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Symmetric matrices represent real self-adjoint maps, i.e. linear maps that have the following property: $$\langle\vec{v},f(\vec{w})\rangle=\langle f(\vec{v}),\vec{w}\rangle$$ where $\langle,\rangle$ donates the scalar (dot) product.

Using this logic:

$$\langle\vec{v},AB\vec{v}\rangle=\langle A\vec{v},B\vec{v}\rangle=\langle BA\vec{v},\vec{v}\rangle$$

Where $A$ and $B$ are symmetric matrices. Using the fact that the real scalar dot product is commutative:

$$\langle BA\vec{v},\vec{v}\rangle=\langle\vec{v},BA\vec{v}\rangle$$

We therefore have the result:

$$\langle\vec{v},AB\vec{v}\rangle=\langle\vec{v},BA\vec{v}\rangle$$

This holds true for any real vector $\vec{v}$ so therefore $AB=BA$.

However, symmetric matrices do not always commute so something is wrong with this proof.

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    $\begingroup$ Aren't you assuming $BA$ is symmetric when you move it back over? $\endgroup$ – Randall Jun 3 '19 at 16:46
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    $\begingroup$ \langle gets a good vector bracket $\langle$ on the left. $\endgroup$ – Randall Jun 3 '19 at 16:49
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    $\begingroup$ i dont understand what happened in the last step of the equalities $\endgroup$ – Milan Jun 3 '19 at 16:52
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    $\begingroup$ @Pancake_Senpai by the way, I don't think commutative is the right word in this context. You're making use of the symmetry of the real inner product: for all $\xi, \eta \in V$, where $V$ is the vector space we are working on, $\langle \xi, \eta\rangle = \langle \eta, \xi \rangle$ $\endgroup$ – peek-a-boo Jun 3 '19 at 16:59
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    $\begingroup$ You would not have this problem if you tried to prove that $\langle v,ABw\rangle=\langle v,BAw\rangle$ for two different vectors $v,w$. (You can't, because it's not true. But if it was, you would then be justified in concluding that $AB=BA$.) $\endgroup$ – user856 Jun 4 '19 at 5:56
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You have proved that $v\mapsto v^TABv$ and $v\mapsto v^TBAv$ are the same quadratic form. However, since $AB$ and $BA$ are not necessarily symmetric, that doesn't mean they are the same matrix.

You can check this by plugging in some matrices where you know commutativity fails, for example $$ A =\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \quad B = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} $$ We then get $$ AB = \begin{pmatrix}0 & 1 \\ 2 & 0 \end{pmatrix} \qquad BA = \begin{pmatrix}0 & 2 \\ 1 & 0 \end{pmatrix} $$ and indeed these define the same quadratic form: $$ (x\;\;y)\begin{pmatrix}0 & 1 \\ 2 & 0 \end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix} = 3xy = (x\;\;y)\begin{pmatrix}0 & 2 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} $$ for all $x,y$, but the matrices are different.

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    $\begingroup$ In particular, the exact logical fallacy is not in the computation, but in the words "so therefore $𝐴𝐵=𝐵𝐴$". $\endgroup$ – Greg Martin Jun 3 '19 at 16:57
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    $\begingroup$ @Pancake_Senpai: They implement different linear maps, but the same quadratic form. (It is more difficult for linear maps to be identical, because they give you an entire vector for each input, whereas a form just gives you a single number). $\endgroup$ – hmakholm left over Monica Jun 3 '19 at 17:05
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    $\begingroup$ @Pancake_Senpai: More concretely, $AB\vec v$ and $BA\vec v$ are generally different, but it so happens that their difference is always orthogonal to $\vec v$, such that the difference disappears when you take the inner product with $\vec v$. $\endgroup$ – hmakholm left over Monica Jun 3 '19 at 17:09
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    $\begingroup$ ^THAT is an amazing explanation. $\endgroup$ – Randall Jun 3 '19 at 17:11
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    $\begingroup$ I'm not sure why, but this seems like a good moment to mention that Henning's answers almost always inspire me to try to improve my own; I learn something new every time I read one, even if it's on a topic that I supposedly know well. :) Advice to new readers: read Henning's answers, even to questions that don't seem important to you right now. $\endgroup$ – John Hughes Jun 3 '19 at 17:25
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With the same reasoning you can prove that for any square matrix $A$ it holds that $$\langle Ax,x\rangle =\langle A^T x,x\rangle.$$ You can't conclude that $A=A^T$ from here (you could if you would have $x$ and $y$ but that does not work).

