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Having gone through the 16-page-list of questions tagged , and going though both the relevant MSE Question and Wikipedia page, I didn't find a single fake proof that involved matrices.

So the question (or challange) here is: what are some fake proof using matrices?

In particular, the fake proof should use a property, an operation, ..., specific to matrices (or at least not present in $\mathbb{R}$ or $\mathbb{C}$), e.g.

  • Noncommutativity
  • (Non-)existence of an inverse
  • Matrix sizes
  • Operations as $\det$, $\text{trace}$, ...
  • Eigenvalues and diagonalization
  • Matrix decompositions and normal forms
  • ...

Note: It does not matter if the result being "proven" is correct or not. The fallacy in the proof itself is what matters.

Examples:


  1. Proof that 1 = 0

Proof: it is a well-known fact that $(x+y)^2 = x^2 + 2xy + y^2$. Now let $$x = \begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix},\;\;y = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}.$$ On the one hand, we have that $$ (x+y)^2 = \begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix}^2 = \begin{pmatrix}1 & 1\\ 0 & 0 \end{pmatrix},$$ on the other hand we have $$x^2 + 2xy + y^2 = \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix} + 2\begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix} + \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}.$$ Since two matrices are equal if and only if all their entries are equal, we conclude that $1 = 0$.

The mistake here is that $x$ and $y$ do not commute. Thus $(x+y)^2 = x^2 + xy + yx + y^2 \neq x^2 + 2xy + y^2$.


  1. Proof that 2 = 0

Proof: We know that $\det (AB) = \det (BA)$, since $$\det (AB) = (\det A) (\det B) = (\det B) (\det A) = \det (BA).$$ Now consider the matrices $$ A = \begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}, \; \; B = \begin{pmatrix}1 & 0\\ 0 & 1\\ 1 & 0 \end{pmatrix}.$$ We have that $$AB = \begin{pmatrix}2 & 0\\ 0 & 1 \end{pmatrix}, \;\; BA = \begin{pmatrix}1 & 0 & 1\\ 0 & 1 & 0 \\1 & 0 & 1\end{pmatrix}.$$ Hence $\det (AB) = 2$ and $\det (BA) = 0$, therefore $2 = 0$.

The mistake here is that $\det$ is defined for square matrices only, and thus $\det AB = \det BA$ only holds in general if $A$ and $B$ are square.

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    $\begingroup$ Why do you want this? $\endgroup$ – Pedro Tamaroff May 27 '16 at 16:57
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    $\begingroup$ First and foremost, from an educational perspective. I like to use "fake proofs" to get a point across to students, e.g. why they can't carelessly switch the order of differentiation, integration, limits, etc. I find that they enjoy the class more if I prove something ridiculous like $1=0$ instead of just warning them about pitfalls, and also that they are more likely to remember to watch out for these pitfalls. And second, because I find it fun to create (and find mistakes in) fake proofs. $\endgroup$ – TastyRomeo May 27 '16 at 17:59
  • $\begingroup$ And as to why this question relates to proofs using matrices in particular: I've taught courses on elementary analysis en calculus so far, and I have a good repository of fake proofs there. But there is a good chance I'll be teaching (linear) algebra in the near future... $\endgroup$ – TastyRomeo May 27 '16 at 18:01
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Here is a fake proof of a true theorem, Cayley-Hamilton.

So the characteristic polynomial of a matrix $A$ is $f_A(t) = \textrm{det}(tI-A)$. Then $f_A(A) = \textrm{det}(AI-A) = \textrm{det}(A-A) = \textrm{det}(0) = 0$.

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    $\begingroup$ "this can be made rigorous" $\endgroup$ – David Bowman Dec 28 '16 at 1:05
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    $\begingroup$ "Is this clear?" $\endgroup$ – Daminark Dec 28 '16 at 1:23
  • $\begingroup$ @Daminark To reply to your chat message yesterday: yes, this is allowed! I've added this in the questionnaire. (Great proof, by the way!) $\endgroup$ – TastyRomeo Dec 29 '16 at 14:47
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To prove: $0=1$.

Certain identities get funky when we pass over to infinite-order matrices. We see such matrices, for example, in representations of operators in quantum mechanics.

Everyone knows that

$Tr(AB-BA)=Tr(AB)-Tr(BA)=0$.

So let

$A_{i,j}=\delta_{i,j-1}, B_{i,j}=A_{j,i}$

Here $\delta$ is the Kronecker delta function, and $i$ and $j$ run through all natural mumbers.

Then $AB-BA$ has a 1 in the upper left position and zeroes everywhere else. The sum of diagonal elments is 1 and yet "$Tr(AB-BA)=0$".

$Tr(AB-BA)=0$ is guaranteed only for finite matrices or, among infinite matrices, when the traces of $AB$ and $BA$ converge absolutely.

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