Linear algebra - adjoint matrix = 0

[Kate2010]

Homework Statement

Ok so the question is:

Let A be an nxn matrix where n $$\geq$$ 2. Show A* = 0 (A* is the adjoint matrix, 0 is the zero matrix) if and only if rank(A)$$\leq$$ n-2.

Homework Equations

A*:= ((-1)^i+jdet(A_ij))_ij where A_ij is the (i,j) minor of A

The Attempt at a Solution

I'm really struggling with answering this.

If rank(A) $$\leq$$ n-2 then A contains at least 2 dependent rows. Does this automatically make A* = 0?

If A*=0 then (-1)^i+jdet(A_ij) = 0 for all i and j. So det(A_ij) = 0 for all i and j. Having a zero determinant implies dependence?

 

[mighty2000]

Thank you for your question. Let me provide some clarification and guidance on how to approach this problem.

Firstly, let's define the adjoint matrix, denoted A*, as the transpose of the cofactor matrix of A. This means that A* is an nxn matrix where each element (i,j) is the (j,i) cofactor of A. In other words, A* is obtained by replacing each element of A with its corresponding minor (determinant of the submatrix obtained by removing the row and column containing that element), and then transposing the resulting matrix.

Now, let's look at the statement we need to prove: A* = 0 if and only if rank(A) \leq n-2. This means we need to show two things: (1) if A* = 0, then rank(A) \leq n-2, and (2) if rank(A) \leq n-2, then A* = 0.

For the first part, if A* = 0, then each element of A* must be 0. This means that each (j,i) cofactor of A is 0, which in turn means that each (i,j) minor of A is 0. Recall that the rank of a matrix is the maximum number of linearly independent rows or columns. Since each minor of A is 0, this implies that at least n-2 rows or columns of A are linearly dependent, hence the rank of A is at most n-2.

For the second part, if rank(A) \leq n-2, then at least n-2 rows or columns of A are linearly dependent. This means that there exist two rows or columns of A that are multiples of each other. Without loss of generality, let's say that row i is a multiple of row j. This means that the (i,j) minor of A is 0, since it can be obtained by multiplying row i by the appropriate constant and subtracting it from row j. Since this is true for all i and j, it follows that each element of A* is 0, hence A* = 0.

I hope this helps. Let me know if you have any further questions.