So the thing is that the quadratic form $x\mapsto \langle Ax,x\rangle$ only determines the symmetric part of the matrix, but not the full matrix.

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  • $\begingroup$ Another way to state this: The map $\psi \colon \mathbb{R}^{n \times n} \to \mathbb{R} \to \mathbb{R}$ given by $\psi(A) := x \mapsto \langle Ax, x \rangle$ is not injective. Namely, we have $\psi(A) = \psi(A^T)$, and certainly if we just pick any non-symmetric $A$, we have a counterexample. The error in the OP precisely attempted to make use of injectivity to conclude the wrong equality. $\endgroup$ – ComFreek Jun 4 '19 at 14:28
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The fallacy in your proof is, as others have observed, in the last step, in which you go from the (correct) equation

$\langle \vec{v}, AB\vec{v} \rangle = \langle \vec{v}, BA\vec{v} \rangle$

to the incorrect conclusion

This holds true for any real vector $\vec{v}$ so therefore $AB = BA$

To see why this is wrong, let's rephrase the underlying assumption here as a question:

Suppose $M, N$ are two matrices such that for all vectors $\vec{v}$, $\langle \vec{v}, M\vec{v} \rangle = \langle \vec{v}, N\vec{v} \rangle$. What can you conclude?

You would like to conclude that $M = N$. Is that true?

Because the inner product is bilinear, the assumption that $\langle \vec{v}, M\vec{v} \rangle = \langle \vec{v}, N\vec{v} \rangle$ is equivalent to $\langle \vec{v}, (M-N)\vec{v} \rangle = 0$. If we introduce a new matrix $K = M-N$, then our question is equivalent to the following:

Suppose $K$ is a matrix such that for all vectors $\vec{v}$, $\langle \vec{v}, K\vec{v} \rangle = 0$. What can you conclude about $K$?

Again, you would like to conclude that $K = 0$. Is that necessarily true?

Stated in this form it is perhaps easier to see where you have made your mistake. It is true that $\langle \vec{v}, \vec{v} \rangle = 0 \Longrightarrow \vec{v} = 0$ (this is one of the properties of any inner product), and the property you wish to assert looks sort of like this, if you don't look too closely. But it is not true in general that if $\langle \vec{v}, K\vec{v} \rangle = 0$ for all $\vec{v}$, then $K = 0$. Indeed, $\langle \vec{v}, K\vec{v} \rangle = 0$ holds for any antisymmetric matrix $K$. In the example of the original question $K = AB - BA$, the commutator of the two given symmetric matrices, which is always antisymmetric.

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    $\begingroup$ It might be worth summarising that $\langle \vec{v}, M\vec{v} \rangle = \langle \vec{v}, N\vec{v} \rangle$ for all $\vec v$ means precisely that $M$ and $N$ have the same symmetric parts, but says nothing about their anti-symmetric parts (in their symmetric+anti-symmetric decompositions). $\endgroup$ – Marc van Leeuwen Jun 5 '19 at 10:03
  • $\begingroup$ This is a brilliant answer - thank you. $\endgroup$ – Pancake_Senpai Jun 5 '19 at 11:10
  • $\begingroup$ the commutator of any two matrices is always antisymmetric or only the commutator of two symmetric matrices is antisymmetric ? $\endgroup$ – Widawensen Jun 5 '19 at 12:19
  • $\begingroup$ For products AB and BA of symmetric matrices we have, it seems, interesting property - they have equal their symmetric parts and their skew-symmetric parts are opposite to each other. $\endgroup$ – Widawensen Jun 5 '19 at 12:33
  • $\begingroup$ @Widawensen Indeed I should have been more careful to emphasize that the given matrices $A$ and $B$ were symmetric, which is why their commutator $[A, B]$ is antisymmetric. Thanks for pointing that out -- I have edited accordingly. $\endgroup$ – mweiss Jun 5 '19 at 20:19
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You have not made explicit why you think you can deduce from what you proved that $AB = BA$ but let me state the following things which are true:

