If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$,

It is possible to prove that $$ \textbf{x}^\top A \textbf{x} = \textbf{x}^\top(\frac{1}{2} A + \frac{1}{2} A^\top)\textbf{x} $$ That could let us think that $$ A = \frac{1}{2} A + \frac{1}{2} A^\top ? $$ However if we take $$A = \begin{bmatrix}1&2\\ 3&4 \end{bmatrix}$$ and compute $$B = \frac{1}{2} A + \frac{1}{2} A^\top$$ we get $$B = \begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}\ne A$$ Although $$ \textbf{x}^\top A \textbf{x} = \textbf{x}^\top B \textbf{x} $$ Here is my question:

How can we explain that the following is not always true? $$ A = \frac{1}{2} A + \frac{1}{2} A^\top $$

  • $\begingroup$ The right hand side is necessarily symmetric, so your equation expresses the requirement that $A$ is symmetric. Obviously not all matrices have this property. The reason your first equality holds is that the antisymmetric part cancels out. $\endgroup$
    – lcv
    Sep 24, 2019 at 16:22
  • $\begingroup$ Shouldn't the explanation be obvious when you subtract 1/2 A from both sides? $\endgroup$
    – Jens
    Sep 24, 2019 at 16:56

4 Answers 4


The equality $x^\top\left(\frac12A+\frac12A^\top\right)x=x^\top Ax$ holds indeed, but you cannot deduce from it that $\frac12A+\frac12A^\top=A$. In fact, you also always have $x^\top Ax=x^\top A^\top x$, but it is not true in general that $A=A^\top$.


In the product $\mathbf{x}^\top A \mathbf{x}$ we get terms like $x_1 A_{12} x_2 + x_2 A_{21} x_1 = (A_{12}+A_{21}) x_1 x_2$ so that for example $\begin{bmatrix}1&2.5\\ 2.5&4 \end{bmatrix}$, $\begin{bmatrix}1&5\\ 0&4 \end{bmatrix}$ and $\begin{bmatrix}1&3\\ 2&4 \end{bmatrix}$ give the same result since $2.5+2.5 = 5+0 = 3+2$. Thus only the symmetric part $\frac12(A+A^\top)$ matters; the antisymmetric part $\frac12(A-A^\top)$ does not matter.


Observe that

$A = \dfrac{1}{2}(A + A^T) \Longleftrightarrow A = A^T, \tag 1$

that is,

$A = \dfrac{1}{2}(A + A^T) \tag 2$

if and only if $A$ is a symmetric matrix. This is easily seen as follows: if

$A = A^T, \tag 3$


$A = \dfrac{1}{2}(2A) = \dfrac{1}{2}(A + A) = \dfrac{1}{2}(A + A^T); \tag 4$

going the other way, if

$A = \dfrac{1}{2}(A + A^T), \tag 5$


$A = \dfrac{1}{2}A + \dfrac{1}{2}A^T, \tag 6$


$\dfrac{1}{2}A = A - \dfrac{1}{2}A = \dfrac{1}{2}A^T, \tag 7$


$A = A^T. \tag 8$

We thus see that (2) is false if $A$ is not symmetric.

Note we need not introduce $\mathbf x \in \Bbb R^n$ to establish this.


It is true that if $$x^T A y = x^T B y$$ for all vectors $x,y$, then $A=B$. The proof is by plugging in basis vectors $e_i$ (with a $1$ in the $i$th entry, and zeroes elsewhere): $$A_{ij} = e_i^T A e_j = e_i^T B e_j = B_{ij}.$$

Now if you are forced to multiply by the same vector on both sides, instead of two arbitrary vectors, you can no longer conclude $$x^TAx = x^TBx \quad \not\Rightarrow \quad A=B,$$ and you have already found a counterexample. You can try to extend the above argument, and easily prove that $A_{ii} = B_{ii}$, but you have no way to isolate the off-diagonal terms.


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