Matrix Operators Here's a sample question for my upcoming Linear Algebra test:
Let $B_1=\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}$ and $B_2=\begin{bmatrix}
\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}
\end{bmatrix}$.
Determine if $B_2$ is a basis in $\mathbb{R}^2$.  If so, find the matrix representation of the operator $T\in Aut(\mathbb{R}^2)$ such that $T(B_1)=(B_2)$.  Compute the corresponding matrix representation for the associated coordinate transformation,under the action of the linear transformation $T$.
Can someone walk me through this?  Thanks!
 A: To determine whether $B$ is a basis in $\mathbb{R}^2$, what we do is to check if $\Big(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\Big)$ and $\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big)$ are linearly  indepdendent as vectors in $\mathbb{R}^2$. For this, we have to take a linear combination of these elements, equate it to zero, and check if all the coefficients are zero.
So our starting point is: Let $c_1,c_2$ be such that:
$$
c_1 \bigg(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\bigg) + c_2 \bigg(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\bigg) = (0,0)
$$
Adding component wise,
$$
 \bigg((c_1+c_2)\frac{1}{\sqrt{2}} ,(c_2-c_1)\frac{1}{\sqrt{2}}\bigg) = (0,0)
$$
Equating components, since $\frac{1}{\sqrt{2}} \neq 0 \neq -\frac{1}{\sqrt{2}}$, we have:
$$
c_1+c_2 = 0, c_2-c_1=0
$$
Finally solving, we get $c_1=c_2=0$. Hence the above two elements are linearly independent, and form a basis on $\mathbb{R}^2$.
To find the representation of $T$ transforming $B_1$ to $B_2$, is child's play. This is because $T(I) = AI=A$ where $A$ is some $2 \times 2$ matrix, hence $T(I) = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$ implies that $A = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}$. Hence this matrix transforms from the basis $B_1$ to the basis $B_2$.
I shall leave you to find the coordinate transformation matrix given this information.
