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# Tag Archives: Linear Algebra

## Determinant

Let and  such that . Prove that

Solution

Note 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

as desired.

## Power of matrix

Let . Prove that .

Solution

The characteristic polynomial of is . This in return means and . Thus,

## Linear Projection

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.

Solution

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.

## Convexity of exponential function

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 .

Solution

Since commute it follows from the properties of the exponential function

(1)

Noting that

(2)

Setting we must prove that . Since commute so are . Hence is symmetric. Thus,

(3)

and that’s all!

## Zero matrix

Let with . If for every matrix then prove that .

Solution

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.

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