# Examples of common false beliefs in mathematics.

The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant; (ii) sin(z) is a bounded function; (iii) sin(z) is defined and analytic everywhere on C; (iv) sin(z) is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of sin(z) to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset U of R must be the whole of R. The "proof" of this statement is that every point x is arbitrarily close to a point u in U, so when you put a small neighbourhood about u it must contain x.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

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I have to say this is proving to be one of the more useful CW big-list questions on the site... –  Qiaochu Yuan May 6 '10 at 0:55
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. –  Unknown May 22 '10 at 9:04
wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. –  Suvrit Sep 20 '10 at 12:39
It's a thought -- I might consider it. –  gowers Oct 4 '10 at 20:13
Meta created tea.mathoverflow.net/discussion/1165/… –  quid Oct 8 '11 at 14:27

In group theory, if $G_1 \cong G_2$ and $H_1 \cong H_2$, then

$G_1 / H_1 \cong G_2 / H_2$.

For example, $\mathbb{Z} / 2\mathbb{Z} \not \cong \mathbb{Z} / \mathbb{Z}$. The point is that the inclusion of $H_j$ into $G_j$ is needed in order to define the quocient.

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False belief: A function being continuous in some open interval implies that it is also differentiable on some point in that interval:

Counterexample:

The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:

$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$

Where $a \in (0, 1)$, $b$ is a positive odd integer, and $ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though.

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Related: if f is continuous on the interval I, there must be an interval J in I on which f is monotone. Easily believed by the beginner. –  Thierry Zell Aug 31 '10 at 2:34

Some things from pseudo-Riemannian geometry are a bit hard to swallow for students who have had previous exposure to Riemannian geometry. Aside from the usual ones arising from sign issues (like, in a two dimensional Lorentzian manifold with positive scalar curvature, time-like geodesics will not have conjugate points), an example is that in Riemannian manifolds, connectedness + geodesic completeness implies geodesic connectedness (every two points is connected by a geodesic). This is not true for Lorentzian manifolds, and the usual example is the pseudo-sphere.

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Two very common errors I see in (bad) statistics textbooks are

(i) zero 3rd moment implies symmetry (though generally stated in terms of "skewness", where skewness has just been defined as a scaled third moment)

(ii) the median lies between the mean and the mode

(I have seen a bunch of related errors as well.)

Another one I often see is some form of claim that the t-statistic goes to the t-distribution (with the usual degrees of freedom) in large samples from non-normal distributions.

Even if we take as given that the samples are drawn under conditions where the central limit theorem holds, this is not the case. I have even seen (flawed) informal arguments given for it.

What does happen is (given some form of the CLT applies) Slutzky's theorem implies that the t-statistic goes to a standard normal as the sample size goes to infinity, and of course the t-distribution also goes to the same thing in the limit - but so, for example, would a t-distribution with only half the degrees of freedom - and countless other things would as well.

The first two errors are readily demonstrated to be false by simple counterexample, and to convince people that they don't have the third usually only requires pointing out that the numerator and denominator of the t-statistic won't be independent if the distribution is non-normal, or any of several other issues, and they usually realize quite quickly that you can't just hand-wave this folk-theorem into existence.

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The fundamental group of the Klein bottle is $D_\infty$, the infinite dihedral group (which is $\mathbb Z \rtimes \mathbb Z_2$).

I believed this for some time, and I seem to recall some others having the same confusion.

The group that has been mistaken for $D_\infty$ is in fact $\mathbb Z \rtimes\mathbb Z$, which can also be written with the presentation $x^2y^2=1$. The former abelianizes to $\mathbb Z_2\oplus \mathbb Z_2$, the latter to $\mathbb Z\oplus \mathbb Z_2$.

A 2-dimensional Lie group is a product of circles and lines, in particular it is abelian.

I don't know if anyone else suffered this one. The mistake is (a) in forgetting that the classification of surfaces doesn't apply since homeomorphic Lie groups are not necessarily isomorphic (e.g., the (bijective, orientation preserving) affine transformations $x\mapsto ax+b$, where $a>0, b\in \mathbb R$ are homeomorphic to $\mathbb R^2$, though not isomorphic) and (b) that Lie groups aren't necessarily connected, in particular $\mathbb R^2$ cross any finite non-abelian group is non-abelian.

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1- A very common mistake that 1st year students (but not even a single mathematician) think that it is true is "a transitive and symmetric relation on a set is reflexive". But as the empty set is a transitive and symmetric relation but not reflexive on any non-empty set. Of course there lots of non-trivial examples also.

2- Another common mistake is that the expression "countable union of countable sets is again countable" is independent of axiom of choice (AC). Many people make the proof of this statement without mentioning axiom of choice. Indeed, in his holly book Algebra, Lang proves this statement just by taking an ordering from each countable set and continues without the mentioning AC.

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For big-list questions, it's usually best to post independent answers as separate answers. –  Nate Eldredge Dec 2 '10 at 15:13
+1 for #2. Baby Rudin is another offender. And many authors use so-called "diagonalization tricks" for proving compactness theorems like Arzela-Ascoli and Prohorov, which typically reduce to the compactness of $[0,1]^\mathbb{N}$. –  Nate Eldredge Dec 2 '10 at 15:21
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Duality reverses inclusions of vector spaces.

