D H said:All you need is one example of a 3x3 matrix with zero as an eigenvalue. Hint: Finding one is a trivial task.
jbunniii said:Do you know any types of matrices for which you can immediately determine all the eigenvalues by inspection?
nicknaq said:The identity matrix, the zero matrix and triangular matrices I suppose?
jbunniii said:OK, let's look at the first two. What are the eigenvalues of those matrices?
Could be. Can you come up with a specific example and check it?nicknaq said:Any 3x3 matrix with a column of zeros? A homogeneous system?
nicknaq said: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 said:That changes everything!
To see if such a matrix can exist, consider the definition: [itex]\lambda[/itex] is an eigenvalue of [itex]M[/itex] if there is a nonzero vector [itex]x[/itex] such that
[tex]Mx = \lambda x[/tex]
If [itex]\lambda = 0[/itex] is an eigenvalue of [itex]M[/itex], what does that imply?
nicknaq said:[tex]Mx[/tex]=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?
Yes, a 3x3 matrix can have 0 as an eigenvalue.
We can prove or disprove this by finding the eigenvalues of the matrix using the characteristic equation and solving for the roots. If any of the roots is 0, then the matrix has 0 as an eigenvalue.
Having 0 as an eigenvalue means that the matrix has a non-trivial kernel, or in other words, there exists a non-zero vector that when multiplied by the matrix results in the zero vector.
Yes, a matrix can have multiple 0 eigenvalues. This means that there are multiple linearly independent vectors that result in the zero vector when multiplied by the matrix.
No, if a matrix has 0 as an eigenvalue, it is not invertible. This is because having 0 as an eigenvalue means that the matrix is not full rank, and an invertible matrix must be full rank.