Eigenvectors - why must det(A-λI)=0?

[RK1992]

if we say that an eigenvector u is a vector whose direction is unchanged by the matrix transformation A:
Au=λu=λIu

rearranging:
Au - λIu = 0
u(A-λI)=0

we know the u=0 is obviously a solution. and i also appreciate that if A-λI=0 then A=λI which is just a scalar multiplied by the identity so is also trivial.

but in class, my teacher said "so obviously, the other solution is when the matrix (A-λI) is singular" but didnt justify it and i don't see why its "obvious" that a matrix being singular means that it must satisfy that equation.

can anyone explain why det(A-λI)=0 is necessary? i am happy with using it but i don't understand why it must work and i find it frustrating not knowing why it does.

BTW we have never learned about vector spaces and things like that, is it possible to explain it without these? i looked on khan academy and they exxplained it using vector spaces and linear independence and that's not on my a level further maths syllabus :/

thanks for reading :)

 

[D H]

The purpose here is to find solutions other than the trivial solution u=0. As you noted, the case A=λI is also not that interesting. Defining B=A-λI, we have the curious case Bu=0 when neither B nor u are zero. Note that if B and u were scalars there are no such solutions.

With B being a matrix and u a vector there are (or can be) non-trivial solutions. Given some matrix B (i.e., not just B=A-λI), all nontrivial solutions to Bu=0 lie in the null space of B. The matrix B only has a non-empty null space if the matrix B is singular -- i.e., if its determinant is zero.

 

[RK1992]

The purpose here is to find solutions other than the trivial solution u=0. As you noted, the case A=λI is also not that interesting. Defining B=A-λI, we have the curious case Bu=0 when neither B nor u are zero. Note that if B and u were scalars there are no such solutions.


With B being a matrix and u a vector there are (or can be) non-trivial solutions. Given some matrix B (i.e., not just B=A-λI), all nontrivial solutions to Bu=0 lie in the null space of B. The matrix B only has a non-empty null space if the matrix B is singular -- i.e., if its determinant is zero.

okay, thanks for that. is it weird for my maths class to be learning about eigenvectors when we have no clue about null spaces? it seems like the Bu=0 where B,u ≠ 0 problem can only be explained with null spaces so it seems odd to try and learn eigenvectors before learning about null spaces :S

 

[Moo Of Doom]

The important thing is that non-singular matrices represent invertible transformations. Thus, if B is non-singular, then Bu = v has a unique solution (i.e. u = B^-1v). Since we know B0 = 0, there can be no other solutions to Bu = 0 unless B is singular.

 

[RK1992]

The important thing is that non-singular matrices represent invertible transformations. Thus, if B is non-singular, then Bu = v has a unique solution (i.e. u = B^-1v). Since we know B0 = 0, there can be no other solutions to Bu = 0 unless B is singular.

sorry for this silly question..
B= A-λI
u is an eigenvector

but what is v?

 

[AlephZero]

If you haven't studied vector spaces etc, think about solving linear equations instead. You probably learned about the conditions for a unique solution, multiple solutions, and no possible solutions to a set of equations.

If the solution to u(A-λI)=0 is unique, then it is zero, which isn't very interesting. What is interesting is the special values of λ where there are non-zero solutions.

In applications to physics and engineering, A will be a matrix representing some properties of a system, and the eigenvalues are the special values of some parameter when "something interesting happens". For example, if you are studying vibrations of an object, the eigenvalues tell you the frequencies at which it can vibrate, and the corresponding eigenvectors tell you the different shapes of the vibration at each frequency.

 

[RK1992]

If you haven't studied vector spaces etc, think about solving linear equations instead. You probably learned about the conditions for a unique solution, multiple solutions, and no possible solutions to a set of equations.

If the solution to u(A-λI)=0 is unique, then it is zero, which isn't very interesting. What is interesting is the special values of λ where there are non-zero solutions.

In applications to physics and engineering, A will be a matrix representing some properties of a system, and the eigenvalues are the special values of some parameter when "something interesting happens". For example, if you are studying vibrations of an object, the eigenvalues tell you the frequencies at which it can vibrate, and the corresponding eigenvectors tell you the different shapes of the vibration at each frequency.

so if Bu=0

then there can be 2 solutions:
u=0 (uninteresting)
or B=0 (also uninteresting)
if B is invertible, then B^-1Bu=B^-10 so u=B^-10=0 which is equivalent to before

i just don't see the logic that says that that says that interesting answers must exist if B^-1 doesn't exist

can it not be proven unequivocally that detB=0 gives non trivial results without using vector spaces? if i want to be fully convinced, will i need to teach myself about vector spaces so that it becomes clearer?