  1. If you have two linear maps $T, S \colon V \rightarrow V$ on an inner product space (real or complex) and if $\left< Tv, u \right> = \left< Sv, u \right>$ for all pairs $u, v \in V$ then $T = S$.
  2. If you have two linear maps $T, S \colon V \rightarrow V$ on a complex inner vector space and if $\left< Tv, v \right> = \left< Sv, v \right>$ then $T = S$. This follows from a polarization identity. Note that this is weaker than the previous item because you require identity only for all vectors and not all pairs. Note also that this fails if $V$ is a real inner vector space (you can take for counterexample $V = \mathbb{R}^2, T = 0$ and $S$ a rotation by $90^{\circ}$s).

Now, let's see what goes wrong with your reasoning. Let's start with two symmetric real matrices $A,B \in M_n(\mathbb{R})$. You can do two things:

  1. Think of $A,B$ as maps from $\mathbb{R}^n$ to $\mathbb{R}^n$. More formally, set $T = L_A$ (the linear map which acts by multiplication by $A$) and $S = L_B$. Then you indeed have shown that $\left< TSv, v \right> = \left< STv, v \right>$ for all $v \in \mathbb{R}^n$. However, by the above, you see that this is not enough to deduce that $TS = ST$ (or $AB = BA$).
  2. Think of $A,B$ as maps from $\mathbb{C}^n$ to $\mathbb{C}^n$. Again, set $T = L_A$ and $S = L_B$. But then you have proved that $\left< TSv, v \right> = \left< STv, v \right>$ only for all $v \in \mathbb{R}^n$ and not for all $v \in \mathbb{C}^n$ (this is where you use "any real vector"). Hence, even though you work on a complex vector space, again you can't deduce that $TS = ST$ (or $AB = BA$).
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Your error lies with these two lines

$$\langle\vec{v},AB\vec{v}\rangle=\langle\vec{v},BA\vec{v}\rangle$$

This holds true for any real vector $\vec{v}$ so therefore $AB=BA$.

The second line is a non sequitur. You seem to have an implicit step in here $(\forall \vec v: \langle\vec{v},M_1\vec{v}\rangle=\langle\vec{v},M_2\vec{v}\rangle) \rightarrow (M_1=M_2) $, however, that is not true. For instance, if $M_1$ rotates $\vec{v}$ ninety degrees clockwise, while $M_2$ rotates ninety degrees counterclockwise, $ \langle\vec{v},M_1\vec{v}\rangle=\langle\vec{v},M_2\vec{v}\rangle=0$.

Your underlying error seems to be affirming the consequent: if the matrices are equal, then the dot product are equal, the dot products are equal, therefore the matrices are equal.

You also may be confused by that fact that $(\forall \vec v: M_1 \vec v = M_2 \vec v ) \rightarrow (M_1=M_2) $ is valid logic. If two matrices define the same map, then they are the same matrix (with the appropriate caveats, such as both being over the same basis, etc.). But here, we aren't defining a map $\vec v \rightarrow M \vec v$, we're defining a map $\vec v \rightarrow \langle\vec{v},M \vec{v}\rangle$. The map involves $M$, but it isn't just $M$.

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Dot product is a relative relationship which summarizes two vectors. A similar relationship is angle. The last statement in your proof amounts to, "because the angle between two vectors is the same, they must be the same vectors," which is wrong.

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    $\begingroup$ That's a bit too simple as an explanation. The equality holds for all vectors! $\endgroup$ – Dirk Jun 4 '19 at 10:33

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