Matrices question (1 Viewer)

[mwseaeagles]

Find all 2 x 2 matrices A for which A^2 = A.

 

[me121]

hmm.. i'm not exactly sure, but if you assume that a matrix A squared, A² is equal to AA, then
AA=A

just guessing, the following work,
0 0
0 0

1 0
0 1

i think that's all.

 

[Forbidden.]

There could be a general form that may help ...

Let's say a matrix is represented by

a (1st row 1st column)
b (1st row 2nd column)
c (2nd row 1st column)
d (2nd row 2nd column)

The matrix multiplication is as follows:

(1st row 1st column) = aa + bc
(1st row 2nd column) = ab + db
(2nd row 1st column) = ca + dc
(2nd row 2nd column) = cb + dd

For AA = A to hold true,

aa + bc = a
ab + db = b
ca + dc = c
cb + dd = d

 

[Affinity]

1 0
0 0

also works.. The methodical way to do this is to consider the jordan decompostion of A.

First, the eigenvalue(s) of A can be nothing else but 1 and 0.

so the jordan canoncial form J of A must be one of

1 0
0 1

0 0
0 0

1 0
0 0

1 1
0 1

0 1
0 0

but the last 2 don't work. (since A^2 = A, J^2 must also equal to J)
so therefore all matrices that satisfy A^2 = A are:

the zero and identity matrices and matrices in the form

S 1 0 S^-1
" 0 0

or written in explicit form

            [ad  -ab]  /
            [cd  -cb] / (ad - bc)

where S ranges over all invertible matrices. (ad-bc =/= 0)
so for example

49 -42
56 -48

works

for general n, the jordan blocks should all be of size 1 and the eigenvalues all 1 or 0

so matrices in the form SDS^-1

where S is invertible and D is diagonal with 0 or 1 down the diagonal

 

[mwseaeagles]

There could be a general form that may help ...

Let's say a matrix is represented by

a (1st row 1st column)
b (1st row 2nd column)
c (2nd row 1st column)
d (2nd row 2nd column)

The matrix multiplication is as follows:

(1st row 1st column) = aa + bc
(1st row 2nd column) = ab + db
(2nd row 1st column) = ca + dc
(2nd row 2nd column) = cb + dd

For AA = A to hold true,

aa + bc = a
ab + db = b
ca + dc = c
cb + dd = d

i got where you did and also have the null and identity matrix.
the answers however go


[a b]
[(a-a^2)/b 1-a]

how did they get that

 

[Affinity]

i got where you did and also have the null and identity matrix.
the answers however go


[a b]
[(a-a^2)/b 1-a]

how did they get that

Well if you insist that your S matrices are normalize so that ad - bc = 1, set ad=x and -ab = y will reduce things to that form

            [ad  -ab]  /
            [cd  -cb] / (ad - bc)

=           [x          y] 
            [x(1-x)/y  1-x]

buy you can't get things like
1 0
0 0
from the form they gave you..
so the answer I gave would be more correct