Question:
M = (I$_n$-$\frac{1}{n}$11$^T$) be a matrix where 1 = (1,1,...,1)$^T$ $\in$ R$^n$ and I$_n$ is the identity matrix of order n. then which of the following is/are true.
M = (I$_n$-$\frac{1}{n}$11$^T$) be a matrix where 1 = (1,1,...,1)$^T$ $\in$ R$^n$ and I$_n$ is the identity matrix of order n. then which of the following is/are true.
Show Hint
The matrix $J=11^T$ (matrix of all ones) has a useful property: $J^2 = nJ$. This is frequently used when working with the centering matrix $M = I - \frac{1}{n}J$ and is worth memorizing.
Updated On: Feb 23, 2026
- M is a projection matrix
- M$^T$ = M
- M$^2$ = I
- trace (M) = n
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The Correct Option is A, B
Solution and Explanation
Step 1: Understanding the Question:
We are given a specific matrix M, known as the centering matrix, and we need to verify several of its properties: whether it's a projection matrix, symmetric, its square, and its trace.
Step 2: Key Definitions:
- Symmetric Matrix: A matrix M is symmetric if $M^T = M$.
- Projection Matrix: A matrix M is a projection matrix if it is symmetric ($M^T = M$) and idempotent ($M^2 = M$).
- Trace: The trace of a matrix is the sum of its diagonal elements.
Step 3: Detailed Explanation:
Let $J = 11^T$. J is an $n \times n$ matrix where every entry is 1.
So, $M = I_n - \frac{1}{n}J$. - Check (B) Symmetry ($M^T = M$): \[ M^T = (I_n - \frac{1}{n}J)^T = I_n^T - (\frac{1}{n}J)^T = I_n - \frac{1}{n}J^T \] The matrix J is symmetric because $J_{ij} = 1$ for all i,j, so $J^T = J$. \[ M^T = I_n - \frac{1}{n}J = M \] Therefore, statement (B) is true. - Check Idempotence ($M^2 = M$): \[ M^2 = (I_n - \frac{1}{n}J)(I_n - \frac{1}{n}J) = I_n^2 - \frac{1}{n}J - \frac{1}{n}J + \frac{1}{n^2}J^2 = I_n - \frac{2}{n}J + \frac{1}{n^2}J^2 \] We need to compute $J^2$. $J^2 = (11^T)(11^T) = 1(1^T1)1^T$. The inner product $1^T1 = \sum_{i=1}^n 1^2 = n$. So, $J^2 = 1(n)1^T = n(11^T) = nJ$. Substitute this back: \[ M^2 = I_n - \frac{2}{n}J + \frac{1}{n^2}(nJ) = I_n - \frac{2}{n}J + \frac{1}{n}J = I_n - \frac{1}{n}J = M \] Since $M^2 = M$, M is idempotent. - Check (A) Projection Matrix: A matrix is a projection matrix if it is both symmetric and idempotent. We have shown that M is both. Therefore, statement (A) is true. - Check (C) $M^2 = I$: We just showed that $M^2 = M$. Since M is not the identity matrix (unless n=1, a trivial case), this statement is false. - Check (D) trace(M) = n: The trace is a linear operator: $\text{trace}(A+B) = \text{trace}(A) + \text{trace}(B)$. \[ \text{trace}(M) = \text{trace}(I_n - \frac{1}{n}J) = \text{trace}(I_n) - \frac{1}{n}\text{trace}(J) \] - $\text{trace}(I_n)$ is the sum of n ones, which is n. - $\text{trace}(J)$ is the sum of n ones (from its diagonal), which is n. \[ \text{trace}(M) = n - \frac{1}{n}(n) = n - 1 \] Since $\text{trace}(M) = n-1$, the statement that $\text{trace}(M) = n$ is false. Step 4: Final Answer:
The true statements are (A) and (B).
We are given a specific matrix M, known as the centering matrix, and we need to verify several of its properties: whether it's a projection matrix, symmetric, its square, and its trace.
Step 2: Key Definitions:
- Symmetric Matrix: A matrix M is symmetric if $M^T = M$.
- Projection Matrix: A matrix M is a projection matrix if it is symmetric ($M^T = M$) and idempotent ($M^2 = M$).
- Trace: The trace of a matrix is the sum of its diagonal elements.
Step 3: Detailed Explanation:
Let $J = 11^T$. J is an $n \times n$ matrix where every entry is 1.
So, $M = I_n - \frac{1}{n}J$. - Check (B) Symmetry ($M^T = M$): \[ M^T = (I_n - \frac{1}{n}J)^T = I_n^T - (\frac{1}{n}J)^T = I_n - \frac{1}{n}J^T \] The matrix J is symmetric because $J_{ij} = 1$ for all i,j, so $J^T = J$. \[ M^T = I_n - \frac{1}{n}J = M \] Therefore, statement (B) is true. - Check Idempotence ($M^2 = M$): \[ M^2 = (I_n - \frac{1}{n}J)(I_n - \frac{1}{n}J) = I_n^2 - \frac{1}{n}J - \frac{1}{n}J + \frac{1}{n^2}J^2 = I_n - \frac{2}{n}J + \frac{1}{n^2}J^2 \] We need to compute $J^2$. $J^2 = (11^T)(11^T) = 1(1^T1)1^T$. The inner product $1^T1 = \sum_{i=1}^n 1^2 = n$. So, $J^2 = 1(n)1^T = n(11^T) = nJ$. Substitute this back: \[ M^2 = I_n - \frac{2}{n}J + \frac{1}{n^2}(nJ) = I_n - \frac{2}{n}J + \frac{1}{n}J = I_n - \frac{1}{n}J = M \] Since $M^2 = M$, M is idempotent. - Check (A) Projection Matrix: A matrix is a projection matrix if it is both symmetric and idempotent. We have shown that M is both. Therefore, statement (A) is true. - Check (C) $M^2 = I$: We just showed that $M^2 = M$. Since M is not the identity matrix (unless n=1, a trivial case), this statement is false. - Check (D) trace(M) = n: The trace is a linear operator: $\text{trace}(A+B) = \text{trace}(A) + \text{trace}(B)$. \[ \text{trace}(M) = \text{trace}(I_n - \frac{1}{n}J) = \text{trace}(I_n) - \frac{1}{n}\text{trace}(J) \] - $\text{trace}(I_n)$ is the sum of n ones, which is n. - $\text{trace}(J)$ is the sum of n ones (from its diagonal), which is n. \[ \text{trace}(M) = n - \frac{1}{n}(n) = n - 1 \] Since $\text{trace}(M) = n-1$, the statement that $\text{trace}(M) = n$ is false. Step 4: Final Answer:
The true statements are (A) and (B).
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