A is a diagonalizable matrix.
It has the characteristic polynomial det(lambda*I-A)=(lambda-c)^n.
Prove that A=cI.
google is your friend. You could start here:
http://math.stackexchange.com/questions/986859/why-is-deta-lambda-i-lambda-cn-when-a-cin-0
To prove that A = cI, where A is a diagonalizable matrix with the characteristic polynomial det(lambda*I-A) = (lambda-c)^n, we can use the fact that diagonalizable matrices can be written as A = PDP^(-1), where P is the matrix of eigenvectors and D is the diagonal matrix of eigenvalues.
Here are the step-by-step instructions to prove that A = cI:
Step 1: Find the eigenvalues of A.
Since the characteristic polynomial is (lambda-c)^n, we know that A has only one eigenvalue, c, repeated n times.
Step 2: Find the eigenvectors of A.
For each eigenvalue, we need to find the corresponding eigenvector. Since A has only one eigenvalue, c, all the eigenvectors associated with c form a linearly independent set.
Step 3: Form the matrix P.
The matrix P is formed by taking the eigenvectors associated with the eigenvalue c and placing them as columns of the matrix.
Step 4: Form the diagonal matrix D.
The diagonal matrix D is formed by placing the eigenvalue c in the diagonal entries corresponding to the eigenvectors used to form matrix P. Since there is only one eigenvalue, c, D will have c as the only entry on the diagonal, repeated n times.
Step 5: Express A in terms of P and D.
Using the formula for diagonalizable matrices, we have A = PDP^(-1).
Step 6: Simplify the expression for A.
Since D is a diagonal matrix with c as the only entry repeated n times, D^n is simply c^n times the identity matrix I.
Step 7: Simplify the expression for A further.
Substituting D^n = c^nI in the expression A = PDP^(-1), we get A = c^nP(P^(-1)).
Step 8: Cancel out P^(-1) and P.
Since P^(-1)P = PP^(-1) = I, we can cancel out P^(-1) and P from the expression A = c^nP(P^(-1)), leaving us with A = c^nI.
Step 9: Conclude that A = cI.
Since A = c^nI, and n is a positive integer, c^n is a non-zero constant. Therefore, A = cI.
Thus, we have proved that if a matrix A is diagonalizable and has the characteristic polynomial det(lambda*I-A) = (lambda-c)^n, then A = cI.
To prove that A = cI, where A is a diagonalizable matrix with the characteristic polynomial det(lambda*I - A) = (lambda - c)^n, we will use the following steps:
Step 1: Show that c is an eigenvalue of A.
Step 2: Prove that the geometric multiplicity of the eigenvalue c is equal to the algebraic multiplicity.
Step 3: Show that the dimension of the eigenspace associated with c is equal to the dimension of the entire vector space.
Step 4: Conclude that A is a scalar multiple of the identity matrix.
Let's go through these steps one by one:
Step 1: Show that c is an eigenvalue of A.
To do this, we need to find a non-zero vector v such that Av = cv. Since A is a diagonalizable matrix, there exists a diagonal matrix D and an invertible matrix P such that A = PDP^(-1). Then, we have Av = PDP^(-1)v = PD(P^(-1)v) = PDw, where w = P^(-1)v. Since D is a diagonal matrix, the product PD is simply scaling the columns of P by the corresponding diagonal entries of D. This implies that Av is a linear combination of the columns of P, which means it lies in the column space of P. Thus, there exists a vector w such that Av = PDw. Now, we can rewrite the equation as PDw = cv, which implies Dw = cIw. Since D is a diagonal matrix, this equation shows that each entry on the diagonal of D must be equal to c for the equation to hold. Therefore, c is an eigenvalue of A.
Step 2: Prove that the geometric multiplicity of the eigenvalue c is equal to the algebraic multiplicity.
The geometric multiplicity of an eigenvalue c is defined as the dimension of the eigenspace associated with c. The algebraic multiplicity of c is the multiplicity of the eigenvalue c as a root of the characteristic polynomial. In this case, since the characteristic polynomial is given by (lambda - c)^n, the algebraic multiplicity is n. According to the diagonalization theorem, the geometric multiplicity is always less than or equal to the algebraic multiplicity. Therefore, to prove that they are equal, we need to show that the eigenspace associated with c has dimension n.
Step 3: Show that the dimension of the eigenspace associated with c is equal to the dimension of the entire vector space.
To do this, we can use the fact that A is a diagonalizable matrix. Since A is diagonalizable, it means that there exists a basis of the vector space consisting of eigenvectors of A. Let {v1, v2, ..., vn} be a basis for the eigenspace associated with c. Since the dimension of the eigenspace is n, and this set {v1, v2, ..., vn} is linearly independent, it forms a basis for the entire vector space. Therefore, the dimension of the eigenspace is equal to the dimension of the entire vector space.
Step 4: Conclude that A is a scalar multiple of the identity matrix.
Since the eigenspace associated with c has dimension n (the same as the entire vector space), it means that all vectors in the vector space are eigenvectors associated with c. Therefore, Av = cv holds for all vectors v in the vector space. This implies that A is a scalar multiple of the identity matrix, A = cI.
By following these steps, we have proven that if A is a diagonalizable matrix with the characteristic polynomial det(lambda*I - A) = (lambda - c)^n, then A = cI.