GATE DA 2026 | Question: 42

Question

Let $M=\left(I_{n}-\frac{1}{n} \mathbf{1 1}^{T}\right)$ be a matrix, where $\mathbf{1}=(1,1,1, \ldots, 1)^{T} \in \mathbb{R}^{n}$ and $I_{n}$ is the identity matrix of order $n$.

Which of the following options is/are correct?

• A. $M^{T}=M$
• B. $M^{2}=I_{n}$
• C. $\operatorname{Trace}(M)=n$
• D. $M$ is a projection matrix

Comments

Def( Orthogonal projection matrix):

• A square matrix $P$ is an orthogonal projection matrix if $P^2 = P $ ( $P$ is idempotent) and $P^T = P$ ( $P$ is symmetric.)

Statement:

• If $ P$ is an orthogonal projection matrix $\implies I - P $ is also orthogonal projection matrix, ($I-P$ will be a orthogonal projection matrix onto $Col(P)^\perp$.

Let  $P = \frac{1}{n}11^T$. Then $P^2 = \frac{1}{n^2}11^T11^T = \frac{1}{n^2}\times n \times 1 1^T = \frac{1}{n}11^T = P$ and $P^T = \frac{1}{n}11^T = P$, hence $P$ is an orthogonal projection matrix onto the line $1^T$.

Then From above statement, $I - P$ is also an orthogonal projection matrix, which is nothing but $M$ is ans orthogonal projection matrix. Option A, D are True. $M$ is idempotent, so option B is False.

 

Then eigen value of $-\frac{1}{n}11^T  = (-\frac{1}{n}1^T1, 0, \ldots, 0) = (-1, 0, \ldots, 0)$. Then the eigen values of $M$ are $(-1 +1, 0+1, \ldots, 0+1)  = (0, 1, \ldots, 1)$.

Then trace of matrix $M$ equals = $0 + 1  + \ldots + 1 = n -1$. So C is false.

 

SO the true options are A, D

 

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Definition of Projection Matrix :

 

Answer 1

Given
\[
M = I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T
\]
where
\[
\mathbf{1} = (1,1,\dots,1)^T \in \mathbb{R}^n
\]

Method 1: Intution based (Take n = 2)

Let n = 2.

\[
\mathbf{1} =
\begin{pmatrix}
1 \\
1
\end{pmatrix}
\]

\[
\mathbf{1}\mathbf{1}^T =
\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}
\]

\[
I_2 =
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\]

\[
M = I_2 - \frac{1}{2}
\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}
=
\begin{pmatrix}
\frac12 & -\frac12 \\
-\frac12 & \frac12
\end{pmatrix}
\]

Clearly \( M^T = M \).

Now compute \( M^2 \):

\[
M^2 =
\begin{pmatrix}
\frac12 & -\frac12 \\
-\frac12 & \frac12
\end{pmatrix}
\begin{pmatrix}
\frac12 & -\frac12 \\
-\frac12 & \frac12
\end{pmatrix}
=
\begin{pmatrix}
\frac12 & -\frac12 \\
-\frac12 & \frac12
\end{pmatrix}
= M
\]

Trace:

\[
\text{Trace}(M) = \frac12 + \frac12 = 1 = 2 - 1
\]

This suggests that in general \( \text{Trace}(M) = n - 1 \).

From this example we observe that (A) and (D) appear true, while (B) and (C) appear false.

This method builds intuition.

Method 2: General approach (time consuming)

(A) Symmetry:

\[
M^T = I_n^T - \frac{1}{n}(\mathbf{1}\mathbf{1}^T)^T
= I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T
= M
\]

Hence (A) is true.

(B) Compute \( M^2 \):

\[
M^2 = \left(I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T\right)^2
\]

\[
= I_n - \frac{2}{n}\mathbf{1}\mathbf{1}^T
+ \frac{1}{n^2}(\mathbf{1}\mathbf{1}^T)(\mathbf{1}\mathbf{1}^T)
\]

Since

\[
(\mathbf{1}\mathbf{1}^T)(\mathbf{1}\mathbf{1}^T)
= \mathbf{1}(\mathbf{1}^T\mathbf{1})\mathbf{1}^T
\]

and

\[
\mathbf{1}^T\mathbf{1} = n
\]

we obtain

\[
M^2 =
I_n - \frac{2}{n}\mathbf{1}\mathbf{1}^T
+ \frac{1}{n}\mathbf{1}\mathbf{1}^T
= I_n - \frac{1}{n}\mathbf{1}\mathbf{1}^T
= M
\]

Thus (B) is false.

(C) Trace:

\[
\text{Trace}(M)
= \text{Trace}(I_n)
- \frac{1}{n}\text{Trace}(\mathbf{1}\mathbf{1}^T)
\]

\[
= n - \frac{1}{n}(n)
= n - 1
\]

Hence (C) is false.

(D) Since

\[
M^T = M
\quad \text{and} \quad
M^2 = M
\]

M is a projection matrix.

Final Answer:

\[
\boxed{(A) \text{ and } (D)}
\]

Answer 2

Solution by Sachin sir - https://www.youtube.com/watch?v=cfBFT01LZ2A&t=9674s

Answer 3

.