Matrix of rank 1, vector, combined use of merit

A matrix of rank 1 has been specifically discussed, with n-1 linear independent vectors from ax=0 Ax = 0, Lenovo to: Aα=0⋅αa\alpha = 0\cdot \alpha, knowing that 0 must be a eigenvalues of a, and n-1-valued eigenvalues.

Such a nature, if examined alone, is too simple. In another article, we summarize the idea that the matrix of rank 1 is exponentiation.

http://blog.csdn.net/u011240016/article/details/52805663

This practice is also common, that is, the matrix of rank 1 can extract the product of two vectors. This product is: column vector x row vector.

You might want to set Α,β\alpha,\beta as a column vector. Then Αβt \alpha\beta^t is a matrix with a rank of 1.

And if a matrix of rank 1 is given, how to extract two vectors.
The row vector is any row of the matrix, and the column vectors are three multiples of this line.

Like what:
A=⎡⎣⎢⎢26−413−2−1−32⎤⎦⎥⎥=⎡⎣⎢⎢13−2⎤⎦⎥⎥⋅[21−1] A = {\LEFT[\BEGIN{ARRAY}{CCC} 2 & 1 &-1 \ 6 & 3 &-3 \-4 & Amp -2 & 2 \end{array}\right]} = {\left[\begin{array}{c} 1 \ 3 \-2 \end{array}\right]} \cdot {\LEFT[\BEGIN{ARRAY}{CCC} 2& 1 &-1 \end{array}\right]}

It is also very fascinating to note that Αtβ\alpha^t\beta is a number. And this number is not a random number. But the performance of the Matrix .
Where the performance of the matrix equals the sum of the eigenvalues.

These points of knowledge will be a very interesting process of thinking through vectors.

To give an example:

Set A=e+αβt A = E + \alpha\beta^t, where Α,β\alpha,\beta are n willi vectors, αtβ=3 \alpha^t\beta = 3, then | a+2e| =

Analysis: This investigation is very interesting, need to firmly grasp the above knowledge points to answer. Αβt \alpha\beta^t is a matrix of rank 1, and 0 is its n-1 -valued eigenvalue. by αtβ=3