Prove that sum of matrices equals zero

Question

The matrix $A$ has size $3 \times 3$ and we know that for any column vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that $A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.

So if $$A = \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}$$

and $$v = \begin{pmatrix} x\\ y\\ z \end{pmatrix}$$

orthogonality of $Av$ and $v$ brought me to the equation

$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0$

and the matrix $A^{T} + A$ equals

$$A^{T} + A = \begin{pmatrix} 2a_{11} & a_{12} + a_{21} & a_{13} + a_{31}\\ a_{12} + a_{21} & 2a_{22} & a_{23} + a_{32}\\ a_{13} + a_{31} & a_{23} + a_{32} & 2a_{33} \end{pmatrix}$$

However I don't see how then prove that $A^{T} + A = 0$.

Answer 1

Following your working,

$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0 \tag{1}$

Note that equation $(1)$ holds for any choice of $v=\begin{pmatrix}x \\ y \\ z \end{pmatrix}$. That is we can choose $x, y, z$.

If you substitute $x=1, y=0, z=0$, we conclude that $a_{11}=0$.

If you substitute $x=0, y=1, z=0$, we conclude that $a_{22}=0$.

If you substitute $x=0, y = 0, z=1$, we conclude that $a_{33}=0$.

By now, we can conclude that the diagonal entries of $A+A^T$ are all zeros.

Hence equation $(1)$ now reduces to

$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) = 0 \tag{2}$

If we substitute $x=1,y=1, z=0$, we conclude that $a_{12}+a_{21}=0$.

If we substitute $x=1, y=0, z=1$, we conclude that $a_{12}+a_{31}=0$.

If we substitue $x=0, y=1, z=1$, we conclude that $a_{23}+a_{32}=0$

Hence we can conclude that the off diagonal of $A+A^T$ are zeros as well.

Hence we have $A+A^T=0$.

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The result actually hold for square matrices $A$ of general size $n \times n$. That is if $A \in \mathbb{R}^{n \times n}$ and for any $v \in \mathbb{R}^n$, if $v$ and $Av$ are orthogonal, then $A+A^T=0$.

From the condition that $v$ and $Av$ are orthogonal, we have $v^TAv=0$.

We let $e_i$ be the standard unit basis with the $i$-th position being $1$ and the other positions being $0$. For example in the case of $n=3$, we have $e_2 = \begin{pmatrix} 0 \\ 1\\ 0\end{pmatrix}$. These matrices are very useful, we have $$e_i^TAe_j = A_{ij}$$

because you can check that $Ae_j$ would be just the $j$-th column of $A$ and then $e_i$ would select the $i$-th row.

We first handle the diagonal entries. We know that $e_iAe_i=0$, this is just saying that $A_{ii}=0$, hence the diagonal part s of $A+A^T=0$.

Also, we know that \begin{align}(e_i+e_j)^TA(e_i+e_j)&=e_i^TAe_i+e_i^TAe_j+e_j^T Ae_i + e_j^TAe_j \\ &= A_{ii} + A_{ij}+A_{ji}+A_{jj}\\ &= A_{ij}+A_{ji}\end{align}

and we know that it is equal to $0$.

Hence $A+A^T=0$.

Answer 2

It suffices to show $\langle (A^\intercal + A) v, v \rangle = 0$ for all $v$. This is true since

\begin{align} \langle (A^\intercal + A) v, v \rangle &= \langle A^\intercal v, v \rangle + \langle Av, v \rangle \\ &= \langle v, Av \rangle + \langle Av, v \rangle \\ &= 2 \cdot \langle Av, v \rangle \\ &= 2 \cdot 0 \\ &= 0.\end{align}

Remark: Note that this holds for all real square matrices $A$.

Answer 3

You know that $A.(1,0,0)$ and $(1,0,0)$ are orthogonal. Since $A.(1,0,0)=(a_{11},a_{21},a_{31})$, this means that $a_{11}=0$. By a similar argument, $a_{22}=a_{33}=0$.

So,$$A=\begin{bmatrix}0&a_{12}&a_{13}\\a_{21}&0&a_{23}\\a_{31}&a_{32}&0\end{bmatrix}.$$And so $A.(1,1,0)=(a_{12},a_{21},a_{31}+a_{32})$. But this vector is orthogonal to $(1,1,0)$, which means that $a_{12}=-a_{21}$. By a similar argument, $a_{13}=-a_{31}$ and $a_{23}=-a_{32}$.

Answer 4

Since all vectors $v$ are orthogonal to $Av$ then $v^TAv=0$ and likewise $(Av)^Tv=v^TA^Tv = 0$ and so adding them together we get $v^TAv + v^TA^Tv = v^T(A+A^T)v=0$ and finally since $v$ was arbitrary we must have $A+A^T=0$.