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That's funny, because I don't imagine this kind of idea would occur to someone who has just learned the definition of a dual space. That would be a strangely sophisticated mistake to make. –  Thierry Zell Apr 7 '11 at 0:21
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A degree $k$ map $S^n\to S^n$ induces multiplication by $k$ on all the homotopy groups $\pi_m(S^n)$.

(Not sure if this is a common error, but I believed it implicitly for a while and it confused me about some things. If you unravel what degree $k$ means and what multiplication by $k$ in $\pi_m$ means, there's no reason at all to expect this to be true, and indeed it is false in general. It is true in the stable range, since $S^n$ looks like $\Omega S^{n+1}$ in the stable range, "degree k" can be defined in terms of the H-space structure on $\Omega S^{n+1}$, and an Eckmann-Hilton argument applies.)

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If $n$ is even and $x \in \pi_{2n-1}(S^n)$ and $f$ a degree $k$ map and $H$ the Hopf invariant, then $H(f_* (x)) = k^2 H(x)$. A related misbelief: if $M$ is a framed manifold and $N\to$M a finite cover, of degree $d$. Then the framed bordism classes satisfy $[N]=d [M]$. Completely wrong. –  Johannes Ebert Apr 14 '11 at 9:04

"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)

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A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.

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Fu... I just "proved" that again as an exercise a few days ago. –  Johannes Hahn Mar 6 '13 at 0:02

A common trap which sometimes I see people fall is that a Hermitian matrix $M$ is negative definite if and only if its leading principal minors are negative.

What is true is the Sylvester's criterion, which says that $M$ is positive definite if and only if its principal minors are positive. Thus, the true statement is that $M$ is negative definite if and only if the principal minors of $-M$ are positive.

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It always confused me as an undergraduate that $\mathbb{Q}\subset\mathbb{R}$ has an open neighborhood $N_\epsilon \supset\mathbb{Q}$ of arbitrarily small measure $\epsilon$, because $\mathbb{Q}$ is dense in $\mathbb{R}$.

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Here's a mistake I've seen from students taking a first course in linear analysis. For a vector $g$ in a Hilbert space $H$, it is true that $\langle f,g\rangle=0$ for every $f\in H$ implies $g=0$. This leads us to the mistaken:

“Let $(g_n)$ be a sequence in $H$. If, for every $f\in H$, $\langle f,g_n\rangle\to0$, then $g_n\to 0$.”

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@Michael: all answers are CW; so if we think some wording needs clarifying, we can do it ourselves! –  Peter LeFanu Lumsdaine Dec 2 '10 at 0:43

I guess you don't want commonly held beliefs of students that for every real number there is a next real number, or that convergent sequences are eventually constant. A version I saw in a book asked whether points on a line "touch." Understanding the topology of a line is a challenge for many students, although presumably not for most mathematicians.

Here is a more esoteric belief that I have even seen in some books:

"The Banach-Tarski Paradox says that a ball the size of a pea can be cut into 5 pieces and reassembled to make a ball the size of the sun."

As a consequence of the Banach-Tarski paradox, a ball the size of a pea can be partitioned (not really "cut") into a finite number of pieces which can be reassembled into a ball the size of the sun, but a simple outer measure argument implies that the number of pieces must be very large (I roughly estimate at least $10^{30}$). The number 5 probably comes from the fact that the basic Banach-Tarski paradox is that a ball of radius 1 can be partitioned into 5 pieces which can be reassembled into two disjoint balls of radius 1. (It can almost, but not quite, be done with four pieces; one of the five pieces can be taken to be a single point.)

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Since points do not touch, this was an objection to the set theoretic view of the geometric continuum as a set of points, for example by Veronese. A decent account of this can be found in Debates about infinity in mathematics around 1890: The Cantor-Veronese controversy, its origins and its outcome, by Detlef Laugwitz. –  Andres Caicedo Dec 2 '13 at 23:22
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When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.

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You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. –  Qiaochu Yuan May 4 '10 at 22:14
In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. –  Leandro May 4 '10 at 22:34

Many people believe that Cantor proved the uncountability of the real line using a diagonal argument. This paper does not that provide that proof; Cantor's stated purpose was to prove the existence of `uncountable infinities' without using the theory of irrational numbers.

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More to the point, I think, is that the paper proves that the power set of any set has greater cardinality than the set itself. This is the first proof that there is no greatest cardinality. (The uncountability of the real line easily follows, even if Cantor does not mention it because he has bigger fish to fry.) –  John Stillwell May 31 '10 at 5:12
Just to fill in some history here: if I remember right, Cantor first proved the uncountability of the reals by other arguments, then later (as you reference) found the diagonal argument, as a proof of the more general statement about power sets. –  Peter LeFanu Lumsdaine Sep 27 '10 at 3:01

True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).

False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.

There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.

But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.

