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Why $\det(A - \lambda I) = 0$ is the eigenvalue equation, the trace and determinant as its coefficients, triangular matrices that give their eigenvalues away, and why similar matrices share it.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to turn $Av = \lambda v$ into $\det(A - \lambda I) = 0$ and say why the identity matrix has to appear, form $A - \lambda I$ by changing the diagonal alone, write the characteristic polynomial of a $2 \times 2$ matrix straight from its trace and determinant, read the eigenvalues of a triangular matrix off its diagonal, use the sum and product identities as a check on any eigenvalue calculation, explain why a rotation has no real eigenvalue, and show that similar matrices share a characteristic polynomial.
A square matrix is singular — it sends some non-zero vector to zero — exactly when its determinant is zero. And $\det(P^{-1}MP) = \det M$ for every invertible $P$, because determinants multiply and $\det(P^{-1})\det(P) = 1$. Those two sentences are the whole of what this lesson borrows; everything else in it is new.
A number $\lambda$ is an eigenvalue of the square matrix $A$ when $Av = \lambda v$ for some $v \ne 0$; such a $v$ is an eigenvector. The polynomial $\chi_A(t) = \det(A - tI)$ is the characteristic polynomial, and its roots are exactly the eigenvalues. The number of times a root is repeated in $\chi_A$ is its algebraic multiplicity. The set of eigenvalues is the spectrum. Two matrices are similar when $B = P^{-1}AP$ for an invertible $P$ — the same map, written in a different basis.
The definition $Av = \lambda v$ asks for two unknowns at once, a number and a vector, and it is not obvious how to search for either. One rearrangement removes the vector from the search:
$$Av = \lambda v \iff Av - \lambda v = 0 \iff (A - \lambda I)v = 0.$$
The middle step is where the $I$ earns its place. $Av - \lambda v$ is a matrix times a vector minus a number times a vector, and those cannot be factored until the number is written as a matrix: $\lambda v = (\lambda I)v$. Then $A - \lambda I$ is one matrix, and the question has become does this matrix kill a non-zero vector?
A square matrix kills a non-zero vector exactly when it is singular, and it is singular exactly when its determinant vanishes. So:
$$\lambda \text{ is an eigenvalue of } A \iff \det(A - \lambda I) = 0.$$
The left-hand side is a polynomial in $\lambda$ of degree $n$ — expanding the determinant of an $n \times n$ matrix whose diagonal carries $-\lambda$ can produce nothing else — so an $n \times n$ matrix has at most $n$ eigenvalues, and exactly $n$ if you count multiplicities and allow complex roots.
For a $2 \times 2$ matrix the polynomial is worth memorising, because both of its coefficients can be read off the matrix without any expansion at all:
$$\chi_A(t) = t^{2} - (\operatorname{tr} A)\,t + \det A.$$
Another way: picture
An eigenvector is a direction the map does not turn. Draw a vector, apply $A$, and draw the result: usually the arrow has swung round as well as changed length. An eigenvector is one of the rare starting arrows whose image lies along the same line through the origin — the map may stretch it, shrink it, or flip it end for end, and that is all. $\det(A - \lambda I) = 0$ is the algebraic test for such a line existing at the particular stretch factor $\lambda$.
Another way: steps
Suppose $\chi_A$ factors completely as $(t - \lambda_1)\cdots(t - \lambda_n)$. Multiplying that out and comparing coefficients with the expansion of $\det(A - tI)$ gives two identities that hold for every square matrix:
$$\lambda_1 + \lambda_2 + \dots + \lambda_n = \operatorname{tr} A, \qquad \lambda_1 \lambda_2 \cdots \lambda_n = \det A.$$
Both are worth having. They are a free check on any eigenvalue calculation — add your answers and compare with the diagonal sum — and for a $2 \times 2$ they are often faster than solving anything, because the eigenvalues are just the pair with this sum and this product. A matrix with trace $7$ and determinant $12$ has eigenvalues $3$ and $4$, found by looking.
They also explain a fact that otherwise looks like a coincidence: a matrix is singular exactly when $0$ is one of its eigenvalues. The determinant is the product of the eigenvalues, and a product is zero exactly when one factor is.
A triangular matrix gives its eigenvalues away completely. Subtracting $t$ from the diagonal leaves it triangular, the determinant of a triangular matrix is the product of the diagonal, so $\chi_A(t) = (a_{11} - t)(a_{22} - t)\cdots(a_{nn} - t)$ and the eigenvalues are the diagonal entries. Everything off the diagonal is irrelevant to them — though not, as the next lesson shows, to the eigenvectors.
If $B = P^{-1}AP$ then
$$\chi_B(t) = \det(P^{-1}AP - tI) = \det\big(P^{-1}(A - tI)P\big) = \det(P)^{-1}\det(A - tI)\det(P) = \chi_A(t),$$
using $tI = P^{-1}(tI)P$ in the second step. So similar matrices have the same characteristic polynomial, hence the same eigenvalues with the same algebraic multiplicities, the same trace and the same determinant.
