Eigenvalues and Eigenvectors

Apply a linear map to a vector and it usually comes back pointing somewhere else. A few directions are exceptions: the map leaves them where they are and only stretches or shrinks them.

Those directions are the eigenvectors, and the stretch factors are the eigenvalues. Finding them turns a matrix into something far easier to work with, because in a basis of eigenvectors the map is just multiplication by $n$ numbers, applying it a thousand times becomes raising those numbers to the thousandth power.

Two things can prevent that. The scale factors may not exist in the field you are working over, which is what happens to a rotation. Or they may exist without enough independent directions to go round, which is what happens to a shear. The first is repaired by moving to the complex numbers; the second cannot be repaired, and telling the two apart is part of the work.

The nonzero requirement is not a technicality. $A0=0=\lambda 0$ holds for every scalar $\lambda$ whatsoever. If the zero vector counted as an eigenvector, every number would be an eigenvalue of every matrix and the definition would carry no information. Excluding it is what makes the concept discriminating.

The asymmetry is: eigenvectors may never be zero; eigenvalues may be. A zero eigenvalue means some nonzero $v$ has $Av=0$, which is a nontrivial kernel, which is singularity. So "$0$ is an eigenvalue of $A$", " $detA=0$", "$A$ is not invertible" and " $\ker A\ne \{0\}$" are four ways of saying one thing.

The eigenspace is a subspace, once zero is put back. The set of eigenvectors for $\lambda$ is not closed under addition in an obvious way. It excludes $0$, so it fails the subspace test at the first condition. What is a subspace is

which includes the zero vector by construction. The eigenvectors for $\lambda$ are exactly the nonzero elements of $E_{\lambda }$. This is why the object worth computing is the eigenspace and its dimension, not a single representative vector.

Eigenvectors are never unique. If $Av=\lambda v$ then $A(cv)=\lambda (cv)$ for every $c\ne 0$, so an eigenvector comes with a whole line of companions, and more when the eigenspace has dimension above 1. Reporting "the eigenvector for $\lambda =5$" is a category error; the answer is a basis for $E_5$.

Why the characteristic polynomial works. $(A-\lambda I)v=0$ with $v\ne 0$ says $A-\lambda I$ has a nontrivial kernel, and by the determinant unit that is exactly $det(A-\lambda I)=0$. The polynomial is therefore not a definition but a consequence. A way of turning the search for a nonzero kernel into root-finding.

Subtracting $\lambda$ correctly. $A-\lambda I$ subtracts $\lambda$ from the diagonal entries only. Subtracting it from every entry is a common slip and produces an unrelated matrix.

1. Form $A-\lambda I$. Subtract $\lambda$ from the diagonal entries only.
2. Compute $det(A-\lambda I)$. The result is a polynomial of degree $n$ in $\lambda$.
3. Find the roots. These are the eigenvalues. Record the multiplicity of each root, that is its algebraic multiplicity.
4. For each eigenvalue, solve $(A-\lambda I)v=0$. Row reduce; the free variables give a basis of the eigenspace $E_{\lambda }$. The number of basis vectors is the geometric multiplicity.
5. Compare the multiplicities. If geometric is less than algebraic for any eigenvalue, the matrix is defective and has no basis of eigenvectors.
6. Check. Verify $Av=\lambda v$ for each pair; confirm the eigenvalues sum to $\mathrm{tr}A$ and multiply to $detA$.

Shortcuts worth taking first.

Where step 4 goes wrong. After substituting a correct eigenvalue, $A-\lambda I$ is singular by construction, so its reduced form must have at least one free variable. If row reduction leaves no free variable, the eigenvalue or the reduction is wrong. The system cannot have only the zero solution. This is the most useful self-check in the procedure, because it catches an arithmetic error before it propagates into a reported eigenvector.

On step 6. The trace and determinant checks cost two additions and one multiplication and catch most sign errors in the characteristic polynomial. Use them before solving for eigenvectors, not after: a wrong eigenvalue makes every subsequent step wasted work.

A note on scale. This procedure is for hand computation on small matrices. Forming and rooting a characteristic polynomial is numerically unstable at size, and production algorithms, the QR iteration and its relatives, compute eigenvalues without ever writing one down.

