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I can't find the missing link between singularity and zero eigenvalues as is stated in the following proposition:

A matrix $A$ is singular if and only if $0$ is an eigenvalue.

Could anyone shed some light?

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    $\begingroup$ Some people define singularity as having a 0 eigenvalue. How does your text define singularity? There are many possible choices, see wiki. You will find some of the solutions below start with the author's favorite definition and end with 0 being an eigenvalue. $\endgroup$ – vadim123 Sep 24 '13 at 14:09
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$A$ singular $\iff\det(A)=0\iff\det(A-0\cdot I)=0\iff 0$ is eigenvalue of $A$.

Michael

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Note that, the determinant of $n\times n$ matrix $A$ can be computed using the eigenvalues as

$$ |A|=\lambda_1\lambda_2\dots\lambda_n ,$$

which is the product of the eigenvalues.

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    $\begingroup$ When $A$ is a matrix over what ring? For example, with the reals $\mathbb{R}$ you can have a $3\times 3$ matrix which represents a rotation of angle $\pi /2$ around the $z$ axis. What are your three $\lambda_i$ in that case? Do you switch to complex numbers then? Edit: That would be first projecting onto the $xy$ plane, and then rotating around the $z$ axis, if the matrix needed to be singular. Depending on what direction, "if" or "only if" we consider. $\endgroup$ – Jeppe Stig Nielsen Sep 24 '13 at 14:56
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We know that $0 \in \lambda(A)$ iff there exists some nonzero solution to the eigenvector equation $Ax = \lambda x = 0\cdot x = 0$. Thus $0$ is an eigenvalue iff $\exists b \in \mathrm{Ker}(A)$ with $b \neq 0$. But since $\mathrm{Ker}(A) \neq \{ 0 \}$ we conclude that $A$ must be singular.

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    $\begingroup$ You mean “0 is an eigenvalue” in you second sentence instead of “0 is an eigenvector”, don't you? $\endgroup$ – Michael Hoppe Sep 24 '13 at 15:06
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Assuming that by “singular”, you mean a square matrix that is not invertible:

Lemma: If $A$ is invertible and $\lambda$ is an eigenvalue of $A$, then $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$.

Let $x$ be the eigenvector of $A$ corresponding to eigenvalue $\lambda$. By definition, $Ax = \lambda x$. Left-multiply by $A^{-1}$, giving $A^{-1}Ax = A^{-1} \lambda x$. The LHS is equal to $x$ ($A^{-1}A = I$ by definition) and the RHS is equal to $\lambda A^{-1} x$ (because matrix × scalar commutes), so $x = \lambda A^{-1} x$. Divide both sides by $\lambda$, giving $\frac{1}{\lambda} x = A^{-1} x$. By definition, $\frac{1}{\lambda}$ is thus an eigenvalue of $A^{-1}$.


So, if $0$ were an eigenvalue of $A$, then $\frac{1}{0}$ would be an eigenvalue of $A^{-1}$. But $\frac{1}{0}$ isn't a number, so $A^{-1}$ can't exist either.

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    $\begingroup$ In your proof you write "Divide both sides by $\lambda$, ...". How can you do that when the problem is you don't know if $\lambda$ is zero or not? $\endgroup$ – Jeppe Stig Nielsen Sep 25 '13 at 9:44
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    $\begingroup$ Good point, Jeppe. At the step before the division, if $\lambda = 0$, then $x = 0$. Which is already a contradiction because an eigenvector must be nonzero. $\endgroup$ – Dan Sep 25 '13 at 12:48
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If 0 is an eigenvalue, then there exists a vector $v$ in your space such that $A.v = 0$. If your matrix size is 4x4 with one 0 eigenvalue and you write the image of the eigenvectors, you get:

$$(v11, v12, v13, 0)$$ $$(v21, v22, v23, 0)$$ $$(v31, v32, v33, 0)$$ $$(v41, v42, v43, 0)$$

You can see it's singular because:

  • the 3 vectors cannot possibly span a 4-dimensional space, it's an hyperplane, a 3 dimensional sub-space in 4 dimensions (so it would be a line in two dimension, a plane in 3 dimension)
  • any point not on the hyperplane cannot be described as a combination of the 3 column vectors described above. So not all points can be reached through multiplication by $A$.
  • so the transformation that is $A$ cannot always be inverted, because for all $y$ there is not always a point $x$ such that $Ax=y$ (for example, the $y$ not on the hyperplane)
  • so $A$ is singular, it cannot be inverted in general

hopes this rolls out the reasoning clearly enough.

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    $\begingroup$ Please consider editing your post to make use of LaTeX for typesetting the math, as it will make the post much more readable. See here for help on how to do it. $\endgroup$ – Mårten W Sep 24 '13 at 22:11
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    $\begingroup$ You: "So if you rewrite you matrix in the space of the eigenvectors, ...". Not all matrices can "be rewritten in the space of eigenvectors". What does that mean e.g. for a non-diagonalizable matrix? Both invertible and non-invertible matrices may be non-diagonalizable. $\endgroup$ – Jeppe Stig Nielsen Sep 25 '13 at 9:48
  • $\begingroup$ Modified to remove 'rewrite...' $\endgroup$ – user96687 Sep 27 '13 at 18:25
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I all depends on your starting definition. Here is one way, suppose $v$ is the eigen vector associated with $\lambda=0$ then $Av=0v=0$. Since $v\ne 0$ by definition then you have a nontrivial vector in the null space of $A$ that makes $A$ singular.

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An $n \times n$ matrix, $\mathbf A$, is singular if and only if there is a non zero column vector $\mathbf x$ such that $\mathbf A \mathbf x = 0 = 0 \mathbf x$, i.e., $0$ is an eigenvalue.

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  • $\begingroup$ This is the simplest explanation that follows from the definition of linear dependency $\endgroup$ – Sayan Pal Feb 19 at 19:43
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very true. can take it like this: any matrix can be diagonalized by using appropriate elementary matrices and we know the eigen values of diagonal matrices are the diagonal elements and so if any of the eigen value is zero then determinant value of matrix is zero and so it is Singular.

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