Question

In general, the eigenvalues of A + B and AB are not known from the eigenvalues of A and B. 1. Use A = [0 1; 0 0] B = [0 0; 1 0] and show that these two matrices verify the statement above. 2. Assume AB = BA, that is, A and B are commuting matrices. It can be proven that A and B share at least one eigenvector, that is, there exists at least one x such that Ax = ?x Bx = ?*x. Show that in this case x is also an eigenvector of A + B and AB and find the corresponding eigenvalues. What if B = A? 3. Assume AB = BA. Suppose all the eigenvalues of A are distinct, that is, the characteristic polynomial of A factors into the product of degree 1 polynomials, which is equivalent to saying that the algebraic multiplicity of each eigenvalue is 1. Therefore, the geometric multiplicity of the eigenvalues is also 1 since it is always less than or equal to the algebraic multiplicity and greater than 0. The geometric multiplicity is the dimension of the space - eigenspace - spanned by all the eigenvectors corresponding to the same eigenvalue. Show that each eigenvector of A is an eigenvector of B.

          In general, the eigenvalues of A + B and AB are not known from the eigenvalues of A and B.

1. Use
A = [0 1; 0 0]
B = [0 0; 1 0]
and show that these two matrices verify the statement above.

2. Assume AB = BA, that is, A and B are commuting matrices. It can be proven that A and B share at least one eigenvector, that is, there exists at least one x such that
Ax = ?x  Bx = ?*x.
Show that in this case x is also an eigenvector of A + B and AB and find the corresponding eigenvalues. What if B = A?

3. Assume AB = BA. Suppose all the eigenvalues of A are distinct, that is, the characteristic polynomial of A factors into the product of degree 1 polynomials, which is equivalent to saying that the algebraic multiplicity of each eigenvalue is 1. Therefore, the geometric multiplicity of the eigenvalues is also 1 since it is always less than or equal to the algebraic multiplicity and greater than 0. The geometric multiplicity is the dimension of the space - eigenspace - spanned by all the eigenvectors corresponding to the same eigenvalue.
Show that each eigenvector of A is an eigenvector of B.

In general, the eigenvalues of A + B and AB are not known from the eigenvalues of A and B.

1. Use
A = [0 1; 0 0]
B = [0 0; 1 0]
and show that these two matrices verify the statement above.

2. Assume AB = BA, that is, A and B are commuting matrices. It can be proven that A and B share at least one eigenvector, that is, there exists at least one x such that
Ax = ?x Bx = ?*x.
Show that in this case x is also an eigenvector of A + B and AB and find the corresponding eigenvalues. What if B = A?

3. Assume AB = BA. Suppose all the eigenvalues of A are distinct, that is, the characteristic polynomial of A factors into the product of degree 1 polynomials, which is equivalent to saying that the algebraic multiplicity of each eigenvalue is 1. Therefore, the geometric multiplicity of the eigenvalues is also 1 since it is always less than or equal to the algebraic multiplicity and greater than 0. The geometric multiplicity is the dimension of the space - eigenspace - spanned by all the eigenvectors corresponding to the same eigenvalue.
Show that each eigenvector of A is an eigenvector of B.

Added by Samuel L.

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