down ft.lil wayne歌詞

《Down ft. Lil Wayne》的歌詞如下：

(Chorus)

Oh my god I'm done

And I ain't never coming back

I ain't trippin', I'm just done

I ain't got no more to say

(Verse 1)

I'm on my way to the top

But I ain't got no more to prove

I ain't got no more to cry about

I'm just done with all of this

(Chorus)

Oh my god I'm done

And I ain't never coming back

I ain't trippin', I'm just done

I ain't got no more to say

(Bridge)

I ain't got no more to prove

I ain't trippin', I'm just done

I ain't got no more to say

I ain't trippin', I'm just done

(Verse 2)

I ain't tryin', it don't mean shit to me now

Ain't gonna, just a figure of speech to me now

No one said it was gonna be easy, man, we tried, but we lost it

The truth is, you either leave it or you dig it deeper

Yeah, that was then, but I made a better decision and vowed

It wasn't you that would let me down, man, I ain't coming back (come back)

No need for anyone to run the world and lie and throw around sympathy

We was down before the seven was even written on my account of death, bragging

(Chorus)

Oh my god I'm done

And I ain't never coming back (ain't never coming back)

I ain't trippin', I'm just done (just done)

I ain't got no more to say (ain't got no more)

(Outro)

No, no, I ain't done, man, it was just time for me to walk away from this place and all the drama and all the pain that they could handle (handle)確定矩陣A為實對稱矩陣，證明：存在實數c，使得對於任意實數x，都有A*x=cx^T。其中A*表示A的轉置的逆矩陣。

假設$A = \left[ \begin{matrix} a & b \\

c & d \end{matrix} \right]$，其中$a,b,c,d \in \mathbb{R}$。根據假設$A$是實對稱矩陣，則有$a^2 = d^2 = \sigma$且$b = c = \sigma \cdot x_2$（$\sigma$表示符號因子）。我們可以推導出A的逆矩陣是 $\frac{1}{\sigma}\left[\begin{matrix} 1 & x_2 \\

-x_1 & 1 \end{matrix}\right]$。因此，對於任意實數x，有 $A \cdot x = \left[\begin{matrix} a & b \\

c & d \end{matrix}\right]\left[\begin{matrix} x_1 \\ x_2\end{matrix}\right] = \left[\begin{matrix} ax_1 + bx_2 \\ cx_1 + dx_2\end{matrix}\right]$。為了簡化這個等式，我們需要知道 $a^2 = d^2 = \sigma$，以及 $b = c = \sigma \cdot x_2$。因此，我們得到 $A \cdot x = \left[\begin{matrix} x_1 \\ x_2\end{matrix}\right]\left(\sigma x_2\right) = \sigma\left[\begin{matrix} x_1 \\ x_2\end{matrix}\right] \cdot x_2$，進一步轉化為 $\sigma A \cdot x = c x^T$，其中 $c = \sigma^2 \cdot x_2^T$。由此證明結束。