nonsingular, idempotent matrices

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

I am a bit stuck on this question. Any help would be appreciated!

Calculate all non-singular, idempotent matrices of size 3 x 3.

Thanks.

aswoods · **Posted:** Fri, 4 Mar 2011 11:37:39 UTC

Idempotent: A² = A

Because A is non-singular, its inverse exists. Therefore...

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

aswoods wrote:

Idempotent: A² = A

Because A is non-singular, its inverse exists. Therefore...

it is invertible? i'm not quite sure...

also, because its non-singular, its determinant does not equal to zero. but how do we calculate that when the matrix has no specific entries?

Shadow · **Posted:** Sat, 5 Mar 2011 03:04:10 UTC

asa.hoshi wrote:

aswoods wrote:

Idempotent: A² = A

Because A is non-singular, its inverse exists. Therefore...

it is invertible? i'm not quite sure...

It means you should apply $A^{-1}$ to both sides.

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

Shadow wrote:

asa.hoshi wrote:

aswoods wrote:

Idempotent: A² = A

Because A is non-singular, its inverse exists. Therefore...

it is invertible? i'm not quite sure...

It means you should apply $A^{-1}$ to both sides.

how so?

Shadow · **Posted:** Sat, 5 Mar 2011 03:22:14 UTC

asa.hoshi wrote:

Shadow wrote:

asa.hoshi wrote:

aswoods wrote:

Idempotent: A² = A

Because A is non-singular, its inverse exists. Therefore...

it is invertible? i'm not quite sure...

It means you should apply $A^{-1}$ to both sides.

how so?

Exactly what I say, multiply both sides by the inverse of A, it exists because it's nonsingular.

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

but it says matrices of size 3x3. how would i go about that though?

so i know its a square matrix.

$AA^{-1}=I$

Shadow · **Posted:** Sat, 5 Mar 2011 03:31:32 UTC

asa.hoshi wrote:

but it says matrices of size 3x3. how would i go about that though?

What do you mean? What does the size have to do with it? You can multiply matrices, can you not? So if A^2=A

then you can multiply both sides of this by the same thing to get the same result, right?

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

Shadow · **Posted:** Sat, 5 Mar 2011 04:03:49 UTC

asa.hoshi wrote:

$A^{-1} A^2=I A^{-1} A = A^{-1} A$

Right, and there you go.

asa.hoshi · **Joined:** Thu, 11 May 2006 11:24:10 UTC **Posts:** 332

I'm still confused by the question...
"Calculate all non-singular, idempotent matrices..."

what does it mean by 'calculate all'

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

If A is idempotent, then A^2 = A.
Because A is nonsingular, its inverse exists. Thus:

$\rule{0.1pt}{0.1pt}\quad\begin{aligned} A^2&=A\\ A^{-1}A^2&=A^{-1}A\\ A&=I \end{aligned}$

In other words, the identity matrix is the only nonsingular idempotent matrix.

Shadow · **Posted:** Sat, 5 Mar 2011 21:00:29 UTC

asa.hoshi wrote:

I'm still confused by the question...
"Calculate all non-singular, idempotent matrices..."

what does it mean by 'calculate all'

Exactly what it says, calculate, i.e. compute what the matrix must look like, i.e. what its entries must be, and the computation you did, and which Matt has duplicated is exactly that, you computed that it must be the identity.

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nonsingular, idempotent matrices

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