Let $L$ be a linear operator on vector space $R^3$ defined with $L(x,y,z)=(-x+2y+(3-c)z,(c+1)x-y+2z,2x+2y-z),$ with $c$ from $R$.

Question

Prove that $L$ is symmetric operator of euclidean space $R^3$ with standard dot product if and only if $c=1$.

So I think i proved that if $c=1$ then $L$ is a symmetric operator. Its matrix in standard basis (which is of course orthonormal) is symmetric, so does that means that $L$ is symmetric operator?

How can I use the fact that $L$ is symmetric with standard dot product to prove that $c=1$? This is my first question here and english is not my native language so I'm not sure I used right terminology.

Answer 1

The matrix representing $L$ with respect to the standard basis $\mathcal{E}$ (which is orthonormal with respect to the standard dot product) is given by

$$ [L]_{\mathcal{E}} = \begin{pmatrix} -1 & 2 & 3 - c \\ c+1 & -1 & 2 \\ 2 & 2 & -1 \end{pmatrix}. $$

Now, use the theorem that $L$ is a symmetric operator iff $[L]_{\mathcal{E}}$ is a symmetric matrix. This clearly occurs iff $c = 1$.