M=(I_n-\frac{1}{n}11^T) be a matrix where 1 = (1,1,...,1)^T ∈ Rⁿ and In is the identity matrix of order n. then which of the following is/are true.

The correct options are:
1. M is a projection matrix
2. M^T = M

To understand why these options are correct, let us analyze the properties of the matrix:
$$M=I_n-\frac{1}{n}11^T$$
where 1 = (1, 1, ..., 1)^T is a column vector of all ones of dimension n, and I_n is the n x n identity matrix.

Step 1: Verify if M is symmetric (M^T = M)
Taking the transpose of M:
$$M^T={(I_n-\frac{1}{n}11^T)}^T$$
Using the properties of matrix transpose, (A - B)^T = A^T - B^T and (A B)^T = B^T A^T:
$$M^T=I_n^T-\frac{1}{n}{\left(11^T\right)}^T$$
Since I_n is symmetric, I_n^T = I_n. Also, for the outer product of the vector 1:
$${\left(11^T\right)}^T={\left(1^T\right)}^T{1}^T=11^T$$
Substituting these back, we get:
$$M^T=I_n-\frac{1}{n}11^T=M$$
Thus, the option M^T = M is correct.

Step 2: Verify if M is a projection matrix (M² = M)
A matrix is a projection matrix if and only if it is idempotent, meaning M² = M. Let us calculate M²:
$$M^2=(I_n-\frac{1}{n}11^T)(I_n-\frac{1}{n}11^T)$$
Expanding the terms:
$$M^2=I_n-\frac{2}{n}11^T+\frac{1}{n^2}\left(11^T\right)\left(11^T\right)$$
Now we simplify the term (1 1^T)(1 1^T). Since matrix multiplication is associative, we can write:
$$\left(11^T\right)\left(11^T\right)=1\left(1^{T}1\right)1^T$$
Since 1^T 1 is the inner product of a vector of all ones of length n with itself, we have:
$$1^{T}1=1+1+...+1=n$$
Therefore:
$$\left(11^T\right)\left(11^T\right)=n\left(11^T\right)$$
Substituting this back into the equation for M²:
$$M^2=I_n-\frac{2}{n}11^T+\frac{1}{n^2}\left(n11^T\right)$$
Simplifying the coefficients:
$$M^2=I_n-\frac{2}{n}11^T+\frac{1}{n}11^T$$
$$M^2=I_n-\frac{1}{n}11^T=M$$
Since M² = M, M is idempotent, and therefore M is a projection matrix.