Thread: Square of a matrix...

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!

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.

Hello, Richard!

If A is an n-by-n matrix, and A² = 0, does this force A = 0? . no

[size=3]

Code:

We have:   | a b | | a b |     | 0 0 |
           |     | |     |  =  |     |
           | c d | | c d |     | 0 0 |


Then:  [1] a² + bc = 0    [2] ab + bd = 0
       [3] ac + cd = 0    [4] bc + d² = 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: .a² .= .-bc

Since a² 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 |

Thanks a million! This has been bugging me all day. Guess I'm back to square one with my original problem though ....Though I did notice that the rows and columns of those two matrices are linearly dependant....hmm, that might be useful....

Originally Posted by Richard Rahl

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.

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???)

Originally Posted by Richard Rahl

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

A bit long winded but OK

RonL