 

[AlephZero]

so if Bu=0

then there can be 2 solutions:
u=0 (uninteresting)
or B=0 (also uninteresting)

Careful!
A is a given (fixed) matrix.
So A-λI can not equal 0, except for the special case where A is a multiple of I.

i just don't see the logic that says that that says that interesting answers must exist if B-1 doesn't exist

Here is an argument usiing equation solving.

Solve the n equations Bu = 0 systematically using Gaussian elimination.
Choose one of the variables, say u1, and eliminate it to give you n-1 equations in the n-1 variables u2, u3, ... un.
Repeat this till you have one equation in one variable, say
c un = 0.

If B is non-singular, then c will be non-zero, so you get un = 0, and then you can back-substitute for get un-1 = ... = u2 = u1 = 0.

If B is singular, the forward elimination will end up with the equation 0 un = 0.
In this case you can choose any non-zero value you like for un, and back-substitute to get a non-zero solution of Bu = 0.

In fact, forward elimination may get to a point where there are several equations left and all the coefficients on the left hand side are zero. In that case, you can choose any values for ALL the variables in the remaining equations, and back substitute.

 

[lurflurf]

If B^-1 exists
to solve Bx=0 we have
x=B^-10=0
Not interesting.
We are looking for solutions other than 0

 

[D H]

so if Bu=0

then there can be 2 solutions:
u=0 (uninteresting)
or B=0 (also uninteresting)

You are missing an entire class of solutions that arise when B is not equal to zero but also is not invertible.

 

[Deveno]

suppose B is invertible. if Bu = v, then u = Iu = (B^-1B)u = B^-1(Bu) = B^-1(v).

so you would expect B(V) to look "just like" V, to be the same dimension, for example (if B is invertible it must be a 1-1 function, and is thus bijective from V to B(V) (don't worry about knowing what a "vector space" is. all you need to know is that u is a vector (nx1 matrix, if you like), and that Bu is another vector (another nx1 matrix, since B is square). so B is just a function (yes, matrices can be functions) that takes the vector u, and spits out another vector Bu (which is a matrix product)).

well, if B is not invertible, then B must not be injective. and if B(u1) = B(u2), then

B(u1-u2) = 0 whether or not u1-u2 = 0 or not. the point being, any non-invertible matrix will map some non-zero vectors to 0.

the flip side of that coin, is that an invertible matrix will map ONLY 0 to 0 (because it's 1-1, because it has an inverse). so if

Au = λu,

then (A - λI)u = 0.

if u is not zero, then A - λI HAS to be singular. non-singular (that is, invertible) matrices always only map the 0-vector to 0 (how could an invertible matrix map 2 elements to 0? if it did, how would you define the inverse? it would have 2 images).

and if A - λI is singular, det(A - λI) = 0.

the point is, if a matrix has any eigenvectors at all (remember, the 0-vector doesn't count), then det(A - λI) = 0.

here's a simple example: suppose we look at the identity matrix I. now the identity matrix has EVERY vector as an eigenvector, because Iv = 1v, for ALL v.

and what is I - I? it's the 0-matrix, which is most definitely singular.

another example, suppose A is the matrix:

[a 0]
[0 b]

it should be clear that (1,0) is an eigenvector corresponding to the eigenvalue a. similarly, (0,1) is an eigenvector corresponding to the eigenvalue b. suppose we look at A - λI for each value of λ. A - aI =

[0 .0. ]
[0 b-a]

while A - bI =

[a-b 0]
[.0.. 0]

clearly, both these matrices are singular. for completeness' sake, let's look at det(A - λI) = (a - λ)(b - λ). see how the roots of this polynomial are a and b?

the following things are equivalent, for a matrix A:

1. A is singular
2. det(A) = 0
3. Au = 0 with u ≠ 0
4. A is not invertible
5. A is not 1-1 from dom(A) to range(A)

the following are also equivalent, for a matrix A:

A. A is invertible
B. det(A) ≠ 0
C. Au = 0 --> u = 0
D. A is a bijection from dom(A) to range(A)

if any of 1,2,3,4 or 5 hold, NONE of A,B,C or D hold, and vice-versa.