That is exactly what should happen. $A$ and $P^{-1}AP$ are the same linear map written in two bases, and an eigenvalue is a fact about the map: a stretch factor along some fixed direction. Changing coordinates renames the direction and cannot change the stretch. Trace and determinant, which looked like arithmetic done on a grid of numbers, turn out to be properties of the map itself.
The converse fails, and it is worth knowing where. $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ both have characteristic polynomial $(t - 1)^{2}$ and they are not similar — the first is the identity, which is $P^{-1}IP$ for every $P$ and so is similar to nothing but itself. The characteristic polynomial is an invariant, not a complete one.
$A - \lambda I$ is not $A$ with $\lambda$ subtracted from every entry. $\lambda I$ is zero off the diagonal. Subtracting it from a $2 \times 2$ changes two entries, not four.
Solve for $t$, do not just expand. $\det(A - tI)$ is a polynomial; the eigenvalues are its roots. Handing in the polynomial is handing in the question in a different form.
A real matrix need not have a real eigenvalue. A rotation moves every direction, so there is nothing for it to fix, and its characteristic polynomial duly has negative discriminant. The theorem that an $n \times n$ matrix has $n$ eigenvalues is a theorem about the complex numbers.
A repeated root is one eigenvalue, not two. The spectrum $\{2, 5\}$ is a set of two numbers whether or not $2$ is a double root. The repetition is recorded as an algebraic multiplicity, and comparing it with the dimension of the eigenspace is the next lesson's business.
$A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$. Trace $7$, determinant $12 - 2 = 10$.
Two numbers read off, no expansion.
So $\chi_A(t) = t^{2} - 7t + 10 = (t - 5)(t - 2)$.
Which pair has sum 7 and product 10?
Eigenvalues $5$ and $2$. Check: $5 + 2 = 7$ and $5 \times 2 = 10$, matching the diagonal sum and the determinant.
The check costs one line and catches sign errors.
$B = \begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}$ is triangular, so $\chi_B(t) = (3 - t)^{2}$ straight away.
Triangular gives it away.
The only eigenvalue is $3$, with algebraic multiplicity $2$. The spectrum is the one-element set $\{3\}$.
A repeated root is still one number.
$C = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ has exactly the same characteristic polynomial and behaves completely differently: every vector is an eigenvector of $C$, and only the multiples of $(1, 0)$ are eigenvectors of $B$.
The polynomial does not see the difference.
Trace $5 + 1 = 6$; determinant $5 \times 1 - 6 \times 2 = -7$.
Read both off the matrix.
So $\chi(t) = t^{2} - 6t - 7$, and we want the pair with sum $6$ and product $-7$.
Sum and product, not expansion.
$(t - 7)(t + 1)$, so the eigenvalues are $7$ and $-1$; they sum to $6$ and multiply to $-7$, as they must.
For $A = \begin{pmatrix} 5 & 8 \\ 3 & 9 \end{pmatrix}$, write the matrix $A - 2I$.
This task has no paper form; do it on a device.
Write the characteristic polynomial $\det(A - tI)$ of $A = \begin{pmatrix} 9 & -3 \\ 6 & 0 \end{pmatrix}$ as a polynomial in $t$.
Answer:
$A$ is a $2 \times 2$ matrix with eigenvalues $3$ and $4$, and $B = P^{-1}AP$ for some invertible $P$. Fill in the trace and the determinant of each.
| Trace | Determinant | |
|---|---|---|
| $A$ | ||
| $B = P^{-1}AP$ |
Match each triangular matrix to its pair of eigenvalues.
| $9$ and $8$ | $9$ and $5$ | $8$ and $5$ | |
|---|---|---|---|
| $\begin{pmatrix} 9 & 3 \\ 0 & 8 \end{pmatrix}$ | |||
| $\begin{pmatrix} 9 & 0 \\ 3 & 5 \end{pmatrix}$ | |||
| $\begin{pmatrix} 8 & 3 \\ 0 & 5 \end{pmatrix}$ |
How many real eigenvalues has $R = \begin{pmatrix} 5 & -6 \\ 6 & 5 \end{pmatrix}$?
Write the set of eigenvalues of $\begin{pmatrix} 9 & 7 & 5 \\ 0 & 9 & 3 \\ 0 & 0 & 1 \end{pmatrix}$.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $A = \begin{pmatrix} 9 & 8 \\ 2 & 4 \end{pmatrix}$, write the matrix $A - 5I$.
This task has no paper form; do it on a device.
You can form $A - \lambda I$, write and solve the characteristic polynomial, and check your roots against the trace and the determinant. Say in your own words why the identity matrix has to appear in $A - \lambda I$ at all. Next: the vectors themselves, and how large a space of them each eigenvalue has.
10. Your turn: the eigenvalues of $\begin{pmatrix} 5 & 6 \\ 2 & 1 \end{pmatrix}$, step 3
Check against the trace and determinant.