Prove or Disprove: a 3x3 matrix A can have 0 as a eigenvalue

[nicknaq]

Homework Statement

Prove or Disprove: a 3x3 matrix A can have 0 as a eigenvalue

Homework Equations

(xI-A)=0

The Attempt at a Solution

I believe it's false just because I've never seen it. I have no idea how to prove it.

 

[D H]

All you need is one example of a 3x3 matrix with zero as an eigenvalue. Hint: Finding one is a trivial task.

 

[nicknaq]

All you need is one example of a 3x3 matrix with zero as an eigenvalue. Hint: Finding one is a trivial task.

Any 3x3 matrix with a column of zeros? A homogeneous system?

 

[jbunniii]

Do you know any types of matrices for which you can immediately determine all the eigenvalues by inspection?

 

[nicknaq]

Do you know any types of matrices for which you can immediately determine all the eigenvalues by inspection?

The identity matrix, the zero matrix and triangular matrices I suppose?

 

[jbunniii]

The identity matrix, the zero matrix and triangular matrices I suppose?

OK, let's look at the first two. What are the eigenvalues of those matrices?

 

[nicknaq]

OK, let's look at the first two. What are the eigenvalues of those matrices?

For the identity matrix the eigenvalue will be 1. For the zero matrix the eigenvalue will be 0.

 

[Hurkyl]

Any 3x3 matrix with a column of zeros? A homogeneous system?

Could be. Can you come up with a specific example and check it?

 

[nicknaq]

WOOPS!
I forgot one vital piece of information guys! I'm sorry.

The matrix must be invertible.

So that rules out the zero matrix or any matrix with a column/row of zeros.

 

[jbunniii]

WOOPS!
I forgot one vital piece of information guys! I'm sorry.

The matrix must be invertible.

So that rules out the zero matrix or any matrix with a column/row of zeros.

That changes everything!

To see if such a matrix can exist, consider the definition: $\lambda$ is an eigenvalue of $M$ if there is a nonzero vector $x$ such that

$$Mx = \lambda x$$

If $\lambda = 0$ is an eigenvalue of $M$, what does that imply?

 

[nicknaq]

That changes everything!

To see if such a matrix can exist, consider the definition: $\lambda$ is an eigenvalue of $M$ if there is a nonzero vector $x$ such that

$$Mx = \lambda x$$

If $\lambda = 0$ is an eigenvalue of $M$, what does that imply?

$$Mx$$=0 so there's only the trivial solution. However x cannot be zero so there's no solution? Is this my proof or is there more to it?

 

[jbunniii]

$$Mx$$=0 so there's only the trivial solution. However x cannot be zero so there's no solution? Is this my proof or is there more to it?

I'm not sure - your wording is a bit unclear.

If 0 is an eigenvalue, then $Mx = 0$ for some nonzero $x$. This is possible if and only if $M$ is singular (not invertible). Do you see why?