$ with eigenvector $(1,0)$. And $F(0,1)=(0,-1)=-1\cdot(0,1)$, so $\lambda=-1$ with eigenvector $(0,1)$. The second line is invariant. The arrow on that line reverses. A negative eigenvalue is still a stretch factor. The stretch includes a flip.

The trace is $0$ and the determinant is $-1$, so the characteristic polynomial is $\lambda^2-1=0$. The roots are $1$ and $-1$. Their sum is the trace. Their product is the determinant. If you only look for arrows that point the same way as the input, you will miss the flip and you will think this reflection has only one eigenvalue. It has two.

Use the lines to compute a power

Return to $B=\begin{pmatrix}4&1\\2&3\end{pmatrix}$, with eigenvalues $5$ and $2$ and eigenvectors $(1,1)$ and $(1,-2)$. Those two directions form a basis. Any input can be written as a combination of them, and $B$ acts by the matching scalar on each piece.

Example Solve $(3,0)=a(1,1)+b(1,-2)$. Then $a+b=3$ and $a-2b=0$, so $b=1$ and $a=2$. Thus $(3,0)=2(1,1)+(1,-2)$. Apply $B$:

$$B(3,0)=2\cdot 5\cdot(1,1)+1\cdot 2\cdot(1,-2)=(10,10)+(2,-4)=(12,6).$$

Direct multiplication gives $B(3,0)=(12,6)$. The two routes agree. For a square, only the scalars change:

$$B^2(3,0)=2\cdot 5^2\cdot(1,1)+1\cdot 2^2\cdot(1,-2)=(50,50)+(4,-8)=(54,42).$$

Check against the first power. $B(12,6)=(48+6,\ 24+18)=(54,42)$. Once the coordinates $a=2$ and $b=1$ are known, the $n$th power is $2\cdot 5^n\cdot(1,1)+2^n\cdot(1,-2)$. The next example is a matrix that refuses to give you a basis of eigenvectors, so this shortcut is not available.

A repeated root with only one line

Example Let $J=\begin{pmatrix}3&1\\0&3\end{pmatrix}$. The trace is $6$ and the determinant is $9$, so the characteristic polynomial is $\lambda^2-6\lambda+9=(\lambda-3)^2$. The only eigenvalue is $3$, with algebraic multiplicity $2$. Now solve $(J-3I)\mathbf{v}=\mathbf{0}$:

$$J-3I=\begin{pmatrix}0&1\\0&0\end{pmatrix}.$$

The equation says the second coordinate is $0$. Every eigenvector is a multiple of $(1,0)$. Check $J(1,0)=(3,0)$. The vector $(0,1)$ is not an eigenvector: $J(0,1)=(1,3)$, which is not a scalar multiple of $(0,1)$.

The eigenspace for $\lambda=3$ is the line through $(1,0)$, together with the zero vector. It is one dimensional inside a two dimensional plane. Algebraic multiplicity counts the root. Geometric multiplicity counts independent eigenvectors. They disagree here. You do not have a basis of eigenvectors, so you cannot diagonalize $J$. A repeated eigenvalue is a warning to check the dimension of the eigenspace. It is not, by itself, a promise of a full set of lines. The linear independence glossary line is the test those two vectors would have needed to pass.

A turn with a stretch, and no real line

Example Let $R=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$. Then $R(1,0)=(0,2)$ and $R(0,1)=(-2,0)$. Every nonzero real vector turns. Lengths double, because $(1,0)$ has length $1$ and its image has length $2$. The trace is $0$ and the determinant is $4$, so $\lambda^2+4=0$, with roots $\pm 2i$. There is no real invariant line. The imaginary pair records the turn. The modulus $2$ records the stretch. Over the reals you describe a rotation together with a scale. You do not invent a real eigenvector to satisfy a habit.

The roots $2i$ and $-2i$ still sum to the trace and multiply to the determinant. A conjugate pair is ordinary data for a real matrix.

Zero is a legal eigenvalue

Example Let $S=\begin{pmatrix}1&2\\2&4\end{pmatrix}$. The second row is twice the first, so the determinant is $0$ and $S$ is not invertible. The trace is $5$. The characteristic polynomial is $\lambda^2-5\lambda=\lambda(\lambda-5)$. The eigenvalues are $0$ and $5$.

For $\lambda=0$, solve $S\mathbf{v}=\mathbf{0}$. The first row says $v_1+2v_2=0$. One eigenvector is $(2,-1)$. Check $S(2,-1)=(0,0)$. That line is flattened to the origin. For $\lambda=5$, one eigenvector is $(1,2)$, and $S(1,2)=(5,10)=5(1,2)$. A zero eigenvalue is the determinant test in geometric language. It is not a sign that the computation failed, and it is not the absence of eigenvalues. The other factor can still be a perfectly ordinary stretch.

A decision framework

  1. Confirm the matrix is square. In a first course, eigenvalues belong to square matrices. A rectangular matrix is a different object. See the matrix glossary line if that distinction is still fuzzy.
  2. Compute the trace and the determinant before you expand a large determinant by hand.
  3. Write $\lambda^2-(\mathrm{tr}\,A)\lambda+\det A=0$ for a $2\times 2$ matrix.
  4. Read the discriminant. If it is positive, hunt two real lines. If it is zero, you have a repeated root, and you must still count independent eigenvectors. If it is negative, stop hunting a real line and describe a rotation and a scale.
  5. For each real $\lambda$, row reduce $A-\lambda I$. Discard the zero vector. Any nonzero vector in the null space is an eigenvector.
  6. Multiply $A\mathbf{v}$ and confirm it equals $\lambda\mathbf{v}$ before you trust the line.
  7. Audit the list. The roots must sum to the trace and multiply to the determinant.
  8. If you need a power, expand the input in an eigenbasis when you have one. If you do not have a basis of eigenvectors, do not pretend a diagonal form exists.

