Joy of Mathematics

Similarity implies equality?

February 7, 2021 Tolaso

Let $A\in \mathbb{C}^{n\times n}$ be similar to $A^2$ . Does $A=A^2$ hold?

Solution

No! Take $A=\begin{pmatrix}1 &0 \\1&1 \end{pmatrix}$ then $A^2=\begin{pmatrix}1 &0\\ 2&1 \end{pmatrix}$ . The matrices are similar but not equal.

Let $V_1, V_2, W_1, W_2, U_1, U_2 \in \; \mathbb{K}$ -Vect, $V_1 \xrightarrow{\;\; \alpha_1 \;\; }W_1 \xrightarrow{\;\; \beta_1 \;\; }U_1$ , $V_2 \xrightarrow{\alpha_2}W_2 \xrightarrow{\beta_2}U_2 \;\; \mathbb{K}$ -linear. Prove that

$(\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2)=(\beta_1 \alpha_1)\otimes(\beta_2 \alpha_2)$

Solution

Recall the general definition of the tensor product of linear maps, we have successively:

$\begin{align*} ((\beta_1 \alpha_1) \otimes (\beta_2 \alpha_2))(v_1 \otimes v_2) &= (\beta_1 \alpha_1)(v_1) \otimes (\beta_2 \alpha_2)(v_2) \\ &=\beta_1(\alpha_1(v_1)) \otimes \beta_2(\alpha_2(v_2)) \\ &= (\beta_1 \otimes \beta_2)(\alpha_1(v_1) \otimes \alpha_2(v_2)) \\ &=((\beta_1 \otimes \beta_2)(\alpha_1 \otimes \alpha_2))(v_1 \otimes v_2) \end{align*}$

Thus, the two linear maps $V_1 \otimes V_2 \rightarrow U_1 \otimes U_2$ are equal when composed with the canonical bilinear map $V_1 \times V_2 \to V_1 \otimes V_2$ , hence equal (by the universal property).

No solution

November 6, 2020 Tolaso

Let $A, B \in \mathcal{M}_{n} \left ( \mathbb{C} \right )$ . Show that

$AB - BA =\mathbb{I}_{n \times n}$

has no solutions.

Solution

Since $\mathrm{tr} (AB) = \mathrm{tr} (BA)$ taking traces on both sides, we have

$\mathrm{tr}(AB - BA) = \mathrm{tr} (\mathbb{I}_{n \times n}) \Rightarrow \mathrm{tr}(AB) - \mathrm{tr}(BA) = n \Rightarrow 0 = n$

Zero determinant

May 24, 2020 Tolaso

Consider the real numbers $x_i, x_j$ for $1 \leq i , j \leq 3$ . Prove that

$\mathcal{D} = \begin{vmatrix} \sin (x_1+y_1) & \sin (x_1+y_2) & \sin (x_1+y_3)\\ \sin (x_2+y_1) & \sin (x_2+y_2) & \sin (x_2+y_3)\\ \sin (x_3+y_1) & \sin (x_3+y_2) & \sin (x_3+y_3) \end{vmatrix}=0$

Solution

Using the identity $\sin (a+b)=\sin a\cos b+\sin b\cos a$ in combination with $\det AB = \det A \det B$ we have:

$\mathcal{D}=\begin{vmatrix} \sin x_1 & \cos x_1 & 0\\ \sin x_2 & \cos x_2 & 0\\ \sin x_3 & \cos x_3 & 0 \end{vmatrix} \cdot \begin{vmatrix} \cos y_1 & \cos y_2 & \cos y_3 \\ \sin y_1 & \sin y_2 & \sin y_3 \\ 0 & 0 & 0 \end{vmatrix} =0$