What eigenvectors are and why you need them separately

An eigenvector is a direction (a column of numbers) that doesn't change direction when you multiply it by a matrix—it only gets stretched or shrunk. An eigenvalue is the number that tells you how much the eigenvector gets stretched. Once you have found an eigenvalue, you cannot read off its eigenvector from that number alone. You have to solve a system of linear equations to find it.

The process is mechanical: for each eigenvalue you found, you substitute it back into the matrix equation and solve for the vector that makes the equation true. This section walks you through the actual steps, using a concrete example so you can follow along with your own matrix.

Key Takeaways

  • For each eigenvalue λ, you solve the equation (A − λI)v = 0, where A is your matrix, I is the identity matrix, and v is the eigenvector you are looking for.
  • Subtract λ from the diagonal entries of your matrix, then row-reduce the result to find which vectors satisfy the equation.
  • The eigenvector is not unique—any nonzero multiple of a solution vector is also an eigenvector for that eigenvalue.
  • If an eigenvalue has multiplicity greater than 1, you may find multiple independent eigenvectors, or you may find only one.

The formula: (A − λI)v = 0

Start with your matrix A and an eigenvalue λ that you have already computed. Create a new matrix by subtracting λ from each diagonal entry of A. This new matrix is called (A − λI). Then solve the equation (A − λI)v = 0 for the vector v.

The vector v you are looking for is in the null space of (A − λI)—the set of all vectors that become zero when you multiply them by that matrix. To find it, row-reduce (A − λI) to row echelon form, identify the free variables, and write the solution as a linear combination of basis vectors.

Here is a worked example. Suppose your matrix is:

A = [4 1] [2 3]

And suppose you have already found that λ = 5 is an eigenvalue. Now compute A − 5I:

A − 5I = [4−5 1 ] = [−1 1] [2 3−5] [2 −2]

Now solve (A − 5I)v = 0. Row-reduce the matrix [−1 1 | 0; 2 −2 | 0]. Multiply the first row by −1 to get [1 −1 | 0; 2 −2 | 0]. Subtract 2 times the first row from the second to get [1 −1 | 0; 0 0 | 0]. The second row is all zeros, so there is one free variable. Set the second entry of v equal to t (a free parameter). Then the first entry must equal t as well, because the first row says 1·v₁ − 1·v₂ = 0. So v = [t; t] = t[1; 1]. Any nonzero multiple of [1; 1] is an eigenvector for λ = 5. Usually you write the answer as v = [1; 1].

Step-by-step process for a 3×3 matrix

Larger matrices follow the same logic but require more arithmetic. Here is the process laid out in order.

Step 1: Write down A − λI. Take your 3×3 matrix A and subtract λ from the three diagonal entries. You now have a 3×3 matrix with λ removed from the main diagonal.

Step 2: Row-reduce to echelon form. Use elementary row operations (swap rows, multiply a row by a nonzero constant, add a multiple of one row to another) to get the matrix into row echelon form. Do not go all the way to reduced row echelon form unless you want to—echelon form is enough to identify free variables.

Step 3: Identify pivot and free variables. A pivot variable corresponds to a column with a leading entry (the first nonzero entry in a row). A free variable corresponds to a column with no leading entry. If you have three columns and two pivots, you have one free variable.

Step 4: Write the general solution. Express each pivot variable in terms of the free variables. If you have one free variable, set it equal to a parameter t and write out the solution vector in terms of t.

Step 5: Extract the eigenvector. Factor out the parameter t. The vector you are left with (the coefficient of t) is your eigenvector. You can scale it by any nonzero constant and still have an eigenvector.

What to do when an eigenvalue has multiplicity greater than 1

An eigenvalue has multiplicity greater than 1 if it appears more than once as a root of the characteristic polynomial. For example, if the characteristic polynomial is (λ − 2)²(λ − 5), then λ = 2 has multiplicity 2.

When you solve (A − λI)v = 0 for a repeated eigenvalue, you may find one eigenvector, or you may find two or more independent eigenvectors. The number of independent eigenvectors you find is called the geometric multiplicity of the eigenvalue. If the geometric multiplicity is less than the algebraic multiplicity (the multiplicity in the characteristic polynomial), the matrix is not diagonalizable, but you can still find the eigenvectors that do exist by following the steps above.

For example, suppose λ = 2 has multiplicity 2, and when you row-reduce (A − 2I) you find two free variables. Then you have two independent eigenvectors for λ = 2. Set the first free variable to 1 and the second to 0, solve for the pivot variables, and write down the first eigenvector. Then set the first free variable to 0 and the second to 1, and write down the second eigenvector. Both are eigenvectors for λ = 2.

Common mistakes and how to avoid them

The most common error is forgetting to subtract λ from the diagonal. You must compute A − λI, not just A. If you skip this step, you will solve the wrong equation and get the wrong answer.

Another frequent mistake is stopping the row reduction too early. You need to reduce all the way to echelon form (or reduced echelon form) so that you can clearly see which variables are free. If you stop partway through, you may miss a free variable and find only part of the solution space.

A third error is confusing the eigenvector with the eigenvalue. The eigenvalue is a single number. The eigenvector is a column of numbers. They are not the same thing, and you cannot substitute one for the other.

Finally, remember that eigenvectors are not unique. If v is an eigenvector, so is 2v, 3v, −v, and any other nonzero scalar multiple. Different textbooks and software may scale eigenvectors differently (for example, to have length 1, or to have the first entry equal to 1). As long as your vector satisfies (A − λI)v = 0 and is nonzero, it is a correct eigenvector.

Checking your work

Once you have found an eigenvector v for eigenvalue λ, verify it by computing Av and checking that Av = λv. Multiply your matrix A by your eigenvector v. Then multiply λ by v. If the two results are equal, your eigenvector is correct. If they are not equal, you made an arithmetic error somewhere—go back and check your row reduction.

This check is fast and catches most mistakes. It also builds confidence that you have the right answer before you move on to the next eigenvalue.

Frequently Asked Questions

Can a matrix have more eigenvectors than eigenvalues?

Yes. Each eigenvalue can have multiple independent eigenvectors. For example, a 3×3 identity matrix has only one eigenvalue (λ = 1), but every nonzero vector in three-dimensional space is an eigenvector for that eigenvalue. In general, the number of independent eigenvectors for an eigenvalue equals the dimension of the null space of (A − λI).

What if (A − λI) row-reduces to the zero matrix?

That means every vector is in the null space, so every nonzero vector is an eigenvector for that eigenvalue. This happens when the eigenvalue has high multiplicity and the matrix has a special structure. You can choose any basis for the space and use those vectors as your eigenvectors.

Do I have to scale eigenvectors to have length 1?

No. Scaling is optional and depends on what you need the eigenvectors for. Some applications (like principal component analysis) use unit eigenvectors. Others do not require scaling. Any nonzero scalar multiple of an eigenvector is still an eigenvector, so use whichever scaling is most convenient for your problem.

What if I get a complex eigenvector?

Complex eigenvectors arise when you have complex eigenvalues, which happen when the characteristic polynomial has complex roots. The process is the same: substitute the complex eigenvalue into (A − λI)v = 0 and solve. Your eigenvector will have complex entries. If you are working with a real matrix and want real eigenvectors only, you can extract them from complex conjugate pairs, but that is a more advanced topic.