Fake proofs using matrices Having gone through the 16-page-list of questions tagged fake-proofs, 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:



*

*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 $(x+y)^2 = x^2 + xy + yx + y^2 \neq x^2 + 2xy + y^2$.




*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.


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.