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What I mean is: People use these formulations as a protective charm against a danger they don't see but intuitively feel is there, although closer inspection shows that it is pure superstition. –  darij grinberg Mar 15 '11 at 17:26
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I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$

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This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! –  Thierry Zell Apr 14 '11 at 15:50

I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

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No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. –  Tom Goodwillie May 4 '11 at 0:16

If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.

False for $\alpha>1$, because this set contains only constant functions.

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"The set A = {a, b} has two elements..."

It's quite simple to notice that a can be the same as b, but after 5 years of university there were people still believing it...

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I'm not sure there is a false belief here, as much as awkward writing. Depending on context, I might very well write "The set $\{a,b \}$ (where $a$ and $b$ might be equal)..." if this issue mattered. –  David Speyer May 6 '10 at 11:16
There are many situations where one needs to speak of a set of two numbers that may or may not be equal. E.g.: "Let x<sub>1</sub>, x<sub>2</sub> &isin; ℝ. Then among all the open intervals containing the set {x<sub>1</sub>, x<sub>2</sub>}, none of them is contained in all the others." If one is addressing mathematicians, there is no need to specify that x<sub>1</sub> might be equal to x<sub>2</sub>. –  Daniel Asimov Jun 17 '10 at 23:34

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:

The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:

There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\\ E_s\\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

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"$\emptyset$ is a basis for {0}" is an immediate consequence of the definitions. There is no false belief in this point. –  Johannes Hahn May 12 '10 at 15:14
Dear Johannes, please reread my post. –  Pierre-Yves Gaillard May 12 '10 at 15:18
Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. –  Robin Chapman May 15 '10 at 14:07
I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. –  Qiaochu Yuan May 17 '10 at 6:19

An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$

$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .

Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.

In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.

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This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction. –  Qiaochu Yuan Feb 24 '11 at 21:08
@Elohemahab: actually, the correct generalization is $\gcd(a, b, c) \text{lcm}(ab, bc, ca) = \text{lcm}(a, b, c) \gcd(ab, bc, ca) = abc$. –  Qiaochu Yuan May 8 '11 at 23:47

This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the $Y^*$ (the dual of $Y$) is isomorphic to a subspace of $X^*$ (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris

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I would also be shocked if this really gets believed often! It seems to be a sort of “mis-dualisation”: they dualise “$X$ is a subobject of $Y$” to “$Y^*$ is a subobject of $X^*$”, where the correct dual is “$X^*$ is a quotient of $Y^*$”. –  Peter LeFanu Lumsdaine Dec 1 '10 at 15:19
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Let $B(r_1) \subset B(r_2)$ be two open balls of radius $r_1$ and $r_2$ respectively. Then $r_1 \leq r_2$.

Bounded metric spaces give trivial counterexamples. Also, $B \left( \frac{1}{6}, \frac{2}{3} \right) \subsetneq B \left( \frac{1}{2}, \frac{1}{2} \right)$ in $(0,+ \infty)$.

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Common false belief: a space that is locally homeomorphic to $\mathbb{R}^n$ must be Hausdorff. More generally, many people forget that the usual definition of a manifold contains the Hausdorff and paracompact conditions.

There are of course examples that show that forgetting this assumption leads to unexpected result, and they are in fact much wilder than I knew a few weeks ago. Notably, among examples of (Hausdorff) non-paracompact "manifolds" are the well-known long line, but also the Prüfer manifold constructed from a closed half-plane by attaching to it a half plane at each boundary point.

Added: Let me give a particular case of this false belief to illustrate what kind of weird things can happen that most people would not realize when they are sloppy with the paracompact hypothesis: there exists a path-connected, locally contractible, simply-connected space that admits non-trivial locally trivial bundles with fiber $[0,1]$. Indeed, the first octant in the product of two long line is not homeomorphic to a product a long ray with an interval, but has a natural bundle structure over a long ray.

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These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $\displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

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@Todd: No i was talking of high school students. –  Chandrasekhar May 4 '11 at 4:08
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Here's another howler some people commit: If m, n are integers such that m divides n^2 then m divides n.

It's true sometimes, for example if m is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers p, q, r with p squarefree such that $m = p q^2$ and $n = p q r$

The usual counterexample is that 8 divides 4^2 but not 4 ;-)

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An even more trivial counterexample is that 4 divides 2^2 but not 2 :-P –  Peter LeFanu Lumsdaine Feb 23 '11 at 9:40

If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.

True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.

False over other fields. For example, over $\mathbb{C}$, $\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.

False for other nondegenerate symmetric bilinear forms: $\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$.

You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

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You seem to have a different definition of "the standard inner product on $\mathbb{C}^n$" than I do. I think that phrase normally refers to the familiar positive definite sesquilinear form, with respect to which self-adjoint matrices are indeed diagonalizable. –  Mark Meckes Jan 28 '11 at 16:46
Of course it's not bilinear -- an "inner product" on a complex vector space is defined to be sesquilinear, not bilinear -- I've spent a lot of time trying to get my linear algebra students to remember that. The failure of such a form to generalize to other fields is indeed sad, but I think the richness of Hilbert space theory helps to make up for that disappointment. :) –  Mark Meckes Jan 28 '11 at 21:24
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