Does this kind of square matrix exist?

[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.

 

[Mark44]

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>.

 

[WWGD]

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 $2 \mathbb \pi/n$

 

[nuuskur]

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