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Similarity implies equality?

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.

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On tensors

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).

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No solution

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

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

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Zero determinant

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:

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

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Zero determinant

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.

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