Check if a scalar is an eigenvalue of a matrix

[danielpanatha]

Homework Statement

We have a matrix Anxn (different than the identity matrix I) and a scalar λ=1. We want to check if λ is an eigenvalue of A.

Homework Equations

As we know, in order for λ to be an eigenvalue of A, there has to be a non-zero vector v, such that Av=λv

The Attempt at a Solution

Av=λv
Av=1v
Av=v
A=I

But we know that A is different than I, so λ is not an eigenvalue of A.
Is my attempt right?

Thanks in advance for your assistance.

 

[Ray Vickson]

Homework Statement

We have a matrix Anxn (different than the identity matrix I) and a scalar λ=1. We want to check if λ is an eigenvalue of A.

Homework Equations

As we know, in order for λ to be an eigenvalue of A, there has to be a non-zero vector v, such that Av=λv

The Attempt at a Solution

Av=λv
Av=1v
Av=v
A=I

But we know that A is different than I, so λ is not an eigenvalue of A.
Is my attempt right?

Thanks in advance for your assistance.

Your last equation is nonsense. From Av = v you cannot conclude that A = I. All you can conclude is that v must be a solution of the homogeneous linear system (A-I)v = 0.

RGV

 

[HallsofIvy]

In particular, the only operations defined on a vector space are addition of vectors and multiplication or division by a scalar. You cannot "divide both sides" by a vector.

(If Av=Bv for all vectors, v, then you can conclude that A= B. But not if Av= Bv for some vector, v.)

 

[Muphrid]

In other words, you can't divide both sides of the equation by the vector [itex]v[/itex]. Linear operators like [itex]\underline A[/itex] may be written down using matrices and said to operate on vectors through matrix multiplication, but what's really going on is that [itex]A[/itex] is a linear function, which is why trying to factor [itex]v[/itex] from both sides is nonsense.

Eigenvalues satisfy the characteristic equation that [itex]\det (\underline A - \lambda \underline I) = 0[/itex].

(PS. In particular, a linear operator acting on a vector has a general form [itex]\underline A(v) = (v\cdot e_1)a + (v \cdot e_2)b + (v \cdot e_3)c + \ldots{}[/itex] where [itex]a,b,c[/itex] are vectors. Unless you know something about the structure of the operator, dividing both sides by a vector doesn't give you anything useful.)

 

[danielpanatha]

Thank you all for your replies.
One last question:
If we think of this theoretically, in the equation Av=v, there has to be a matrix A that will be multiplied by v and the result will be the same vector v. I understand that we cannot divide both parts of the equation by v, but could you please give me an example with a non-identity matrix A and a non-zero vector v that satisfy the equation Av=v?

 

[Muphrid]

Sure. Consider a linear operator on [itex]\mathbb R^4[/itex].

[tex]\underline A(e_1) = e_1 \\
\underline A(e_2) = -e_2 \\
\underline A(e_3) = e_3 \\
\underline A(e_4) = -e_4[/tex]

Any linear combination of [itex]e_1, e_3[/itex] is an eigenvector of this operator with eigenvalue 1.

 

[danielpanatha]

Sure. Consider a linear operator on [itex]\mathbb R^4[/itex].

[tex]\underline A(e_1) = e_1 \\
\underline A(e_2) = -e_2 \\
\underline A(e_3) = e_3 \\
\underline A(e_4) = -e_4[/tex]

Any linear combination of [itex]e_1, e_3[/itex] is an eigenvector of this operator with eigenvalue 1.

Thank you!

 

[Mark44]

I understand that we cannot divide both parts of the equation by v, but could you please give me an example with a non-identity matrix A and a non-zero vector v that satisfy the equation Av=v?

$$A = \begin{bmatrix}1 & 1\\ 0 & 1 \end{bmatrix}$$
$$\lambda = 1, v = \begin{bmatrix}1 \\ 0 \end{bmatrix}$$

 

[danielpanatha]

$$A = \begin{bmatrix}1 & 1\\ 0 & 1 \end{bmatrix}$$
$$\lambda = 1, v = \begin{bmatrix}1 \\ 0 \end{bmatrix}$$

Thanks!

 

[Chestermiller]

If (A - I)v = 0, what condition needs to be satisfied in order for v not to have the trivial solution v = 0?

Chet

 

[Mark44]

If (A - I)v = 0, what condition needs to be satisfied in order for v not to have the trivial solution v = 0?

Chet

If A - I is invertible, the only solution is v = 0.
If A - I does not have an inverse, then v = 0 is still a solution, but there are also nonzero solutions.

So in both cases, v = 0 is a solution. The only difference is whether that is the unique solution or it is one of an infinite number of solutions.

You can tell whether a square matrix is invertible -- its determinant is nonzero. If an inverse does not exist, the determinant is zero.

 

[Chestermiller]

If A - I is invertible, the only solution is v = 0.
If A - I does not have an inverse, then v = 0 is still a solution, but there are also nonzero solutions.

So in both cases, v = 0 is a solution. The only difference is whether that is the unique solution or it is one of an infinite number of solutions.

You can tell whether a square matrix is invertible -- its determinant is nonzero. If an inverse does not exist, the determinant is zero.

Thanks Mark. That was the response I was trying to elicit from the OP.

Chet

 

[HallsofIvy]

Yes, but you phrased it poorly- by leaving out a single word! You said, "If (A - I)v = 0, what condition needs to be satisfied in order for v not to have the trivial solution v = 0?"

v= 0 is, as Mark44 said, always a solution. What you meant to say was, "If (A - I)v = 0, what condition needs to be satisfied in order for v not to have only the trivial solution v = 0?"

Notice, by the way, that the "eigenvalue" question is "existence and uniqueness" turned on its head. A solution, the trivial solution, always exists for the equation [itex]Av=\lambda v[/itex]. [itex]\lambda[/itex] is an eigenvalue if and only if that solution is NOT unique.