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# Example solving for the eigenvalues of a 2x2 matrix

Example solving for the eigenvalues of a 2x2 matrix. Created by Sal Khan.

## Want to join the conversation?

• At the end of the video Sal gets the factors (λ-5)(λ+1), but says that the eigenvalues are 5 and -1. Why did the signs change?
(1 vote)
• To find the eigenvalues you have to find a characteristic polynomial P which you then have to set equal to zero. So in this case P is equal to (λ-5)(λ+1). Set this to zero and solve for λ. So you get λ-5=0 which gives λ=5 and λ+1=0 which gives λ= -1
• Hi Sal, in my notes I have the characteristic equation : | A-hI |= 0 which is the reverse of yours. It doesn't seem to matter in this case which way they go A-hI or hI-A do you know if this is always the case?
• Why are there only two eigenvalues for this matrix?

For any reflection transformation surely there are infinite eigenvalues, because all of the vectors along the line of reflection would not be changed, nor would those orthogonal to it. And there are infinitely many vectors along these lines, thus infinitely many eigenvalues for these vectors.

It seems weird to me that there would only be two eigenvalues for a transformation - I would have thought there would either be 0 (i.e. rotation) or infinite.

Is there some limit to eigenvectors (thus eigenvalues) along a line? i.e. can they only be unit vectors before the transformation?

This is ignoring imaginary eigenvalues of course.
• We only count eigenvectors as separate if one is not just a scaling of the other. Otherwise, as you point out, every matrix would have either 0 or infinitely many eigenvectors. And we can show that if v and cv (for some scalar c) are eigenvectors of a matrix A, then they have the same eigenvalue.

Suppose vectors v and cv have eigenvalues p and q. So
Av=pv, A(cv)=q(cv)
A(cv)=c(Av). Substitute from the first equation to get A(cv)=c(pv)
So from the second equation, q(cv)=c(pv)
(qc)v=(cp)v
Since v is an eigenvector, it cannot be the 0 vector, so qc=cp, or q=p. The eigenvalues are the same.
• What we have do when characteristic equation does not have any roots?
• When working with only the real numbers, you are not guaranteed roots, so you could factor the characteristic polynomial as much as possible, until it is no longer factor-able.
However, the Fundamental Theorem of Algebra tells us that if we shift our perspective from real eigenvalues to complex eigenvalues (i.e. eigenvalues with imaginary components), then our characteristic polynomial is guaranteed to have all roots. By all roots, I mean that an nth degree characteristic polynomial has exactly n complex roots.

An extension/generalization to this answer (if you know some field theory), is if your matrix A is nxn in a field F, then since every field F has an algebraic closure F', if we work in the field F', A has n eigenvalues. In other words, the characteristic polynomial under F' has n roots.
• Can you tell me the basic applications of eigenvalues and eigenvectors?
• Machine Learning. Advanced use case but worthy of a mention.
• what is trivial and non trivial?
• If a solution is trivial, there is only one solution. For example, we have equation y=x. If y=0, then x=0.

Non-trivial refers to many possible solutions. Look at y=sin(x). For y=0, there is more than one solution. In fact, there is an infinite amount of solutions: x=0, pi, 2*pi, 3*pi...n*pi
• Sum of eigenvalues is equal to trace of matrix
• how is this λI matrix with 1's and 0's generated? (I mean how I determine when to put 1 or 0)
(1 vote)
• It's the identity matrix multiplied by λ. So the 1's are all on the diagonal from top-left to bottom-right.