If T : Rn → Rn is given by T (x1, x2, ..., xn) = (x2 + x3, x3, ..., xn, 0) then write downthe matrix of T w.r.t. the standard basis of Rn

The transformation T : Rn → Rn is given by T (x1, x2, ..., xn) = (x2 + x3, x3, ..., xn, 0). We want to find the matrix representation of this transformation with respect to the standard basis of Rn.

The standard basis of Rn is {e1, e2, ..., en}, where ei is a vector with 1 in the i-th position and 0s elsewhere.

We can find the matrix representation by applying the transformation T to each basis vector and writing the result as a column in the matrix.

1. Apply T to e1: T(e1) = (0, 0, ..., 0) which is the zero vector. So the first column of the matrix is all zeros.

2. Apply T to e2: T(e2) = (1, 0, ..., 0). So the second column of the matrix is (1, 0, ..., 0).

3. Apply T to e3: T(e3) = (1, 1, 0, ..., 0). So the third column of the matrix is (1, 1, 0, ..., 0).

4. For i > 3, apply T to ei: T(e

The transformation T is defined as T(x1, x2, ..., xn) = (x2 + x3, x3, ..., xn, 0). This means that the first element of the output vector is the sum of the second and third elements of the input vector, the second element of the output vector is the third element of the input vector, and so on, until the last element of the output vector, which is always 0.

The standard basis of R^n is the set of vectors e1, e2, ..., en, where each vector has a 1 in one position and 0s everywhere else. For example, e1 = (1, 0, ..., 0), e2 = (0, 1, ..., 0), and so on.

The matrix of T with respect to the standard basis is obtained by applying T to each basis vector and writing the output as a column of the matrix.

When we apply T to e1, we get (0, 0, ..., 0), because there are no x2 or x3 in e1.

When we apply T to e2, we get (1, 0, ..., 0), because x2 is 1 in e2 and all other xi are 0.

When we apply T to e3, we get (1, 1, 0, ..., 0), because x3 is 1 in e3 and all other xi are 0.

For i > 3, when we apply T to ei, we get (0, ..., 0, 1, 0, ..., 0), where the 1 is in the (i-1)th position, because xi is 1 in ei and all other xi are 0.

Finally, when we apply T to en, we get (0, ..., 0, 1, 0), because xn is 1 in en and all other xi are 0.

So, the matrix of T with respect to the standard basis is:

0 1 1 0 ... 0 0
0 0 1 0 ... 0 0
0 0 0 1 ... 0 0
. . . . ... . .
. . . . ... . .
. . . . ... . .
0 0 0 0 ... 1 0
0 0 0 0 ... 0 0

This is a n x n matrix, where n is the dimension of the vector space.