I am self studying linear algebra and trying quiz questions and I am struck on this particular question whose image I am adding.
Let $$A=\begin{bmatrix}1&2\\4&3\end{bmatrix}\in\mathbf{M}_2(\mathbb{R})$$ and $\phi:\mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}$ be the bilinear map defined by $$\phi(v,w)=v^TAw$$
I need help in verifying this particular part which is to be proved true / false.
There exists a $2\times2$ symmetric matrix B such that $\phi(v, v) $ =$ v^{T}Bv$ for all $v\in \mathbb{R}^{2} $ .
I have solved other parts but I am really struck on this . Can someone please tell reasoning behind this or a counterexample.
Best Answer
You know that $$\phi(v,v)=(v_1, v_2)^T(v_1+2v_2, 4v_1+3v_2)=v_1^2+6v_1v_2+3v_2^2$$
Let $B=\begin{bmatrix}a & b \\ b & c\end{bmatrix}$. Then $v^TBv=av_1^2+2bv_1v_2+cv_2^2$. Equate this to $\phi(v,v)$ above to get
$$v_1^2+6v_1v_2+3v_2^2=av_1^2+2bv_1v_2+cv_2^2$$
Since $v$ is arbitrary, you can adjust values of $v_1,v_2$ to find $a,b,c$. Put $v_2=0$ to get $a=1$, $v_1=0$ to get $c=3$, and $v_1=v_2=1$ to get $10=4+2b\implies b=3$.