矩阵转置–矩阵| 12年级数学

矩阵知识对于数学的各个分支都是必不可少的。矩阵是数学中最强大的工具之一。现在，请参阅本文中矩阵的功能之一。

转置矩阵

这是矩阵的主要属性之一。转置的意思是交换两个或多个事物的位置。在矩阵的情况下，转置含义会更改元素的索引。在这种情况下，我们将行元素与列元素交换，反之亦然。

设A是大小为m×n的矩阵，A ^t 是矩阵A的转置
其中A的[a(ij)] = A ^t的[a(ji)]，这里1≤i≤m和1≤j≤n

例子：

设矩阵A的大小为2×3 ，

$A = \begin{bmatrix} 2 & 5 & 3\\ 4 & 7 & 0 \end{bmatrix}$

所以，

转置A或

$A ^{t} = \begin{bmatrix} 2 & 4 \\ 5 & 7 \\ 3 & 0 \end{bmatrix}$

转置后变为3×2

转置属性

矩阵乘积的转置

该属性表示， (AB) ^t = B ^t A ^t

证明

Here A and B are two matrices of size m × n and n × p respectively

and At and Bt are their transpose form of size n × m and p × n respectively (from the product rule of matrices).

It implies, if A = [a(ij)], and At = [c(ji)]

Then, [c(ji)] = [a(ij)]

and,

If B = [b(jk)], and Bt = [d(kj)]

Then, [d(kj)] = [b(jk)]

Now, from the product rule of matrices we can write,

AB is m × p matrix and (AB)t is p × m matrix.

Also, Bt is a p × n matrix and At is a n × m matrix.

This implies that,

(Bt)(At) is a p × m matrix.

Therefore,

(AB)t and (Bt)(At) are both p × m matrices.

Now we can write,

(k, i)th element of (AB)t = (i, k)th element of AB 

$\sum_{j=1}^{n} a_{ij} b_{jk}$
$\sum_{j=1}^{n} c_{ji} d_{kj}$

$\sum_{j=1}^{n} d_{kj} c_{ji}$

(k, i)th element of (B^t)(A^t)

Therefore, 

the elements of (AB)^t and (B^t)(A^t) are equal.

Therefore,

(AB)^t = (B^t)(A^t)

例子：

让，

$A = \begin{bmatrix} -2 & 1 & 3\\ 0 & 4 & -1 \end{bmatrix}$

和

$B = \begin{bmatrix} 2 & 1 \\-3 & 0 \\ 4 & -5 \end{bmatrix}$

因此证明对于这些矩阵，(AB) ^t =(B ^t )(A ^t )

解决方案：

Here A and B are 2 × 3 and 3 × 2 matrices respectively. So, by product rule of a matrix, we can find their product
and the final matrices would be of 2 × 2 matrix.

Find L.H.S –

Now, 

$AB= \begin{bmatrix} -2 & 1 & 3\\ 0 & 4 & -1 \end{bmatrix} \times \begin{bmatrix} 2 & 1 \\-3 & 0 \\ 4 & -5 \end{bmatrix}$
$=\begin{bmatrix} (-2)*2+1*(-3)+3*4 & (-2)*1+1*0+3*(-5) \\ 0*2+4*(-3)+(-1)*4 & 0*1+4*0+(-1)*(-5) \end{bmatrix}$

$= \begin{bmatrix} 5 & -17 \\ -16 & 5 \end{bmatrix}$

So, Transpose of AB or 

$AB^{t} = \begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}$

$\begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}$

Find R.H.S –

$A^{t} = \begin{bmatrix} -2 & 0 \\ 1 & 4 \\ 3 & -1 \end{bmatrix}$

and

$B^{t} = \begin{bmatrix} 2 & -3 & 4 \\ 1 & 0 & -5 \end{bmatrix}$

So,

$B^{t}.A^{t} = \begin{bmatrix} 2 & -3 & 4\\ 1 & 0 & -5 \end{bmatrix} \times \begin{bmatrix} -2 & 0 \\ 1 & 4 \\ 3 & -1 \end{bmatrix}$
$= B^{t}.A^{t} = \begin{bmatrix} 2*(-2)+(-3)*1+4*3 & 2*0+(-3)*4+4*(-1) \\ 1*(-2)+0*1+(-5)*3 & 1*0+0*4+(-5)*(-1) \end{bmatrix}$
$= \begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}$

Therefore,

(AB)^t = B^t.A^t

矩阵加法的转置

该属性表示(A + B) ^t = A ^t + B ^t

证明：

Here A and B are two matrices of size m × n

Let, A = [a(ij)] and B = [b(ij)] of size m × n.

So, (A + B) is also an m × n matrix

Also, A^t and B^t are n × m matrices.

So that, the Transpose of (A + B) or (A + B)^t is an n × m matrix.

Now we can say, A^t + B^t is also an n × m matrix.

Now, from the transpose rule,(j, i)th element of (A + B)^t = (i, j)th element of (A + B)

= (i, j)th element of A + (i, j)th element of B= (j, i)th element of A^t + (j, i)th element of B^t= (j, i)th element of(A^t + B^t)

Therefore,

(A + B)^t = A^t + B^t

例子：

让，

$A= \begin{bmatrix} -1 & 5 \\ 3 & 2 \end{bmatrix}$

和

$B= \begin{bmatrix} 3 & -2 \\5 & 4 \end{bmatrix}$

证明对于这些矩阵，(A + B) ^t = A ^t + B ^t

解决方案：

Find L.H.S,

$(A+B)= \begin{bmatrix} -1 & 5 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\5 & 4 \end{bmatrix}$
$= \begin{bmatrix} (-1)+3 & 5+(-2) \\ 3+5 & 2+4 \end{bmatrix}$
$= \begin{bmatrix} 2 & 3 \\ 8 & 6 \end{bmatrix}$

So, 

$(A+B)^{t} = \begin{bmatrix} 2 & 8 \\ 3 & 6 \end{bmatrix}$

Find R.H.S

$A^{t} = \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix}$
$\begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix}$

 and, 

$B^{t} = \begin{bmatrix} 3 & 5 \\ -2 & 4 \end{bmatrix}$

Now, 

$A^{t} + B^{t} = \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 5 \\ -2 & 4 \end{bmatrix}$
$= \begin{bmatrix} (-1)+3 & 3+5 \\ 5+(-2) & 2+4 \end{bmatrix}$
$A^{t} + B^{t} = \begin{bmatrix} 2 & 8 \\ 3 & 6 \end{bmatrix}$

Therefore, 

(A + B)^t = A^t + B^t