- #1

- 106

- 11

Now I manipulate the equation to give this

A^2-A=0. -->A(A-I)= 0

So A can be I or 0

Are there any other values A can take?

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- #1

- 106

- 11

Now I manipulate the equation to give this

A^2-A=0. -->A(A-I)= 0

So A can be I or 0

Are there any other values A can take?

- #2

Gold Member

- 2,002

- 1,063

For an example

[tex]A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}[/tex]

satisfies the relation. It is I for some dimension, 0 for others. In general it is named as projection operator.

[tex]A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}[/tex]

satisfies the relation. It is I for some dimension, 0 for others. In general it is named as projection operator.

Last edited:

- #3

Mentor

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- 9,127

Yes, lots more. Since ##A^2 - A = 0## (i.e., the zero matrix), then ##|A^2 - A| = |0|## or ##|A||A - I| = |0|##.

Now I manipulate the equation to give this

A^2-A=0. -->A(A-I)= 0

So A can be I or 0

Are there any other values A can take?

The last equation is satisfied for any square matrix A whose determinant is zero, or any square matrix for which ##|A - I| = 0##. Any non-invertible matrix A will satisfy ##|A| = 0##, and similar for ##A - I##.

This is a somewhat confusing way to say that your example maps a vector <x, y, z> to <x, y, 0>.It is I for some dimension, 0 for others. In general it is named as projection operator.

- #4

Science Advisor

Gold Member

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Notice the ring of matrices is not an integral domain, i. e., ##AB=0 ## does not imply either A or B is zero. And notice there is always a solution to ##A^n=I##: Rotation by an angle ##2 \mathbb \pi/n ##

Now I manipulate the equation to give this

A^2-A=0. -->A(A-I)= 0

So A can be I or 0

Are there any other values A can take?

Last edited by a moderator:

- #5

Science Advisor

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- 728

These are called idempotent matrices. Even in dimension ##2##, there is a whole family of examples.

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