Idempotent matrix

Square matrix

The square matrix is the matrix in which the number of rows = number of columns. So matrix $A=\left[a_{i j}\right]_{m \times n}$ is said to be a square matrix when $m=n$.

$
\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]_{3 \times 3} \text { or, }\left[\begin{array}{cc}
2 & -4 \\
7 & 3
\end{array}\right]_{2 \times 2}$

Idempotent matrix

An idempotent matrix is defined as a square matrix that remains unchanged when multiplied by itself.

A square matrix is said to be an idempotent matrix if it satisfies the condition A² = A.

Properties of Idempotent Matrix

1) Every identity matrix is also an idempotent matrix, as the identity matrix gives the same matrix when multiplied by itself.

2) All idempotent matrices are singular matrices, except for the identity matrix.

3) The determinant of an idempotent matrix is either one or zero. If A is an idempotent matrix then |A| = 1 or 0.

4) The non-diagonal entries of an idempotent matrix can be non-zero entries.

5) The trace of an idempotent matrix is always an integer and equal to the rank of the matrix

6) The relationship between idempotent and involuntary matrices is if A is a square matrix “A” is said to be an idempotent matrix if and only if P = 2A − I is an involuntary matrix.

Recommended Video Based on Ídempotent Matrices

Solved Examples Based on Idempotent Matrices

Eaxmple 1: What is an index of the idempotent matrix

$
A=\left[\begin{array}{ccc}
2 & 5 & 14 \\
1 & 3 & 8 \\
-1 & -2 & -6
\end{array}\right]
$

1) 2
2) 3
3) 4
4) 5

Solution: We know that the idempotent matrix - $A^2=A$

$
A^2=\left[\begin{array}{ccc}
2 & 5 & 14 \\
1 & 3 & 8 \\
-1 & -2 & -6
\end{array}\right]\left[\begin{array}{ccc}
2 & 5 & 14 \\
1 & 3 & 8 \\
-1 & -2 & -6
\end{array}\right]=\left[\begin{array}{ccc}
-5 & -3 & 16 \\
-3 & -2 & -10 \\
2 & 1 & 6
\end{array}\right]
$

now if we multiply $A^2 \times A^2$
we get $A^4=I$.
Thus A is an idempotent matrix of order $=4$
Hence, the answer is the option (3).

Example 2: Which of the following matrices is idempotent?

1) \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
2) \( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \)
3) \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \)
4) \( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \)

Solution:
A matrix \( A \) is idempotent if \( A^2 = A \).

1. Matrix A: \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)

$A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

\( A^2 = A \), so Matrix A is idempotent.

2. Matrix B: \( \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \)

$B^2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$

\( B^2 \neq B \), so Matrix B is not idempotent.

3. Matrix C: \( \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \)

$ C^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

\( C^2 \neq C \), so Matrix C is not idempotent.

4. Matrix D: \( \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \)

$D^2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$

\( D^2 = D \), so Matrix D is idempotent.

Hence, matrices that are idempotent are 1 and 4.

Example 3: Which of the following is a property of an idempotent matrix \( A \)?

1) \( A \) is invertible.
2) \( A \) has a trace equal to its rank.
3) The eigenvalues of \( A \) are purely imaginary.
4) \( A \) is always diagonalizable.

Solution:
1. False. An idempotent matrix is not necessarily invertible. For instance, a matrix with eigenvalue 0 is not invertible.
2. True. For an idempotent matrix, the trace (sum of eigenvalues) equals the rank because the eigenvalues are either 0 or 1.
3. False. The eigenvalues of an idempotent matrix are 0 and 1, which are not purely imaginary.
4. False. An idempotent matrix is not necessarily diagonalizable, though it can be.

Hence, the answer is option 2.

Example 4: Find the value of a for which the matrix $\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ a & -2 & -3\end{array}\right]$ is idempotent.

1) 2

2) -1

3) 1

4) 0

Solution: We know that, For Idempotent A²=A

$\begin{aligned} & A^2=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ a & -2 & -3\end{array}\right] \times\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ a & -2 & -3\end{array}\right]=\left[\begin{array}{ccc}4+2-4 a & -4-6+8 & -8-8+12 \\ -2-3+4 a & 2+9-8 & 4+12-12 \\ 2 a+2-3 a & -2 a-6+6 & -4 a-8+9\end{array}\right] \\ & A^2=\left[\begin{array}{ccc}6-4 a & -2 & -4 \\ 4 a-5 & 3 & 4 \\ -a+2 & -2 & -3\end{array}\right]=A=\left[\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ a & -2 & -3\end{array}\right]\end{aligned}$

Solving any element we get a=1

Hence, the answer is the option 3.

Example 5: Consider the matrix \( B \):
$B = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}$
Is \( B \) an idempotent matrix?
1) Yes
2) No

Solution:
Compute \( B^2 \):
$B^2 = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \times \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} = \begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}$
Since \( B^2 \neq B \), \( B \) is not idempotent.