Show that a triangular matrix is normal if it is diagonal
Answer:
Given:
A triangular matrix is normal if it is diagonal.
For n=2, Let A= "\\begin{bmatrix}\n a & 0 \\\\\n b & c\n\\end{bmatrix}" such that
A∗A=AA∗= "\\begin{bmatrix}\n \\overline a & b \\\\\n b & \\overline c\n\\end{bmatrix}\n\n\u200b""\\begin{bmatrix}\n a & 0 \\\\\n b & c\n\\end{bmatrix}" -"\\begin{bmatrix}\n a & 0 \\\\\n b & c\n\\end{bmatrix}""\\begin{bmatrix}\n \\overline a & \\overline b \\\\\n 0 & \\overline c\n\\end{bmatrix}\n\n\u200b"
= "\\begin{bmatrix}\n ||a||^2+||b||^2\u2212||a||^2 & \\overline bc\u2212a \\overline b \\\\\n \\overline cb\u2212b \\overline a & |c||^2\u2212(||b||^2+||c||^2)\n\\end{bmatrix}"
= "\\begin{bmatrix}\n b||^2 & \\overline bc\u2212a \\overline b \\\\\n \\overline cb\u2212b \\overline a & \u2212||b||^2\n\\end{bmatrix}"
So b=0 and A is diagonal ,if the result is true.For n≥2, Let A= "\\begin{bmatrix}\n T & 0 \\\\\n v & a\n\\end{bmatrix}" where T is a n x n triangular matrix,v a 1 x n matrix and a is a complex number . Since A∗A=AA∗ we have,
"\\begin{bmatrix}\n 0 & 0 \\\\\n 0 & 0\n\\end{bmatrix}"= "\\begin{bmatrix}\n T^* & v* \\\\\n 0 & \\overline a\n\\end{bmatrix}""\\begin{bmatrix}\n T & 0 \\\\\n v & a\n\\end{bmatrix}"-"\\begin{bmatrix}\n T & 0 \\\\\n v & a\n\\end{bmatrix}" "\\begin{bmatrix}\n T^* & v* \\\\\n 0 & \\overline a\n\\end{bmatrix}"
= "\\begin{bmatrix}\n T^*T + vv*& v*a \\\\\n \\overline a v & |a|^2\n\\end{bmatrix}"- "\\begin{bmatrix}\n TT^* & Tv* \\\\\n vT* & |v|^2 +|a|^2\n\\end{bmatrix}"
= "\\begin{bmatrix}\n T^*T- TT*+vv* & v*a-Tv* \\\\\n \\overline a v-vT & -|v|^2\n\\end{bmatrix}"
Hence v=0 and T is normal .since T is lower triangular,then T is normal if A is diagonal.
Comments
Leave a comment