Let the sum of the elements of any row of matrix A of order n be a and prove that a is a eigenvalue of matrix A to find the eigenvector corresponding to a Let a be the sum of the elements of any matrix A of order n be a and prove that a is a eigenvalue of matrix A to find the eigenvector corresponding to a

Let the sum of the elements of any row of matrix A of order n be a and prove that a is a eigenvalue of matrix A to find the eigenvector corresponding to a Let a be the sum of the elements of any matrix A of order n be a and prove that a is a eigenvalue of matrix A to find the eigenvector corresponding to a

Consider column vector x =(1,1,...,1)
The product of it and the matrix is (a, a,..., a)
It satisfies Ax = ax, so a is the eigenvalue and x is the eigenvector

Any non-zero n-dimensional vector is the eigenvector of n--order quantity matrix A

The quantity matrix A is a square matrix in which the elements on the main diagonal are the same and the other elements are 0
I.e. kE.
For any non-zero n-dimensional vector x, Ax = kEx = kx
So x is the eigenvector of A which belongs to the eigenvalue k.

The problem of linear algebra: If any n-dimensional non-zero vector is the eigenvector of matrix A of order n, why does A have n linearly independent eigenvectors? Ask the relatives for an explanation. On the Problem of Linear Algebra: If any n-dimensional nonzero vector is the eigenvector of n-order matrix A, why does A have n linearly independent eigenvectors? Ask the relatives for an explanation.

Since any n-dimensional nonzero vector is the eigenvector of A,
Then take out each column of the n-order unit matrix, and these n vectors are linearly independent and are all eigenvectors of A

When m > n, the vector group of m n dimensions must be linearly related or this corollary There's a theorem here: r sets of n-dimensional row vectors, when r

You're confusing the set of row and column vectors.
In the theorem, A row full rank, A row vector group linearly independent
But its set of column vectors is not necessarily
If r

You're confusing the set of row and column vectors.
In the theorem, A row full rank, A row vector group linearly independent
But its column vector set is not necessarily
If r

Any N+1 vector in n-dimensional vector space must be linearly related. I don't understand this concept. Can anyone explain it to me? I'd better have a specific example to show that I' m confused by it, Any N+1 vector in n-dimensional vector space must be linearly related. I don't understand this concept. Who can explain it to me? I'd better have a specific example to show that I' m confused by it,

To take the simplest example: x1+x2+x3+x4=02*x1+3x2=0 How many solutions do you have for this system of equations? The answer is that any N+1 vector in countless n-dimensional vector spaces must be linearly related, that is to say, in this n+1 n-dimensional vector, we must find a vector that can be linearly represented by the remaining vectors, such as two-dimensional vector [1,...

How many solutions do you: x1+x2+x3+x4=02*x1+3x2=0 How many solutions do you have for this system of equations? The answer is that any N+1 vector in countless n-dimensional vector spaces must be linearly related, that is to say, in this n+1 n-dimensional vector, we must find a vector that can be linearly represented by the remaining vectors, such as two-dimensional vector [1,...

What is the square of the module of the n-dimensional column vector β

Real vector
Let β=(β1,β2,β3...βn)'
||β||^2=β'β=β1^2+β2^2+..+βN^2
The literal representation is defined as the sum of the squares of each term.
Complex vector
The square of each term becomes the product of its conjugate complex number or the square of its module.