What an eigenvalue is not

  • It is not a diagonal entry unless the matrix is triangular. The matrix $B$ above has diagonal entries $4$ and $3$. Its eigenvalues are $5$ and $2$.
  • It is not a property of a vector by itself. An eigenvector is a vector for a specific matrix. Change the matrix and the special lines move.
  • A zero eigenvalue is not a failed search. It means the linear map flattens some direction, which is useful information.
  • A scalar multiple of an eigenvector is not a new direction. $(1,1)$ and $(4,4)$ span the same line.

How the numbers show up in differential equations

For $\mathbf{x}'=A\mathbf{x}$, motion along an eigenvector looks like a scalar exponential times that vector. A positive real eigenvalue grows. A negative one decays. A complex pair oscillates. This page does not solve systems. It names the numbers those solutions use. The scalar equation $y'=ky$ is the one dimensional version, and it is separable. The walkthrough is separable differential equations, on the differential equations pillar. When a later course linearizes a nonlinear system, the matrix is often a Jacobian, and its eigenvalues describe the linear approximation. That warning lives on the Jacobian explainer.

How to know you understood

  1. For the diagonal stretch with entries $2$ and $3$, say why $(1,1)$ is not an eigenvector, and compute the image $(2,3)$.
  2. For $\begin{pmatrix}4&1\\2&3\end{pmatrix}$, get eigenvalues $5$ and $2$ from trace $7$ and determinant $10$, and check that the image of $(1,1)$ is $(5,5)$.
  3. Give a negative eigenvalue as a flip along a line, using the reflection with diagonal entries $1$ and $-1$.
  4. For $\begin{pmatrix}3&1\\0&3\end{pmatrix}$, show there is only one independent eigenvector even though $3$ is a double root.
  5. For a quarter turn scaled by $2$, explain why there is no real invariant line, and why a conjugate pair can still be the correct spectrum.
  6. State the two audits in one sentence. The eigenvalues sum to the trace and multiply to the determinant.

Why anyone cares

  • Diagonalization. If you have a basis of eigenvectors, $A$ becomes a diagonal matrix in those coordinates. Powers and matrix exponentials become lists of scalar powers.
  • Discrete dynamics. For $\mathbf{x}_{n+1}=A\mathbf{x}_n$, eigenvalues decide growth, decay, and oscillation. A factor with $|\lambda|>1$ amplifies that mode. A factor with $|\lambda|<1$ damps it.
  • Continuous dynamics. For $\mathbf{x}'=A\mathbf{x}$, signs of real parts govern growth or decay. The section above names that link. Jacobians are where the same numbers describe a linear approximation.
  • Quadratic forms and variance. Symmetric matrices have an orthonormal eigenbasis. Principal axes of an ellipse, and principal components of a covariance, are the same geometry.

The surrounding vocabulary lives on the linear algebra pillar. A study order is the linear algebra first course pathway.

Try it with the matching tools

After you can say the sentence “$A\mathbf{v}$ is a scalar multiple of $\mathbf{v}$,” compute a few $2\times 2$ examples and audit the output.

Try it. Open the 2 by 2 eigenvalue calculator and enter $a=4$, $b=1$, $c=2$, $d=3$. You should see roots $5$ and $2$. Confirm that they sum to the trace $7$ and multiply to the determinant $10$. Then try $a=0$, $b=-1$, $c=1$, $d=0$ and watch a conjugate pair appear.

If you still need the determinant, the inverse, or a matrix product before you hunt for $\lambda$, start on the 2 by 2 matrix calculator. That page is the arithmetic bench. The 2 by 2 eigenvalue page is the spectrum bench, and it does not find eigenvectors. Once you have $\lambda$, you still solve $(A-\lambda I)\mathbf{v}=\mathbf{0}$ by hand on a 2 by 2 matrix. When the matrix is 3 by 3 and you need the determinant or the inverse, use the 3 by 3 matrix calculator. When you want the eigenvalues of that 3 by 3 matrix, with the characteristic polynomial written out, use the 3x3 eigenvalue calculator. A determinant of zero is the same fact as a zero eigenvalue: the map flattens some direction. The reduced row echelon form calculator can check a row reduction with exact fractions.

Compute the stretch factors for a 2 by 2 matrix after you can say what those numbers mean.

Open the eigenvalue calculator

How to practice

  1. Compute $\mathrm{tr}\,A$ and $\det A$ first. Write the quadratic. Solve it.
  2. Check sum and product before you believe the roots.
  3. For each real $\lambda$, row reduce $A-\lambda I$ and name a spanning vector for the null space.
  4. Multiply $A\mathbf{v}$ by hand and confirm it equals $\lambda\mathbf{v}$.
  5. If the discriminant is negative, stop hunting for a real line. Describe a rotation and scale instead.

The same $2\times 2$ algebra appears on the quadratic formula calculator if you want a second arithmetic check of the characteristic polynomial. Local linear maps in several variables are Jacobians. Their eigenvalues, when they are defined, describe the linear approximation, not the full nonlinear map.

Citations and further reading

Last reviewed: 21 September 2026.