Let V be the vector space of real 2 × 2 matrices with inner product
(A|B) = tr(B
tA).
Let U be the subspace of V consisting of the symmetric matrices. Find an orthog-
onal basis for U
⊥ where U
⊥ = {A ∈ V | (A|B) = 0 ∀B ∈ U}.
Let
"B=\\begin{pmatrix}\n a & b \\\\\n c & d\n\\end{pmatrix}\\isin U^\\bot"
Then "B" is orthogonal to any element in "U" and, in particular, to the basis elements. Thus,
"Tr(\\begin{pmatrix}\n a & b \\\\\n c & d\n\\end{pmatrix}\\cdot \\begin{pmatrix}\n 1 & 0 \\\\\n 0 & -1\n\\end{pmatrix})=Tr\\begin{pmatrix}\n a & -b \\\\\n c & -d\n\\end{pmatrix}=a-d=0"
Thus, "a=d" . Similarly, dotting with the other basis elements, we get the two following equations:
"Tr(\\begin{pmatrix}\n a & b \\\\\n c & d\n\\end{pmatrix}\\cdot \\begin{pmatrix}\n 0 & 1 \\\\\n 0 & 0\n\\end{pmatrix})=Tr\\begin{pmatrix}\n 0 & a \\\\\n 0 & c\n\\end{pmatrix}=c=0"
"Tr(\\begin{pmatrix}\n a & b \\\\\n c & d\n\\end{pmatrix}\\cdot \\begin{pmatrix}\n 0 & 0 \\\\\n 1 & 0\n\\end{pmatrix})=Tr\\begin{pmatrix}\n b & 0 \\\\\n d & 0\n\\end{pmatrix}=b=0"
Thus, using the fact that "a=d" and "c=b=0" , we get that every matrix in "U^\\bot" is of the form
"\\begin{pmatrix}\n a & 0 \\\\\n 0 & a\n\\end{pmatrix}"
Thus, "U^\\bot" is spanned by the identity matrix.
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