Make a matrix defective if possible

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In summary: All you need to do is check whether or not the matrix has a zero eigenvalue. If it does, then the matrix is defective.Otherwise, the matrix is not defective.
  • #1
Dustinsfl
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[tex]\begin{bmatrix}
1 & 1 & 0\\
1 & 1 & 0\\
0 & 0 & \alpha
\end{bmatrix}[/tex] and [itex]det(A-\lambda I)=\lambda(\alpha-\lambda)(\lambda-2)=0[/itex]

Therefore, [itex]\lambda_1=0[/itex], [itex]\lambda_2=\alpha[/itex], and [itex]\lambda_3=2[/itex].

In order to make this matrix defective, I need to make the left two column vectors dependent[tex]\begin{bmatrix}
1-\alpha & 1 & 0\\
1 & 1-\alpha & 0\\
0 & 0 & 0
\end{bmatrix}[/tex].

Is there an easy way to find or determine if this can be done for a given matrix?
 
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  • #2
I am thinking the only way this will work if alpha=0 or 2? Then I need to check those values to see if the eigenspace has only one eigenvector. However, is that necessarily true that it will only work at those values or is it possible that another value besides 0 or 2 will work?
 
  • #3
well to be defective it must have an eigenvalue with >1 algebraic multiplicity, but looking at
http://en.wikipedia.org/wiki/Defective_matrix
says normal matricies are not defective, and your matrix is real & symmetric as far as I can tell...
 
Last edited:
  • #4
If a matrix has an eigenvalue, then it has at least one eigenvector corresponding to that eigenvalue. So, yes, the only way it can be defective is if alpha=0 or alpha=2.
 
  • #5
Dick said:
If a matrix has an eigenvalue, then it has at least one eigenvector corresponding to that eigenvalue. So, yes, the only way it can be defective is if alpha=0 or alpha=2.


Thanks
 
  • #6
Careful, what Dick said was that the matrix can be defective if [itex]\alpha=0[/itex] or [itex]\alpha=2[/itex]...In order to determine if it actually is defective for either of those two values, you need to check whether or not they yield an incomplete basis of eigenvectors.
 
  • #7
gabbagabbahey said:
Careful, what Dick said was that the matrix can be defective if [itex]\alpha=0[/itex] or [itex]\alpha=2[/itex]...In order to determine if it actually is defective for either of those two values, you need to check whether or not they yield an incomplete basis of eigenvectors.

Right. I should have emphasized the 'CAN'. In fact, for those two values, if you think about lanedance's general observation, you don't actually have to compute the eigenvectors.
 

1. What is a defective matrix?

A defective matrix is a square matrix that cannot be inverted, meaning that it does not have a unique solution. In other words, the matrix is not full rank and some of its eigenvalues are equal to zero.

2. How can a matrix be made defective?

A matrix can be made defective by either removing one or more of its rows or columns, or by setting one or more of its eigenvalues to zero. This will result in a matrix that does not have a unique inverse.

3. Why would someone want to make a matrix defective?

Making a matrix defective can be useful in certain mathematical calculations, such as finding the Jordan canonical form of a matrix. It can also be used to demonstrate certain mathematical concepts or to test the robustness of algorithms.

4. Can any matrix be made defective?

Yes, any square matrix can be made defective by following the methods mentioned above. However, not all matrices will have a practical application for being made defective.

5. How does a defective matrix differ from a singular matrix?

A defective matrix is a square matrix that does not have a unique inverse, while a singular matrix is a square matrix that is not invertible. In other words, a singular matrix has a determinant of zero, while a defective matrix can have a non-zero determinant but still not be invertible.

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