Given that the column vectors of 4 × 3 matrix A are linearly independent, what is the rank of transpose matrix A

A transpose matrix rank = number of columns = 3

Why is the rank of a less than or equal to the rank of B when the group of column vectors of a matrix can be represented by the group of column vectors of B?

When the column vector group of matrix A can be represented by the column vector group of matrix B
There must be C with a = BC
R（A)=R(AB)

Why can we write the matrix rank of row vector multiplied by column vector less than or equal to 1?

According to the property of rank, R (AB)

A. B and C are MXM, NxN and mxn matrices respectively (M > n), and AC = CB. The rank of C is R. it is proved that a and B have at least r identical eigenvalues

There's a proof with block matrix, I made a picture version
In fact, the language of linear transformation, invariant subspace and quotient space can give a more graceful proof, which is only relatively abstract
We use the following lemma:
Let a be a linear transformation on V and W be an invariant subspace of a, then the characteristic polynomial of a is equal to the characteristic polynomial restricted by a on W multiplied by the reduced characteristic polynomial of a on the quotient space V / W
In essence, it is the same as the | λ E & # 8722; a | & nbsp; = & nbsp; | λ E & # 8722; R | ·| λ E & # 8722; u |, which is proved above
It is proved that: let the dimensions of linear space m and n be m, N and m respectively
The linear transformation with matrix A on M is still denoted as a, and the linear transformation with matrix B on N is still denoted as B
The linear mapping from n to m with matrix C is still denoted as C, then the equation AC = CB of linear mapping is established by the condition
The kernel Ker (c) & nbsp; (all x satisfying CX = 0) of C is an N-R dimensional subspace of N and an invariant subspace of B
B-invariant is because if x ∈ Ker (c), & nbsp; that is CX = 0, & nbsp; then C (BX) = ACX = 0, & nbsp; BX ∈ Ker (c) can be obtained
On the other hand, the image set im (c) & nbsp; (i.e. C (n)) of C is an r-dimensional subspace of M and an invariant subspace of A
If y ∈ im (c), & nbsp; there is x ∈ n such that CX = y, & nbsp; then ay = ACX = C (BX), & nbsp; there is ay ∈ im (c)
From the fundamental theorem of homomorphism, & nbsp; C gives an isomorphism of quotient space n / Ker (c) to im (c)
The condition AC = CB shows that under the isomorphism given by C, the reduction of B in N / Ker (c) exactly corresponds to the restriction of a in im (c)
If we take the corresponding bases on N / Ker (c) and im (c) (Z before and CZ after), & nbsp; then they have the same matrix
Then there are the same characteristic polynomials with degree R (i.e. the dimensions of N / Ker (c) and im (c))
But B is divisible by the characteristic polynomial reduced by N / Ker (c) and a is divisible by the characteristic polynomial restricted by im (c)
Then the characteristic polynomials of a and B have a common factor of degree R and at least r common eigenvalues

It is known that a is an mxn matrix whose m row vectors are the fundamental solution system of the homogeneous linear equations CX = 0, and B is an invertible matrix of order m. It is proved that the row vectors of Ba are the fundamental solution system of Cx = 0

Knowledge point: the vector system which is equivalent to the basic solution system of homogeneous linear equations and contains the same number of vectors is still the basic solution system of equations
It is proved that because B is invertible, the row vector group of Ba is equivalent to that of A
And the number of rows of Ba and a is m
So the row vector of Ba is also the basic solution system of Cx = 0

How to judge whether this is distributive lattice or modular lattice by hastur
It is known that this is a lattice. How can we quickly determine whether this is a distributive lattice or a modular lattice through hastur

Look at the shelf. If there is a pentagonal lattice, it is not

chart
For graph G =, where | V | = n, | e | = n + 1, it is proved that the degree of at least one node in G is ≥ 3

It is proved that if the degree of all nodes in G is less than 3, or no more than 2, then the sum of the degree of N nodes is no more than 2n, and the sum of the degree of nodes is equal to 2 times of the number of edges, that is, the sum of the degree of nodes = 2 | e | = 2n + 2, so there is a contradiction of 2n ≥ 2n + 2, n ≥ n + 1

G is an undirected connected graph with n nodes. It is proved that G has at least n-1 edges, and the undirected connected graph with n-1 edges is a tree

By using the extended path method, a point is randomly selected. Every time it needs to be connected with another point, it needs at least one edge. Because it is a connected graph, there are at least n-1 edges. When there are only n-1 edges, each edge is a bridge, so it can be seen that it is a tree

Let a = {1,2,3,4,5,6,7,8,10,12,24} and R be the integer division relation on A. please draw the Haas diagram of poset,
The maximum element, minimum element, maximum element, minimum element, upper bound, lower bound, supremum and infimum of B = {2,4,6} are obtained

The maximum element 4 and 6, the minimum element 2, the minimum element 2 without the maximum element, the upper bound 4, 6, 8, 12, 24, the lower bound 2, the infimum 2 without the infimum;

Draw the Haas diagram of set a = {1,2,3,4,5,6} under the partial order relation of "integral division", and find out: (1) the largest element, the smallest element and the largest element of set a respectively

the bottom layer is 1, the middle layer is 2,3,5, and the upper layer is 4,6. Then 1 connected 2,1 connected 3,1 connected 5,2 connected 4,2 connected 6,3 connected 6 draw lines. The maximum element is 4,6, the minimum element is 1, and the maximum element is 4,6

Given that the column vectors of 4 × 3 matrix A are linearly independent, what is the rank of transpose matrix A