Is This Matrix Idempotent?

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In summary, we have discussed the concept of an idempotent matrix and how to prove that a specific matrix is idempotent. We have also shown that the associative property of multiplication extends to any choice of bracketing in longer expressions. The proof of this property is not particularly difficult or interesting, but it involves using induction to show that any bracketing gives the same result as the standard bracketing where operations are performed from left to right.
  • #1
mathmari
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Hey! :eek:

We have that a matrix $A$ is idempotent if it holds that $A^2=A$.

We suppose that $X$ is a $m\times n$-matrix and that $(X^TX)^{-1}$ exists.

I want to show that $A=I_m-X(X^TX)^{-1}X^T$ is idempotent. I have done the following:

$$A^2 =A\cdot A=(I_m-X(X^TX)^{-1}X^T)\cdot (I_m-X(X^TX)^{-1}X^T) \\ =I_m(I_m-X(X^TX)^{-1}X^T)-X(X^TX)^{-1}X^T(I_m-X(X^TX)^{-1}X^T) \\ =I_mI_m-I_mX(X^TX)^{-1}X^T-X(X^TX)^{-1}X^TI_m+(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T) \\ =I_m-X(X^TX)^{-1}X^T-X(X^TX)^{-1}X^T+X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+XI_n(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+X(X^TX)^{-1}X^T \\ =I_m-X(X^TX)^{-1}X^T =A$$

Is this correct? (Wondering)

Is the step from the $3$th to the $4$th line correct? Or doesn't it hold that $(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T)=X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T$ ? (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

We have that a matrix $A$ is idempotent if it holds that $A^2=A$.

We suppose that $X$ is a $m\times n$-matrix and that $(X^TX)^{-1}$ exists.

I want to show that $A=I_m-X(X^TX)^{-1}X^T$ is idempotent. I have done the following:

$$A^2 =A\cdot A=(I_m-X(X^TX)^{-1}X^T)\cdot (I_m-X(X^TX)^{-1}X^T) \\ =I_m(I_m-X(X^TX)^{-1}X^T)-X(X^TX)^{-1}X^T(I_m-X(X^TX)^{-1}X^T) \\ =I_mI_m-I_mX(X^TX)^{-1}X^T-X(X^TX)^{-1}X^TI_m+(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T) \\ =I_m-X(X^TX)^{-1}X^T-X(X^TX)^{-1}X^T+X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+XI_n(X^TX)^{-1}X^T \\ =I_m-2X(X^TX)^{-1}X^T+X(X^TX)^{-1}X^T \\ =I_m-X(X^TX)^{-1}X^T =A$$

Is this correct? (Wondering)

Is the step from the $3$th to the $4$th line correct? Or doesn't it hold that $(X(X^TX)^{-1}X^T)(X(X^TX)^{-1}X^T)=X(X^TX)^{-1}(X^TX)(X^TX)^{-1}X^T$ ? (Wondering)
Your solution is correct. The associative property of multiplication $A(BC) = (AB)C$ extends to any choice of bracketing in longer expressions. For example $(AB)((CD)E) = A((BC)(DE)).$ The proof of this "general associative law" is messy but not particularly hard or interesting, and consists essentially of using induction to show that any bracketing gives the same result as the standard bracketing where the operations are performed in sequence from left to right, like $(((AB)C)D)E$.
 
  • #3
Opalg said:
Your solution is correct. The associative property of multiplication $A(BC) = (AB)C$ extends to any choice of bracketing in longer expressions. For example $(AB)((CD)E) = A((BC)(DE)).$ The proof of this "general associative law" is messy but not particularly hard or interesting, and consists essentially of using induction to show that any bracketing gives the same result as the standard bracketing where the operations are performed in sequence from left to right, like $(((AB)C)D)E$.

Ah ok... Thanks a lot! (Happy)
 

FAQ: Is This Matrix Idempotent?

1. What does it mean for a matrix to be idempotent?

When a matrix is multiplied by itself, the result is the same matrix. In other words, the matrix does not change when it is raised to a power greater than one.

2. How can I tell if a matrix is idempotent?

A matrix is idempotent if and only if it satisfies the equation A^2 = A, where A is the matrix in question. This can be checked by multiplying the matrix by itself and seeing if the result is equal to the original matrix.

3. What are the properties of an idempotent matrix?

Some properties of an idempotent matrix include:

  • The matrix has at least one eigenvalue of 0 or 1.
  • The rank of the matrix is equal to its trace (sum of diagonal elements).
  • The matrix is symmetric if it is also orthogonal.

4. Can a non-square matrix be idempotent?

No, a non-square matrix cannot be idempotent. The definition of an idempotent matrix requires the matrix to be square (same number of rows and columns).

5. How is an idempotent matrix used in mathematics and science?

Idempotent matrices have various applications in mathematics and science, including:

  • In linear algebra, idempotent matrices are used in projections, which are used to project a vector onto a subspace.
  • In probability and statistics, idempotent matrices are used to calculate the variance of a random variable.
  • In computer science, idempotent matrices are used in algorithms for data compression and image processing.

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