Let $A: R^2 \to R^2$ be a linear operator. Its matrix representation in the standard basis $(e_1, e_2)$ is not given explicitly. We are given a new basis $(t_1, t_2)$ where $t_1 = e_1 + e_2$ and $t_2 = e_1 - e_2$. The problem asks us to find the matrix of the operator $A$ with respect to the basis $(t_1, t_2)$. We are also told to provide the sum of all elements of this matrix as a single integer number as the answer.

1. Problem Description

Let $A: R^2 \to R^2$ be a linear operator. Its matrix representation in the standard basis $(e_1, e_2)$ is not given explicitly. We are given a new basis $(t_1, t_2)$ where $t_1 = e_1 + e_2$ and $t_2 = e_1 - e_2$. The problem asks us to find the matrix of the operator $A$ with respect to the basis $(t_1, t_2)$. We are also told to provide the sum of all elements of this matrix as a single integer number as the answer.

2. Solution Steps

First, we express $e_1$ and $e_2$ in terms of $t_1$ and $t_2$.

We have the following system of equations:

$t_1 = e_1 + e_2$

$t_2 = e_1 - e_2$

Adding these two equations gives:

$t_1 + t_2 = 2e_1$

So,

$e_1 = \frac{1}{2}(t_1 + t_2)$.

Subtracting the second equation from the first gives:

$t_1 - t_2 = 2e_2$

So,

$e_2 = \frac{1}{2}(t_1 - t_2)$.

Let the matrix of A with respect to the standard basis $(e_1, e_2)$ be $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. This means that

$A(e_1) = ae_1 + ce_2$

$A(e_2) = be_1 + de_2$.

We want to find the matrix of $A$ with respect to the basis $(t_1, t_2)$. Let this matrix be $M' = \begin{pmatrix} p & q \\ r & s \end{pmatrix}$. Then we must have:

$A(t_1) = pt_1 + rt_2$

$A(t_2) = qt_1 + st_2$

Now, we express $A(t_1)$ and $A(t_2)$ in terms of $t_1$ and $t_2$:

$A(t_1) = A(e_1 + e_2) = A(e_1) + A(e_2) = (ae_1 + ce_2) + (be_1 + de_2) = (a+b)e_1 + (c+d)e_2$.

Substituting $e_1 = \frac{1}{2}(t_1 + t_2)$ and $e_2 = \frac{1}{2}(t_1 - t_2)$, we get:

$A(t_1) = (a+b)\frac{1}{2}(t_1+t_2) + (c+d)\frac{1}{2}(t_1-t_2) = \frac{1}{2}[(a+b)+(c+d)]t_1 + \frac{1}{2}[(a+b)-(c+d)]t_2$.

Thus,

$p = \frac{1}{2}(a+b+c+d)$

$r = \frac{1}{2}(a+b-c-d)$.

Similarly,

$A(t_2) = A(e_1 - e_2) = A(e_1) - A(e_2) = (ae_1 + ce_2) - (be_1 + de_2) = (a-b)e_1 + (c-d)e_2$.

Substituting $e_1 = \frac{1}{2}(t_1 + t_2)$ and $e_2 = \frac{1}{2}(t_1 - t_2)$, we get:

$A(t_2) = (a-b)\frac{1}{2}(t_1+t_2) + (c-d)\frac{1}{2}(t_1-t_2) = \frac{1}{2}[(a-b)+(c-d)]t_1 + \frac{1}{2}[(a-b)-(c-d)]t_2$.

Thus,

$q = \frac{1}{2}(a-b+c-d)$

$s = \frac{1}{2}(a-b-c+d)$.

The matrix $M'$ is

$M' = \begin{pmatrix} \frac{1}{2}(a+b+c+d) & \frac{1}{2}(a-b+c-d) \\ \frac{1}{2}(a+b-c-d) & \frac{1}{2}(a-b-c+d) \end{pmatrix}$

The sum of the elements of the matrix is

$\frac{1}{2}(a+b+c+d) + \frac{1}{2}(a-b+c-d) + \frac{1}{2}(a+b-c-d) + \frac{1}{2}(a-b-c+d) = \frac{1}{2}(4a) = 2a$.

Since we need the sum of all elements of the matrix as a *single integer number*, we need to consider the case when $A$ is the sum operator. Then $A(x,y) = x+y$. Thus $A(e_1) = A(1,0) = 1+0 = 1$, and $A(e_2) = A(0,1) = 0+1 = 1$. So $Ae_1 = 1e_1 + 0e_2$ and $Ae_2 = 1e_1 + 0e_2$. Then $a=1, b=1, c=0, d=0$.

The sum of the matrix elements is then $2a = 2(1) = 2$.

3. Final Answer

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