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I believe I'm missing an important concept and I need your help.

I have the following question:

"If $A^2 - A = 0$ then $A = 0$ or $A = I$"

I know that the answer is FALSE (only because someone told me) but when I try to find out a concrete matrix which satisfies this equation (which isn't $0$ or $I$) I fail.

Can you please give me a direction to find a concrete matrix? What is the idea behind this question?

Guy

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2  
Every diagonal matrix that has $0$ and $1$ on the main diagonal. –  user26857 Oct 28 '12 at 17:54
    
As for why it's false, you can factor the left side into $A(A-I)=0$. Certainly $A=0$ and $A=I$ are solutions, but since nonzero matrices can multiply to zero, other options still exist. –  Robert Mastragostino Oct 28 '12 at 18:05
1  
One should perhaps mention that there is a name for matrices with this property (or better, for the linear mappings they induce)---they are called projections. Geometrically, they are projections onto a subspace of the ambient space along a complementary subspace. This implies that there is a plethora of examples. –  jbc Oct 28 '12 at 18:44

5 Answers 5

up vote 1 down vote accepted

If for a polynomial $p$ and a matrix $A$ you have $p(A)=0$ then for every invertible matrix $W$ you have $$p(W^{-1}AW)=W^{-1}p(A)W=0 . $$

Here $p=x^2-x$, you can take $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $W$ any invertible matrix to make a lot of examples.

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Certainly, it is true that $A^2=A$ if either $A=0$ or $A=I$. So it should hold if you join the two together: $$\begin{pmatrix}1&0\\0&0\end{pmatrix}.$$ That is, a very small $I$ ($=1$) in the upper left, and an equally small $0$ in the lower right. The off-diagonal zeroes keep them from interfering with each other.

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Very nicely put. –  Brian M. Scott Oct 28 '12 at 17:57

You can take, for example, $\text{diag}(1,0)$.

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I don't like people that rushes to downvote...anyway, your answer is wrong as $$\begin{pmatrix}1&0\\0&-1\end{pmatrix}^2\neq\begin{pmatrix}1&0\\0&-1\end{pmatrix} $$ –  DonAntonio Oct 28 '12 at 17:55

$$A=\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$$ Read about Idempotent Matrices.

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Any projection operator obeys this relation. It should be intuitive that when you apply this operator again, the projected vector should not change.

One can prove this concretely. For a unit vector $u$, let $\underline A(a) = a - (u \cdot a)u$. This projects the vector $a$ onto the subspace orthogonal to $u$. Clearly $\underline A^2(a) = a - (u \cdot a) u - (u \cdot a) u + (u \cdot u)(u \cdot a) u = a - (u \cdot a)u = \underline A(a)$.

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