Answer to Question #159352 in Linear Algebra for Abhipsita Das

Question

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}.

Answer 1

Let

$B=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\isin U^\bot$

Then $B$ is orthogonal to any element in $U$ and, in particular, to the basis elements. Thus,

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

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

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

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

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

$\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$

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