upper triangular and diagonal If $A$ is real, upper triangular $n\times n$ matrix such that $AA^t=A^tA$. Then show that $A$ is diagonal.
We know that every upper triangular, symmetric matrix is diagonal. But I have problem to show that $A=A^t$ from the given condition. How would we show that?
 A: Let $A = \begin{bmatrix}v_1 & v_2 & \cdots & v_n\end{bmatrix}$ and $A^T = \begin{bmatrix}h_1 & h_2 & \cdots & h_n\end{bmatrix}$.
Then, $AA^T = \begin{bmatrix}
h_1 \cdot h_1 & h_1 \cdot h_2 & \cdots & h_1 \cdot h_n\\
h_2 \cdot h_1 & h_2 \cdot h_2 & \cdots & h_2 \cdot h_n\\
\vdots & \vdots & & \vdots\\
h_n \cdot h_1 & h_n \cdot h_2 & \cdots & h_n \cdot v_n\\
\end{bmatrix}$ and $A^TA = \begin{bmatrix}
v_1 \cdot v_1 & v_1 \cdot v_2 & \cdots & v_1 \cdot v_n\\
v_2 \cdot v_1 & v_2 \cdot v_2 & \cdots & v_2 \cdot v_n\\
\vdots & \vdots & & \vdots\\
v_n \cdot v_1 & v_n \cdot v_2 & \cdots & v_n \cdot v_n\\
\end{bmatrix}$.
Comparing them gives $\|h_1\|=\|v_1\|$ and so on.
Since $\|h_1\|=\|v_1\|$, we have $a_{11}^2 + a_{12}^2 + \cdots a_{n2}^2 = a_{11}^2$.
Since the entries are real, the only component of the first row which can be non-zero is $a_{11}$.
Rinse and repeat to prove that the elements on the diagonal are the only elements which can be non-zero.
A: Let A be:
$$ \left[
    \begin{array}{ccc}
      a_{11} & a_{12} & \cdots & a_{1n}\\
      0&a_{22} & \cdots & a_{2n}\\
      \vdots & \vdots & \ddots & \vdots \\
      0 & 0 & \cdots & a_{nn} 
    \end{array}
\right] $$
Then A^T will be a lower triangular matrix such as :
 $$ \left[
    \begin{array}{ccc}
      a_{11} & 0 & \cdots & 0\\
      a_{12}&a_{22} & \cdots & 0\\
      \vdots & \vdots & \ddots & \vdots \\
      a_{1n} & a_{2n} & \cdots & a_{nn} 
    \end{array}
\right] $$
Then AA^T=A^TA only if A is diagonal. 
