# Does this kind of square matrix exist?

• B
• Trollfaz

#### Trollfaz

I had a homework question that gives A as an arbitrary matrix. Then the question states that A^2=A
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?

• anuttarasammyak

For an example
$$A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}$$
satisfies the relation. It is I for some dimension, 0 for others. In general it is named as projection operator.

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• mathwonk, jasonRF and PeroK
I had a homework question that gives A as an arbitrary matrix. Then the question states that A^2=A
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?
Yes, lots more. Since ##A^2 - A = 0## (i.e., the zero matrix), then ##|A^2 - A| = |0|## or ##|A||A - I| = |0|##.
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##.
It is I for some dimension, 0 for others. In general it is named as projection operator.
This is a somewhat confusing way to say that your example maps a vector <x, y, z> to <x, y, 0>.

I had a homework question that gives A as an arbitrary matrix. Then the question states that A^2=A
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?
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 ##

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