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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.
Consider the matrices and . If is invertible prove that is also invertible.
So we have to answer the question if is a zero of the essentially same characteristic polynomials. and have quite similar characteristic polynomials. In fact if denotes the polynomial of , then the polynomial of will be . It is easy to see that cannot be an eigenvalue of the matrix, otherwise it wouldn’t be invertible. Now, let us assume that is not invertible. Then it must have an eigenvalue of and let be the corresponding eigenvector. Hence:
meaning that has an eigenvalue of which is a contradiction. The result follows.