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Let $ A: \mathbb{C}^2 \to \mathbb{C}^2$ be given by

$ A =\begin{bmatrix} 1 & 1 \\ -1 & -1 \\ \end{bmatrix} $

(with respect to the standard basis). Find a basis of $\mathbb{C}^2$ with respect to which the matrix reprsentation of $A$ is

$D = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$

So far, I have set up the problem as such:

$ D = C^{-1}AC$ and substituted it as

$\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} = C^{-1}\begin{bmatrix} 1 & 1 \\ -1 & -1 \\ \end{bmatrix}C$, in which $C$ is the basis. Thus far, I have only seen examples in which I was given the basis and asked to solve for the matrix representation. I am not sure what the problem means by "with respect to the standard basis" either. How do I go about solving this problem?

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$A =\begin{bmatrix} 1 & 1 \\ -1 & -1 \\ \end{bmatrix}$ with respect to standard basis means that the columns of the matrix are precisely $Ae_1$ and $Ae_2$, written in the standard basis:

$$Ae_1 = e_1 - e_2 = (1,-1)$$ $$Ae_2 = e_1 - e_2 = (1,-1)$$

We get that $$A(x,y) = A(xe_1 + ye_2) = xAe_1 + yAe_2 = (x+y,-x-y), \quad\forall (x,y)\in\mathbb{C}^2$$

Now, let's try to find a vector $(x,y) \in \mathbb{C}^2$ such that $A(x,y) = 0$, as this will be our first basis vector for the matrix $\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$.

We have:

$$0 = A(x+y) = (x+y,-x-y)\implies x+y = 0$$

so we can for example take $(1,-1)$.

Now let's try to find $(u,v)\in\mathbb{C}^2$ such that $A(u,v) = (u+v,-u-v)= (1,-1)$. We see that $(u,v) = \left(\frac12,\frac12\right)$.

Furthermore, $(1,-1)$ and $\left(\frac12,\frac12\right)$ are linearly dependent since they are not proportional, so $\left\{(1,-1), \left(\frac12,\frac12\right)\right\}$ is a basis for $\mathbb{C}^2$.

Now you can once again verify that $A(1,-1) = 0$ and $A\left(\frac12,\frac12\right) = (1,-1)$ so the matrix representation of $A$ in this basis is indeed $\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$.

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