Correct me if anything is wrong what I say, I just start to study linear algebra
The question that I encounter is as following:
Suppose that S, the subspace of $\mathbb{R}^{2 \times 2}$ formed by symetric matrices of 2nd order ($A \in \mathbb{R}^{2 \times 2}: A^T = A)$. We also know that $\mathcal{B} = \{A_1,A_2,A_3\}$. $A_1 = \begin{pmatrix} 1& -2 \\ -2 & 1\end{pmatrix}, A_2 = \begin{pmatrix} 1& 3 \\ 3 & 6\end{pmatrix}, A_3 = \begin{pmatrix} -1& 1 \\ 1 & -3 \end{pmatrix}$ forms a basis of $S$. Suppose we have $x = \begin{pmatrix} a & b \\ b & c\end{pmatrix} \in S$, find the the coordinate relative to the basis, $[x]_{\mathcal{B}}$
We know that a linear transformation like following $T: \mathbb{R}^{2 \times 2} \to \mathbb{R}^4$ should be isomorhpism which means:
$$\displaystyle T(\begin{pmatrix} a & b \\ c& d\end{pmatrix}) = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}$$
From the textbook I have learnt that, if we have a basis $\mathcal{B} = \{b_1,b_2,\dots,b_n\}$, then the vector equation $x = c_1b_1 + c_2b_2 + \dots + c_nb_n = P_{\mathcal{B}}[x]_{\mathcal{B}}$ and $[x]_{\mathcal{B}}$ is the coordinate of x rleative to the basis.
Also, from the textbook, I learned that since $b_1,b_2,\dots,b_n$ form a basis, then they are inverivle and so is $P_{\mathcal{B}}$. As a result, we have $[x]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}x$
This is what I plan to do with the question I have encountered. However, if I try to write $A_1,A_2,A_3$ into column form and combine them, I would have a matrix of $4 \times 3$ which is not invertible since it is not a square matrix.
I am not sure where I did wrong
Best Answer
$S$ is a $3$-dimensional space, so the change of coordinate matrix is $3\times3$. $S$ is isomorphic to $\Bbb R^3$, for instance via $\begin{pmatrix}a&b\\b&c\end{pmatrix}\mapsto\begin{pmatrix}a\\b\\c\end{pmatrix}$. Then the change of basis matrix would be $\begin{pmatrix}1&1&-1\\-2&3&1\\1&6&-3\end{pmatrix}$.