Talk:Defective matrix

Singular matrix?

How does this relate to a matrix being singular? —Ben FrantzDale 22:22, 21 December 2006 (UTC)

Not. All four combinations (invertible + diagonalizable, invertible + defective, singular + diagonalizable, singlar + defective) are possible. -- Jitse Niesen (talk) 14:48, 5 January 2007 (UTC)

Perhaps I'm overlooking something simple, but isn't a matrix singular iff its determinate is zero? The same appears from the examples to be true of a defective matrix. —Ben FrantzDale 14:53, 5 January 2007 (UTC)

It's dangerous to generalize from one example :) Here are some others:

• ${\begin{bmatrix}1&1\\0&1\end{bmatrix}}$ is invertible and defective
• ${\begin{bmatrix}0&0\\0&1\end{bmatrix}}$ is singular and not defective

A matrix is singular iff one of its eigenvalues is zero. Whether a matrix is defective has to do with the multiplicities of the eigenvalues, and not with the location of the eigenvalues. This explains why the properties are independent. -- Jitse Niesen (talk) 15:48, 5 January 2007 (UTC)

Aah. thanks. —Ben FrantzDale 16:16, 5 January 2007 (UTC)

Merge

I think this page should be merged with its antonym, diagonalizable matrix. -- Jitse Niesen (talk) 14:48, 5 January 2007 (UTC)

Singular example

Hi Plastikspork. I see you reverted my change to the simple example. Could you explain your reasoning? You said "It was already a simple example". My objection to the original example was that it was singular (not simple), which I thought might be confusing.--catslash (talk) 00:18, 21 April 2010 (UTC)

I didn't revert your example entirely. I just changed it to something a bit more simple. I agree that the singular one wasn't so good. This article actually needs a bit of help. There is no mention of the connection to the Jordan canonical form and the defective property of a general Jordan block. Thanks for your help. Plastikspork _―Œ^(talk) 01:58, 21 April 2010 (UTC)

Sorry, I didn't look very carefully at what you'd done - too hasty with my indignation. --catslash (talk) 09:07, 21 April 2010 (UTC)

Square matrix?

Does a defective matrix truly need to be square? In other words, non-diagonalizable matrices can be found which are not square (all of these, in fact), and also not diagonalizable, which seems to be the word sense of "defective". Since failing diagonalization is a trait not limited to square matrices, it seems artificial to limit these matrices to being square. Are there different conventions on this, or is there some other terminology which helps clarify the status of these non-square matrices with respect to the definition of defective matrix? — Preceding unsigned comment added by 75.139.254.117 (talk) 06:07, 11 January 2017 (UTC)

For a matrix to have eigenvectors, it must represent a linear transformation of a finite-dimensional vector space into itself and hence must be a square matrix. Note also from the lead that a defective matrix "is therefore not diagonalizable." This is a result—not the definition. The inference does not hold in the other direction, that is, a matrix which is not diagonalizable is not necessarily a defective matrix. For more information about eigenvectors, see most any linear algebra textbook. For more information about defective matrices, see Golub and Van Loan.—Anita5192 (talk) 07:41, 11 January 2017 (UTC)

"Double eigenvalue" in the example

I do not understand the claim that the 2x2 matrix in the example has "a double eigenvalue of 3". To me, the characteristic polynomial has two roots: +3 and -3. The +3 solution is an eigenvalue with the stated eigenvector. But the -3 solution has no eigenvector, the only solution to Av = -3v is v=0. Is this always what happens for a defective matrix? And if it is why do we say that the 'algebraic multiplicity' of the eigenvalue 3 is m=2 when it does not really recur as a root of the characteristic polynomial? ~2026-24859-88 (talk) 18:11, 22 April 2026 (UTC)

The characteristic polynomial in this example is

$\det \left({\begin{pmatrix}3&1\\{\color {Gray}0}&3\end{pmatrix}}-\lambda {\begin{pmatrix}1&{\color {Gray}0}\\{\color {Gray}0}&1\end{pmatrix}}\right)=\det \left({\begin{pmatrix}3-\lambda &1\\{\color {Gray}0}&3-\lambda \end{pmatrix}}\right)=(3-\lambda )^{2}=\lambda ^{2}-6\lambda +9$

The solutions for $\lambda$ in

$(3-\lambda )^{2}=0$

are $\lambda =3$ and $\lambda =3$. It is not the case that $\lambda =-3$ is a solution (which is why there is no corresponding eigenvector). catslash (talk) 22:17, 22 April 2026 (UTC)

Whoops! I got confused and wrongly thought the eigenvalues were the solution of $\lambda ^{2}=\det(A)$ for some reason. Thanks for the correction ~2026-24859-88 (talk) 07:23, 23 April 2026 (UTC)