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# Tag Archives: Linear Algebra

## Zero determinant

Consider the real numbers for . Prove that Solution

Using the identity in combination with we have: ## Zero determinant

Let such that Prove that if is odd.

Solution

Let . Then Taking conjugate transpose we also have that Hence . However it also holds . Combiming these two we get that If we are done. Otherwise is real. In that case we have since is odd. Hence as wanted.

## Determinant of linearly independent vectors

Let be unitary linearly independent vectors. Evaluate the determinant where denotes the outer product.

## 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, ### Donate to Tolaso Network 