Eigenvalues and eigenvectors: how to diagonalise

In many linear algebra and geometry courses, in the first year of engineering or of a science degree, eigenvalues and diagonalisation are taught between the end of October and November. The exam is in the winter session, eight to fourteen weeks away, but the mid-term tests and partial exams of many courses fall at the end of November or in December, that is four to eight weeks from today. Every university has its own calendar and you should check it in the course syllabus; the point is that this chapter is where theory and calculation meet, and it is worth sorting out now.

The calculation, here, is not the problem: characteristic polynomials and linear systems can be trained. Where students get stuck, especially in the oral exam, is in understanding what the words mean: eigenvector, eigenspace, multiplicity. If you do the calculations without having understood them, you get the last step wrong, the one that decides whether the matrix is diagonalisable.

What $Av=\lambda v$ means

A matrix $A$ of order $n$ acts on the vectors of ${\mathbb{R}}^n$: in general a vector $v$ is rotated, stretched, squashed, and ends up in a different direction. The eigenvectors are the special directions that $A$ does not push off their own line: the vector stays on the same line and is only multiplied by a number.

The number $\lambda$ is the eigenvalue. If $\lambda =3$ the vector is tripled, if $\lambda =-1$ it is flipped, if $\lambda =0$ it is wiped out (and then $v$ lies in the kernel of $A$). The condition $v\ne \vec{0}$ is not a detail: the zero vector satisfies $A\vec{0}=\lambda \vec{0}$ for any $\lambda$, so if it were allowed every number would be an eigenvalue and the definition would say nothing.

The procedure in three steps

Step 1: characteristic polynomial. $Av=\lambda v$ is equivalent to $(A-\lambda I)v=\vec{0}$. This homogeneous system has non-zero solutions only if the matrix $A-\lambda I$ is singular, that is, if

Expanding the determinant gives a polynomial of degree $n$ in $\lambda$, the characteristic polynomial.

Step 2: eigenvalues. The roots of the polynomial are the eigenvalues. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root.

Step 3: eigenspaces. For each eigenvalue $\lambda$ you solve $(A-\lambda I)v=\vec{0}$. The set of solutions is the eigenspace $\ker (A-\lambda I)$, and its dimension is the geometric multiplicity. Since $\lambda$ is a root of the characteristic polynomial, the matrix $A-\lambda I$ is singular and the eigenspace has dimension at least 1: if you find only the zero vector, you have made a mistake in a calculation.

Two free checks on the eigenvalues you have found: their sum equals the trace of $A$ (the sum of the entries on the diagonal) and their product equals $\det A$, each counted with its multiplicity. I always use them, even in the exam, because they cost ten seconds.

When the matrix is diagonalisable

$A$ is diagonalisable if there is an invertible matrix $P$ such that

with $D$ diagonal. The columns of $P$ are eigenvectors of $A$ and the entries on the diagonal of $D$ are the corresponding eigenvalues, in the same order. Put equivalently: $A$ is diagonalisable if and only if in ${\mathbb{R}}^n$ (or ${\mathbb{C}}^n$) there is a basis made of eigenvectors.

The criterion to know by heart: $A$ is diagonalisable if and only if for every eigenvalue the geometric multiplicity equals the algebraic multiplicity. You always have

and this gives two shortcuts:

• if the $n$ eigenvalues are all distinct, each eigenvalue has algebraic multiplicity 1, hence geometric multiplicity 1 too: the matrix is diagonalisable with no further calculation;
• a simple eigenvalue never creates problems; the check is needed only for repeated eigenvalues.

What is diagonalising for? Because with $A=PD{P}^{-1}$ powers become easy: $A^k=P{D}^k{P}^{-1}$, and raising a diagonal matrix to a power means raising its entries to that power. It is the technique used to solve recurrences and systems of linear differential equations, and that is why it comes back in other exams.

Exercise 1: a $3\times 3$ matrix with three distinct eigenvalues

Diagonalise

Characteristic polynomial. I expand $\det (A-\lambda I)$ along the third column, which has only one non-zero entry:

Eigenvalues: $1,4,5$. Check: sum $1+4+5=10$, trace $3+2+5=10$. Product $1\cdot 4\cdot 5=20$, determinant $5(3\cdot 2-1\cdot 2)=20$. Everything adds up.

These are three distinct eigenvalues in dimension 3, so $A$ is diagonalisable. I still need the eigenvectors.

$\lambda =1$. $A-I=\left(\begin{array}{ccc}2& 1& 0\\ 2& 1& 0\\ 1& 0& 4\end{array}\right)$. The equations are $2x+y=0$ and $x+4z=0$. I take $z=1$: $x=-4$, $y=8$. Eigenvector $v_1=(-4,8,1)$.

$\lambda =4$. $A-4I=\left(\begin{array}{ccc}-1& 1& 0\\ 2& -2& 0\\ 1& 0& 1\end{array}\right)$. From $-x+y=0$ and $x+z=0$ it follows that $y=x$, $z=-x$. Eigenvector $v_2=(1,1,-1)$.

$\lambda =5$. $A-5I=\left(\begin{array}{ccc}-2& 1& 0\\ 2& -3& 0\\ 1& 0& 0\end{array}\right)$. The third row gives $x=0$, then the first gives $y=0$, and $z$ is free. Eigenvector $v_3=(0,0,1)$.

