Answer to Question #159352 in Linear Algebra for Abhipsita Das

Question #159352

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}.


1
Expert's answer
2021-02-01T05:42:06-0500

Let

B=(abcd)UB=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\isin U^\bot

Then BB is orthogonal to any element in UU and, in particular, to the basis elements. Thus,

Tr((abcd)(1001))=Tr(abcd)=ad=0Tr(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\cdot \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix})=Tr\begin{pmatrix} a & -b \\ c & -d \end{pmatrix}=a-d=0

Thus, a=da=d . Similarly, dotting with the other basis elements, we get the two following equations:

Tr((abcd)(0100))=Tr(0a0c)=c=0Tr(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\cdot \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix})=Tr\begin{pmatrix} 0 & a \\ 0 & c \end{pmatrix}=c=0

Tr((abcd)(0010))=Tr(b0d0)=b=0Tr(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\cdot \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix})=Tr\begin{pmatrix} b & 0 \\ d & 0 \end{pmatrix}=b=0

Thus, using the fact that a=da=d and c=b=0c=b=0 , we get that every matrix in UU^\bot is of the form

(a00a)\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}

Thus, UU^\bot is spanned by the identity matrix.


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