Step 1–2: the characteristic polynomial. For a $2\times 2$, $\mathrm{tr}A=4+3=7$ and $detA=12-2=10$, so

Directly, , note $\lambda$ is subtracted from the diagonal only.

Step 3: the roots. ${\lambda }^2-7\lambda +10=(\lambda -5)(\lambda -2)$, so $\lambda =5$ and $\lambda =2$, each of algebraic multiplicity 1.

Check before going further. $5+2=7=\mathrm{tr}A$ and $5\times 2=10=detA$. Two seconds, and it would have caught a sign error in the polynomial.

Step 4: the eigenspaces.

For $\lambda =5$:

Row reducing, the second row is $-2$ times the first, leaving $-v_1+v_2=0$, so $v_2=v_1$ and

There is one free variable, as there must be: $A-5I$ is singular by construction.

For $\lambda =2$:

The rows are equal, leaving $2v_1+v_2=0$, so $v_2=-2v_1$ and

Step 5: multiplicities. Both eigenvalues have algebraic and geometric multiplicity 1, so they agree. The matrix is not defective, and $\{(1,1),(1,-2)\}$ is a basis of ${\mathbb{R}}^2$ consisting of eigenvectors.

Step 6: verify each pair.

In the basis $\{(1,1),(1,-2)\}$ the map is multiplication by 5 in one coordinate and by 2 in the other. Applying $A$ ten times scales those coordinates by $5^{10}$ and $2^{10}$. No matrix multiplication required. And since both eigenvalues exceed 1 in magnitude, every nonzero vector grows under repetition, with the $\lambda =5$ component eventually dominating: the direction $(1,1)$ is where any long run ends up pointing.

Theorem (multiplicity bound). For every eigenvalue $\lambda$ of $A$,

Lower bound. $\lambda$ being an eigenvalue means $det(A-\lambda I)=0$, so $A-\lambda I$ is singular and its kernel contains a nonzero vector. Hence $\dim E_{\lambda }\ge 1$: an eigenvalue always has at least one direction to act on.

Upper bound. Let $k=\dim E_{\lambda }$ and take a basis $\{u_1,\dots ,u_k\}$ of $E_{\lambda }$, extended to a basis of the whole space. In that basis the matrix of the map has the block form , because each $u_i$ is sent to $\lambda u_i$. Its characteristic polynomial is therefore $({\lambda }^{\prime }-\lambda {)}^k\cdot det(C-{\lambda }^{\prime }I)$, which is divisible by $({\lambda }^{\prime }-\lambda {)}^k$. So $\lambda$ is a root of multiplicity at least $k$, giving $k\le$ algebraic multiplicity. $◼$

The argument uses that similar matrices share a characteristic polynomial, which holds because $det(P^{-1}AP-\lambda I)=det(P^{-1}(A-\lambda I)P)=det(A-\lambda I)$ by multiplicativity.

Theorem (trace and determinant). Over a field where $p$ splits, listing the eigenvalues with algebraic multiplicity,

Proof. The characteristic polynomial factors as $p(\lambda )=\prod _i({\lambda }_i-\lambda )$. Setting $\lambda =0$ gives $p(0)=detA=\prod _i{\lambda }_i$. Comparing the coefficient of ${\lambda }^{n-1}$ on both sides, expanding $det(A-\lambda I)$, that coefficient comes only from the product of diagonal entries, gives the trace identity. $◼$

Why these matter in practice. The multiplicity bound is what makes "defective" a meaningful category: the geometric multiplicity can fall short but never exceed, so a total count of independent eigenvectors below $n$ is the only failure mode. A matrix is diagonalisable exactly when geometric equals algebraic for every eigenvalue.

The invariants are the cheapest available check. Computing eigenvalues $3$ and $4$ for a matrix of trace $8$ means an error has already occurred, and finding it now costs less than solving two eigenspace systems first.

A corollary. $n$ distinct eigenvalues force diagonalisability: each contributes at least one dimension of eigenspace, the algebraic multiplicities are all 1 so the geometric ones are too, and the total is $n$. Distinct eigenvalues are sufficient but not necessary. The identity matrix has one eigenvalue of algebraic multiplicity $n$ and is already diagonal.

