Joy of Mathematics

Zero determinant

April 16, 2020July 19, 2020 Tolaso

Let $A \in \mathcal{M}_n \left( \mathbb{C} \right)$ such that

$A \det A + A^* \det A^*=i \left( A+ A^* \right)$

Prove that $\det A=0$ if $n$ is odd.

Solution

Let $\det A = z$ . Then

$zA + \bar{z}A^{\ast} = i(A+A^{\ast})$

Taking conjugate transpose we also have that

$\bar{z}A^{\ast} + zA = -i(A^{\ast} + A)$

Hence $A + A^{\ast} = 0$ . However it also holds $zA + \bar{z}A^{\ast} = 0$ . Combiming these two we get that

$(z-\bar{z})A = 0$

If $A = 0$ we are done. Otherwise $z$ is real. In that case we have

$z = \det A = \det(-A^{\ast}) = (-1)^n\det A^\ast = -\bar{z} = -z$

since $n$ is odd. Hence $\det{A} = z = 0$ as wanted.

Determinant of linearly independent vectors

April 14, 2020 Tolaso

Let $w_1, w_2, \dots, w_n$ be unitary linearly independent vectors. Evaluate the determinant

$\mathcal{D} = \det \left(n \mathbb{I}-\sum_{i=1}^n w_i \otimes w_i \right)$

where $w_i \otimes w_i$ denotes the outer product.

Determinant

February 22, 2020 Tolaso

Let $x>0$ and $A \in \mathbb{R}^{2 \times 2}$ such that $\det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = 0$ . Prove that

$\det \left( A^2 + A + x \mathbb{I}_{2 \times 2} \right) = x$

Solution

Note that

$\det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = \det \left(A + i \sqrt {x} \mathbb{I}_{2 \times 2} \right) \det \left(A - i\sqrt{x} \mathbb{I}_{2 \times 2} \right)$

Since $A$ is real, its complex eigenvalues come in conjugate pairs. Thus, in this case we conclude that $A$ has eigenvalues $\pm i \sqrt x$ .

Now, if $\lambda$ is an eigenvalue of $A$ , then $\lambda^2 + \lambda + x$ is an eigenvalue of $A^2 + A + x \mathbb{I}_{2 \times 2}$ . Thus, the matrix $A^2 + A + x \mathbb{I}_{2 \times 2}$ has eigenvalues $(i\sqrt {x} )^2 + i\sqrt {x} + x = i\sqrt {x}$ and $(-i\sqrt {x})^2 - i\sqrt {x} + x = -i\sqrt {x}$ .

Now, $\det \left(A^2 + A + x \mathbb{I}_{2 \times 2} \right)$ is the product of these eigenvalues, which is to say

$\det \left(A^2 + A + x\mathbb{I}_{2 \times 2} \right) = (i\sqrt{x})(-i\sqrt{x}) = x$

as desired.