I worked it out but not in full detail and I think the answer is that A=0.
For example say,
[a b]
[c d] =A
Then solve,
[0 0]
[0 0] = A^2
Meaning,
a^2+bc=0
ad+bd=0
ac+cd=0
bc+d^2=0
I think the only solution is the trivial one.
Hey, I did this last semister in Matrix Algebra, and now I need it for a proof in Lineary, but can't remember how it goes (and annoyingly can't find my notes).
If A is an n by n matrix, and A^2 = 0, does this force A = 0? My gut tells me no...but I can't come up with a counter example.
This isn't the actual question I'm facing, but I think it might help in my serach for the answer. The question I'm doing says "Show A^2 = 0 iff col A is contained within null A.
I'm not sure if I'm on the right track with the whole "Is A forced to be 0 thing", but it was the best I could think of to start it off.
Thanks!
Hello, Richard!
[size=3]If A is an n-by-n matrix, and AČ = 0, does this force A = 0? . noCode:We have: | a b | | a b | | 0 0 | | | | | = | | | c d | | c d | | 0 0 | Then: [1] aČ + bc = 0 [2] ab + bd = 0 [3] ac + cd = 0 [4] bc + dČ = 0
From [2], we have: .b(a + d) .= .0
From [3], we have: .c(a + d) .= .0
If b,c ≠ 0, then: .d = -a
From [1], we have: .aČ .= .-bc
Since aČ is positive, b and c must have opposite signs.
And a is the mean proportional of |b| and |c|.
Some examples:Code:| 4 -2 | | 3 0 -1 | | 8 -4 | | 0 0 0 | | 9 0 -3 |
Let A be a square nxn matrixm the null(A) is the space of col.
vectors x, such that Ax=0.
col(A) is the space spanned by the cols of A.
Let A_i be the i'th col of A, then:
A^2=A[A_1,A_2, .., A_n]=0
so for each i, A A_i =0, so A_i is in the null space of A. Hence col(A) is
a subspace of null(A)
Similarly if col(A) is a subspece of null(A), then A A_i=0 for each col A_i of A,
but:
A^2=AA=A[A_1,A_2... A_n]=0.
Which completes the proof.
RonL
Okay, just to recap and make sure I understand:
A^2 = A[C1....Cn] = 0
In order for the two matricies to equate (A^2 = the zero matrix), when we multiply each column of A by A, we get:
[AC1 AC2 .... ACn] (distributive)
since this equals the zero matrix, each ACi = 0
So, since null A is the set of solution vectors to AX=0, which by default always has the trivial solutionh, given that every element of the resulting matrix is 0, it's colums must be contained within null A (its always a subset, since they are all zeros, which is always in null A).
Given that it's an IFF statement, just turn this around...
col A c null A -> A^2 = 0
Given that col A is a subset of null A:
each Ci is an element of null A
if so, each column of ACi = 0
which is the same as
0 = A[C1....Cn]
which is
0 = AA
0 = A^2
QED
(PS...can a non trivial solution exist such that such that col B c null A?, I don't think so???)