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Tag Archives: Linear Algebra
Let and such that . Prove that
Since is real, its complex eigenvalues come in conjugate pairs. Thus, in this case we conclude that has eigenvalues .
Now, if is an eigenvalue of , then is an eigenvalue of . Thus, the matrix has eigenvalues and .
Now, is the product of these eigenvalues, which is to say
Let . Prove that .
The characteristic polynomial of is . This in return means and . Thus,
Let be a linear space over such that and be a linear projection such that any non zero vector of is an eigenvector of . Prove that there exists such that where is the identity endomorphism.
Let . Since any non zero vector is an eigenvector it follows that every basis of is also an eigenbasis. Let be such a basis and be the respective , not necessarily distinct , eigenvalues of the eigenvectors of . For the vector which also happens to be eigenvector with eigenvalue , it holds that:
But then for each it holds that
The result follows.
Remark: This proof also works in the case is infinite.
If are symmetric real matrices we write if-f the matrix is non negative definite. Examine if
for each pair real symmetric matrices such that .
Since commute it follows from the properties of the exponential function
Setting we must prove that . Since commute so are . Hence is symmetric. Thus,
and that’s all!
Let with . If for every matrix then prove that .
Suppose that , say for some . Let be any permutation matrix with and let be the matrix obtained from by changing its -entry to . Finally let where .
We have that and that is a polynomial in . Furthermore, the coefficient of of this polynomial is depending on the sign of the corresponding permutation. So the polynomial is not identically zero, a contradiction.