I wonder if anyone can share wrong but useful and/or fun proofs in linear algebra. It can not only be fun, but also useful for someone who learns the subject. You are welcome to add explanations but please hide them in order not to spoil the fun.

Let me start from my own "proof" that every square matrix has zero determinant (I am sure that "proof" was discovered many many times).

"Theorem". Every square matrix $A$ over any field $K$ has zero determinant.

"Proof". If $A$ has two equal rows or columns we are done. Otherwise, we will construct them. First let add all the rows except the first row to the first row. Next let add all the rows except the last row to the last row. Clearly, now first and last row are equal since both or them are sum of every row in $A$, so $\det(A) = 0$. $\Box$

After we added rows to the first row of $A$ it changed, but the "proof" assumes that the first row is the same.


5 Answers 5


I personally like the following wrong "proof" of the Cayley-Hamilton theorem (which is actually a true statement, just the proof is wrong).

Theorem: (Cayley-Hamilton). Let $K$ be a field and $A \in M_n(K)$ be an $n \times n$-matrix with entries in $K$. Then $\mathrm{charpol}_A(A)=0$ (this is $A$ is a root of its own characteristic polynomial).

"Proof": By definition $\mathrm{charpol}_A(x)= \det(xI_n-A)$ so plugging in $x=A$ we obtain $\mathrm{charpol}_A(A) = \det(AI_n-A) = \det(A-A) = \det(0_{n \times n}) = 0$. $\square$

What is wrong with this proof?

The "equality" $\mathrm{charpol}_A(A)= \det(AI_n-A)$ is not actually an equality because the left hand side of it is a matrix in $M_n(K)$ while the right hand side is a scalar/a number in $K$. In the statement of the theorem the expression $\mathrm{charpol}_A(A)=0$ means that $\mathrm{charpol}_A(A)$ is the $n\times n$ matrix with just zeros and not the number zero.

  • 4
    $\begingroup$ One of my co-alumni actually submitted this tentative proof to the mathematics department of our college, after attending the lesson on the Cayley-Hamilton theorem, where the teacher spent almost the entire hour writing down the actual 3-page long proof. That was fun, the teacher actually sought counsel among his peers on this one. $\endgroup$ Apr 17 at 6:00
  • $\begingroup$ A former student of mine (who is now a researcher) rediscovered this “proof”. $\endgroup$ Apr 19 at 8:31
  • $\begingroup$ (+1) One nice thing about this is that whilst $$A\otimes I-I\otimes A=0$$ (as suggested by the "proof") is clearly false, it is suggestive of the true statement that contracting indices on $$A^T\otimes I- I\otimes A$$ gives $0$. In general this contraction of $$B^T\otimes I- I\otimes A$$ is a (tensorial) factor of ${\rm charpol}_B(A)$, yielding one of the standard proofs that ${\rm charpol}_A(A)=0$. $\endgroup$
    – tkf
    Apr 24 at 11:02

Let $A,B,C$ be the infinite matrices (with rows and columns indexed by the natural numbers) given by:\begin{eqnarray*} A_{ij}&=&1,\qquad \textrm{for all } i,j,\\ B_{ij}&=&0,\qquad \textrm{for all } i,j,\\ C_{ij}&=&\begin{cases} 1, &\textrm{if } i=j, \\ -1, &\textrm{if } i=j+1,\\ 0, &\textrm{otherwise.} \end{cases} \end{eqnarray*}

Matrix multiplication, is defined in the usual way, whenever the sums for each entry of the product converge.

Then $A=B$.

Proof: We have $$ACA=(AC)A=BA=B,$$ but also $$ACA=A(CA)=A.$$

Thus $A=B$. $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box$


Matrix multiplication is not in general associative. Note that composition of linear maps is associative, even for infinite-dimensional vector spaces. Therefore infinite matrices need not represent linear maps in any 'natural' way.

  • $\begingroup$ +1 for posting one I understand $\endgroup$ Apr 17 at 23:15

"Theorem": If $T:V \rightarrow V$ and $U: V \rightarrow V$ are linear operators such that $TU=I$, then $U=T^{-1}$

"Proof:" Since $U$ is a right-inverse, it is injective. Since $T$ is a left-inverse and $U$ is injective, $T$ is surjective. Surjectivity is equivalent to invertibility, so $T^{-1}$ exists and we have $TT^{-1}=I=TU \implies T^{-1}=U$.

This neglects cases where $V$ is not finite-dimensional. Invertibility and surjectivity are only equivalent if $V$ is finite-dimensional.


If $A$ and $B$ are linear maps from a vector space to itself and $I-AB$ is invertible, then $I-BA$ is invertible with inverse $I+BCA$ where $C=(I-AB)^{-1}$. Proof: $(I-BA)^{-1} = I+ BA + (BA)^2 + (BA)^3 + \cdots = I + B(I + AB + (AB)^2 + \cdots)A = I+B(I-AB)^{-1}A$


The "infinite series" makes no sense in general, but the result is correct and can be checked by composing the claimed inverse !with $I-BA$.



Let $V$ an $n$-dimensional vector space and $f:v \mapsto V$ a non-injective linear application. Then there exists $m<n$ such that $f^m=0$.


Since $f$ is not injective, $\dim(f(V)) < \dim(V)$, therefore $\dim(f^2(V)) < \dim(f(V)) < dim(V)$ and so on. Since $n$ is finite, there must exist some $m$ such that $\dim(f^m(V))= 0$.

  • $\begingroup$ Can you please add spoiler section? I edited it but my edit was rejected by a moderator for some reason. $\endgroup$ Apr 21 at 11:52

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