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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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Determinant of linearly independent vectors

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

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.

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Power of matrix

Let \displaystyle A=\begin{pmatrix} -2 & 4 &3 \\ 0 & 0 & 0\\ -1 &5 &2 \end{pmatrix}. Prove that A^{593}-2A^{15}+A=0.

Solution

The characteristic polynomial of A is p(x)=x-x^3. This in return means A=A^3 and A^3=A^5. Thus,

    \begin{align*} A^{593} -2 A^{15} +A &= A^{591} \cdot A^2 - 2 \left ( A^3 \right )^5 + A\\ &=\left ( A^3 \right )^{197} \cdot A^2 - 2 A^5 + A \\ &= A^{197} \cdot A^2 - 2 A^3 + A \\ &=A^{199} - 2 A +A\\ &=A^{198} \cdot A- A\\ &=\left ( A^3 \right )^{66} \cdot A - A \\ &=A^{66} \cdot A - A\\ &= \left ( A^3 \right )^{22} \cdot A - A\\ &= A^{22} \cdot A -A \\ &= A^{23} - A\\ &= \left ( A^3 \right )^{7} \cdot A^2 -A\\ &= A^7 \cdot A^2 - A\\ &= A^9 - A\\ &=\left ( A^3 \right )^3 - A\\ &=A^3 -A\\ &=A -A\\ &=\mathbb{O} \end{align*}

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