Question #160284

Show that a triangular matrix is normal if it is diagonal


1
Expert's answer
2021-02-09T01:35:04-0500

Answer:


Given:

A triangular matrix is normal if it is diagonal.


For n=2, Let A= [a0bc]\begin{bmatrix} a & 0 \\ b & c \end{bmatrix} such that

AA=AA= [abbc]\begin{bmatrix} \overline a & b \\ b & \overline c \end{bmatrix} ​[a0bc]\begin{bmatrix} a & 0 \\ b & c \end{bmatrix} -[a0bc]\begin{bmatrix} a & 0 \\ b & c \end{bmatrix}[ab0c]\begin{bmatrix} \overline a & \overline b \\ 0 & \overline c \end{bmatrix} ​

= [a2+b2a2bcabcbbac2(b2+c2)]\begin{bmatrix} ||a||^2+||b||^2−||a||^2 & \overline bc−a \overline b \\ \overline cb−b \overline a & |c||^2−(||b||^2+||c||^2) \end{bmatrix}


= [b2bcabcbbab2]\begin{bmatrix} b||^2 & \overline bc−a \overline b \\ \overline cb−b \overline a & −||b||^2 \end{bmatrix}


So b=0 and A is diagonal ,if the result is true.For n≥2, Let A= [T0va]\begin{bmatrix} T & 0 \\ v & a \end{bmatrix} where T is a n x n triangular matrix,v a 1 x n matrix and a is a complex number . Since AA=AA we have,


[0000]\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}= [Tv0a]\begin{bmatrix} T^* & v* \\ 0 & \overline a \end{bmatrix}[T0va]\begin{bmatrix} T & 0 \\ v & a \end{bmatrix}-[T0va]\begin{bmatrix} T & 0 \\ v & a \end{bmatrix} [Tv0a]\begin{bmatrix} T^* & v* \\ 0 & \overline a \end{bmatrix}


=  [TT+vvvaava2]\begin{bmatrix} T^*T + vv*& v*a \\ \overline a v & |a|^2 \end{bmatrix}-  [TTTvvTv2+a2]\begin{bmatrix} TT^* & Tv* \\ vT* & |v|^2 +|a|^2 \end{bmatrix}


=  [TTTT+vvvaTvavvTv2]\begin{bmatrix} T^*T- TT*+vv* & v*a-Tv* \\ \overline a v-vT & -|v|^2 \end{bmatrix}


Hence v=0 and T is normal .since T is lower triangular,then T is normal if A is diagonal.


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