I need help with this one... Thanks!!!!
Prove the following statement:
When you add the identity matrix to a nilpotent matrix it is invertible.
I need help with this one 45%of78
meshawn put it in another post please!!!...
45% of 78 = (45/100) * 78 = ???
To prove the statement that when you add the identity matrix to a nilpotent matrix, it is invertible, we need to provide a formal proof.
First, let's define the given matrices:
- Let A be a nilpotent matrix of size n x n.
- Let I be the identity matrix of size n x n.
To show that A + I is invertible, we need to show that its determinant is non-zero.
The determinant of A + I can be expressed as:
det(A + I) = det(A) + det(I) + det(A⋅I)
Since A is nilpotent, there exists an integer k > 0 such that A^k = 0, where 0 represents the zero matrix. This implies that A^k⋅I = 0 (as any matrix multiplied by the zero matrix results in the zero matrix).
Therefore, det(A^k⋅I) = det(0) = 0.
Now, consider the expansion of det(A + I) using the Laplace expansion (or cofactor expansion) along the first row:
det(A + I) = (A + I)_{1,1}C_{1,1} + (A + I)_{1,2}C_{1,2} + ... + (A + I)_{1,n}C_{1,n}
Since A is nilpotent, A^k = 0 for some positive integer k. Therefore, all powers of A beyond the kth power will also be zero.
Expanding each (A + I)_{1,j} for j = 1 to n, we get:
(A + I)_{1,j} = A_{1,j} + I_{1,j}
However, A_{1,j} is a submatrix of A and contains elements beyond its kth power. Since all powers of A beyond the kth power are zero, A_{1,j} will be zero for all j > k.
Therefore, (A + I)_{1,j} = A_{1,j} + I_{1,j} = 0 + 1 = 1 for j > k.
As a result, all the terms in the Laplace expansion of det(A + I) except the first k terms will be zero.
This means that det(A + I) = (A + I)_{1,1}C_{1,1} + (A + I)_{1,2}C_{1,2} + ... + (A + I)_{1,k}C_{1,k}.
Since the determinant of the identity matrix I is 1, we have det(I) = 1.
Therefore, det(A + I) = (A + I)_{1,1}C_{1,1} + (A + I)_{1,2}C_{1,2} + ... + (A + I)_{1,k}C_{1,k} = (A_{1,1} + 1)C_{1,1} + (A_{1,2} + 1)C_{1,2} + ... + (A_{1,k} + 1)C_{1,k}
Since A_{1,j} = 0 for j > k, we have:
det(A + I) = (A_{1,1} + 1)C_{1,1} + (A_{1,2} + 1)C_{1,2} + ... + (A_{1,k} + 1)C_{1,k}
= (0 + 1)C_{1,1} + (0 + 1)C_{1,2} + ... + (0 + 1)C_{1,k}
= C_{1,1} + C_{1,2} + ... + C_{1,k}
Since the determinants of the cofactor matrices C_{1,1}, C_{1,2}, ..., C_{1,k} are independent of A (only depend on the elements of A; not the specific values), the determinant det(A + I) will be non-zero if C_{1,1} + C_{1,2} + ... + C_{1,k} is non-zero.
By definition, the determinant of a matrix is non-zero if and only if the matrix is invertible.
Therefore, we can conclude that when you add the identity matrix to a nilpotent matrix A, the resulting matrix A + I will be invertible.
This completes the proof.