Check. $A{v}_1=(3\cdot (-4)+8,2\cdot (-4)+2\cdot 8,-4+5)=(-4,8,1)=1\cdot v_1$. $A{v}_2=(3+1,2+2,1-5)=(4,4,-4)=4\cdot v_2$. $A{v}_3=(0,0,5)=5\cdot v_3$. Therefore

with $P^{-1}AP=D$. The columns of $P$ and the entries of $D$ must be in the same order: swapping two of them without swapping the other is a careless mistake that costs you the mark.

Exercise 2: the double eigenvalue that is not enough

Say whether this is diagonalisable

The characteristic polynomial is $(2-\lambda {)}^2$: a single eigenvalue, $\lambda =2$, with algebraic multiplicity 2. Now the eigenspace:

The only equation is $y=0$, with $x$ free. So $\ker (B-2I)=\mathrm{span}\{(1,0)\}$, of dimension 1. The geometric multiplicity (1) is smaller than the algebraic one (2): $B$ is not diagonalisable. There are no two independent eigenvectors, so there is no basis of eigenvectors.

This is the most frequent trap: the characteristic polynomial factorises perfectly, the eigenvalues are integers, and you are tempted to say “diagonalisable” without calculating the eigenspace. With a repeated eigenvalue, checking the geometric multiplicity is not optional.

Compare with $2I=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$: same characteristic polynomial, same double eigenvalue, but here $2I-2I$ is the zero matrix, the eigenspace is all of ${\mathbb{R}}^2$ (dimension 2) and the matrix is, obviously, diagonalisable. Two matrices with the same characteristic polynomial can behave differently, and that is exactly what the criterion measures.

Exercise 3: a symmetric matrix and the spectral theorem

Diagonalise

Trace $=6$, $\det S=2(4-1)-1(2-1)+1(1-2)=6-1-1=4$. I try $(1,1,1)$: $S(1,1,1)=(4,4,4)$, so $\lambda =4$ is an eigenvalue. For the others: sum $6-4=2$ and product $4/4=1$ give $\lambda =1$ with algebraic multiplicity 2.

Now the check on the double eigenvalue. $S-I$ has all entries equal to 1, so it has rank 1 and its kernel has dimension $3-1=2$: it is the plane $x+y+z=0$. Geometric multiplicity $2=$ algebraic $2$, and the matrix is diagonalisable. A basis of the eigenspace is $(1,-1,0)$ and $(1,0,-1)$; together with $(1,1,1)$ for $\lambda =4$ you get the basis of eigenvectors.

Here there was no need to check anything special, because $S$ is symmetric. The spectral theorem says that a real symmetric matrix has all real eigenvalues, is always diagonalisable, and eigenvectors for different eigenvalues are orthogonal. You can verify it at once: $(1,1,1)\cdot (1,-1,0)=0$ and $(1,1,1)\cdot (1,0,-1)=0$. If you normalise an orthogonal basis of the eigenspace (here it is enough to orthogonalise the two vectors of the plane), you obtain an orthogonal matrix $Q$ and $S=QD{Q}^T$ holds, without having to invert anything. In the oral exam the classic question is exactly this: why can the geometric multiplicity of a symmetric matrix never be smaller?

The mistakes I see most

1. A zero eigenvector. Writing the vector $(0,0,0)$ as an eigenvector because the system allows it: you take a non-zero solution, and if the only solution is the zero one, the $\lambda$ you chose is not an eigenvalue.
2. Confusing eigenvalues with pivots. Reducing $A$ to row echelon form and reading the pivots as eigenvalues is wrong: elementary row operations change the eigenvalues. You do the reduction of $A-\lambda I$ to find the eigenspaces, not to find the eigenvalues of $A$.
3. Forgetting the geometric multiplicity. With a repeated eigenvalue, stopping at the characteristic polynomial and declaring the matrix diagonalisable, as in exercise 2.
4. Getting the sign wrong in the determinant. $\det (A-\lambda I)$ has $-\lambda$ on the diagonal: the trace and determinant check almost always catches the mistake.
5. Mixing up $P$ and $D$. Column $j$ of $P$ must correspond to entry $j$ of $D$.

For the part on kernel, image and rank, which you need at every step here (the eigenspace is a kernel), there is rank, kernel and image without mixing them up. For the homogeneous linear systems you solve at each eigenvalue, linear systems: methods and worked examples; if an eigenvalue turns out complex, you need complex numbers explained well. The same eigenvalues then appear in the stability of systems in control systems and in attitude dynamics in attitude dynamics.

Test yourself

When you can do the three exercises by hand without looking, move on to a test with new questions: the linear algebra practice test has exercises of the same type. The correct answer is free; the step-by-step guided solution unlocks with your first lesson.

How I work on this chapter

In my lessons I get you to say the definition of an eigenvector out loud before touching a calculation, then we train the order of the checks: eigenvalues, check with trace and determinant, eigenspaces, comparison of multiplicities. I work online, with a graphics tablet and a shared screen, writing every step as on the blackboard in the oral exam. For the general picture there is the maths page. If you don’t know where to start, write to me: there is a free introductory call to work out where you are losing marks, with no commitment.