# If $A \in \mathbb{R}^{n\times n}$ and $\textbf{x}\in \mathbb{R}^{n}$why $A = \frac{1}{2} A + \frac{1}{2} A^\top$ is not always true?

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? $$A = \frac{1}{2} A + \frac{1}{2} A^\top$$

• 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.
– lcv
Commented Sep 24, 2019 at 16:22
• Shouldn't the explanation be obvious when you subtract 1/2 A from both sides?
– Jens
Commented Sep 24, 2019 at 16:56

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

$$A = A^T, \tag 3$$

then

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

going the other way, if

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

then

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

whence

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

so

$$A = A^T. \tag 8$$

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.