Triangular: read the diagonal. is lower triangular, so its eigenvalues are $2$, $3$ and $-1$. Checking one: $det(D-2I)=0$. The invariants agree, trace $2+3-1=4$ and $detD=2\cdot 3\cdot (-1)=-6$, matching the sum and product of the eigenvalues.

Defective: a shear. is triangular, so $\lambda =3$ with algebraic multiplicity 2. But

has rank 1, so its kernel has dimension 1: $E_3=\mathrm{span}\{(1,0)\}$, geometric multiplicity 1. Checking: $B(1,0{)}^{\mathsf{T}}=(3,0{)}^{\mathsf{T}}=3(1,0{)}^{\mathsf{T}}$.

Geometric algebraic $2$, so $B$ is defective: there is no basis of ${\mathbb{R}}^2$ made of its eigenvectors, and no change of basis makes it diagonal. Geometrically it fixes the horizontal direction and shears everything else, never settling any second direction.

Complex: a rotation. rotates the plane by a quarter turn. Trace $0$, determinant $1$, so $p(\lambda )={\lambda }^2+1$, whose roots are $\pm i$, no real eigenvalues at all, which the picture predicts: a quarter turn leaves no real direction fixed.

Over $\mathbb{C}$ it is diagonalisable. For $\lambda =i$, solving $(C-iI)v=0$ gives $v=(1,-i)$, and

since $i(-i)=1$. The other eigenvalue $-i$ has eigenvector $(1,i)$, the entrywise conjugate. That pairing is general: a real matrix has real characteristic polynomial, so its non-real roots come in conjugate pairs, with conjugate eigenvectors.

Singular: zero as an eigenvalue. has $detS=0$, so $0$ is an eigenvalue with $E_0=\ker S=\mathrm{span}\{(2,-1)\}$; indeed $S(2,-1{)}^{\mathsf{T}}=(0,0{)}^{\mathsf{T}}$. The trace is 5, so the other eigenvalue is 5, with eigenvector $(1,2)$: $S(1,2{)}^{\mathsf{T}}=(5,10{)}^{\mathsf{T}}=5(1,2{)}^{\mathsf{T}}$.

Eigenvalues can be read off structure (triangular), can exist without enough eigenvectors (defective), can require a larger field (rotation), and can be zero (singular). Only the second is an obstruction that no change of setting removes.

Subtracting $\lambda$ from every entry. Writing $A-\lambda$ rather than $A-\lambda I$ produces a matrix unrelated to the problem, and a characteristic polynomial of the wrong degree. Only the diagonal is affected.

Accepting the zero vector as an eigenvector. After reducing $(A-\lambda I)v=0$, reporting $v=0$ as the answer. That solution always exists and says nothing; the eigenspace is the kernel, and the eigenvectors are its nonzero elements. If reduction produced no free variable, the eigenvalue or the arithmetic is wrong, $A-\lambda I$ is singular by construction.

Reporting one vector as "the" eigenvector. Every nonzero multiple of an eigenvector is an eigenvector, and an eigenspace of dimension 2 has a whole plane of them. The answer is a basis for $E_{\lambda }$, not a representative.

Concluding non-diagonalisability from a repeated root. Algebraic multiplicity above 1 does not by itself mean defective: the identity matrix has $\lambda =1$ with algebraic multiplicity $n$ and geometric multiplicity $n$, and is already diagonal. The verdict needs the geometric multiplicity, which means solving for the eigenspace rather than stopping at the polynomial.

Concluding a real matrix has no eigenvalues. A rotation has no real eigenvalues, which is a statement about the field rather than about the matrix. Over $\mathbb{C}$ every $n\times n$ matrix has $n$ eigenvalues with multiplicity, because a degree-$n$ polynomial always splits there. Saying "no eigenvalues" without naming the field leaves out the part that decides it.

The first is mechanical. The next two mistake one solution for the solution set. The last two draw a conclusion from insufficient evidence. A repeated root without the eigenspace, or an absence of roots without the field. Each produces a confident answer that is wrong in a different way, and only the first is visible in the arithmetic.