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# Tag Archives: Linear Algebra

## 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.

## Invertible matrix

Consider the matrices and . If is invertible prove that is also invertible.

Solution

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.

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