Let a be mxn real matrix and ATA be positive definite matrix. It is proved that the linear equation AX = 0 has only zero solution No one can do it

Let a be mxn real matrix and ATA be positive definite matrix. It is proved that the linear equation AX = 0 has only zero solution No one can do it

Let a be mxn real matrix and a ^ TA be positive definite matrix,
So | a ^ TA | > 0, so the rank of (a ^ TA) is n
So the equation (a ^ TA) x = 0 has only zero solution
Now we just need to prove that equation (a ^ TA) x = 0 and equation AX = 0 have the same solution
1) Let α be the solution of the equation AX = 0, then a α = 0
So (a ^ TA) α = a ^ t (a α) = a ^ t * 0 = 0, that is, α is the solution of the equation (a ^ TA) x = 0
2) Let α be the solution of the equation (a ^ TA) x = 0, then (a ^ TA) α = 0
So α ^ t (a ^ TA) α = (a α) ^ t (a α) = 0
Let a α = (x1, X2,..., XM) ^ t
So α ^ t (a ^ TA) α = (a α) ^ t (a α) = X1 ^ 2 + x2 ^ 2 +.. + XM ^ 2 = 0
Since x1, X2,..., XM are real numbers, X1 = x2 =... = XM = 0
So a α = 0
So α is the solution of the equation AX = 0,
Therefore, the equation (a ^ TA) x = 0 has the same solution as the equation AX = 0, so AX = 0 has only zero solution

A. It is proved that when the real number T is sufficiently large, Ta + B is also positive definite

As long as it is proved that every principal subformula of TA + B is > 0 after t is sufficiently large, every principal subformula of TA + B can be regarded as a polynomial about t, and the coefficient of the highest degree term is the corresponding principal subformula of A. A is positive definite, so every principal subformula of a is > 0, so the coefficient of the highest term of the polynomial is positive, and t is sufficiently large and constant > 0

Let a be a real symmetric matrix and t be a real number. It is proved that when t is sufficiently large, the matrix te + A is a positive definite matrix

Let A1,..., an be the eigenvalue of A
Then t + A1,..., t + an is the eigenvalue of te + a
And a is a real symmetric matrix
So when t + A1,..., t + an are all greater than 0, te + A is a positive definite matrix
So when t is sufficiently large, the matrix te + A is positive definite

Let a be a real symmetric matrix. It is proved that te + a must be a positive definite matrix as long as the real number T is large enough

Because a is real symmetric, there is an orthogonal matrix P such that p'ap is a diagonal matrix, denoted as C, where p'p = E. so p '(TE + a) P = te + C. note that te + C is a diagonal matrix, so long as t is large enough, all elements on the diagonal can be positive

Let the rank r (a) = 3, η 1, η 2, η 3 of the coefficient matrix A of the system AX = b be the solution of the system, and η 1 + η 2 = (2,0,4,6) t
And η 1 + η 2 = (2,0,4,6) t, η 2 + η 3 = (1, - 2,1,2) t, then the general solution of the equation AX = B is?

Ax = B is a system of linear equations with four variables, and the rank of coefficient matrix A is R (a) = 3
So the number of vectors in the solution is 4-3 = 1,
η1+η2=(2,0,4,6)T,η2+η3=(1,-2,1,2)T
So η 1 - η 3 = η 1 + η 2 - (η 2 + η 3) = (1,2,3,4) t
And a (η 1 - η 3) = B-B = 0,
So η 1 - η 3 = (1,2,3,4) t is the solution vector of the homogeneous equation AX = 0,
And a (η 1 + η 2) = a η 1 + a η 2 = 2B,
So (η 1 + η 2) / 2 is a special solution of AX = B,
That is, (η 1 + η 2) / 2 = (1,0,2,3) t is a special solution of AX = B,
So the general solution of the equation AX = B is as follows:
X = (1,0,2,3) t + K (1,2,3,4) t, where k is a constant

The rank of coefficient matrix of two equations with the same solution is the same
On the other hand, if two matrices with the same number of columns have the same rank, do the equations composed of these two matrices have the same solution
I think that the same rank is a necessary and insufficient condition for the same solution of the equations

That's right
The same solution of two equations
If and only if the row vectors of their augmented matrices are equivalent,
The same rank does not mean that two vector groups are equivalent

η 1 and η 2 are the solutions of AX = B and Ax = 0

∵ η 1, η 2 are solutions of non-homogeneous linear equations AX = B
∴Aη1=b Aη2=b
∴Aη1-Aη2=b-b=0
A(η1-η2)=0
∴X=η1-η2

A is a square matrix of order 6. A * is the adjoint matrix of A. if R (a) = 3, then the basic solution system of homogeneous linear equations a * x = 0 contains the number of solution vectors

According to the theorem, the number of solution vectors in the basic solution system of homogeneous linear equations a * x = 0 is N-R (a *), because R (a) = 3, that is, the highest order of the non-zero subformula in a is 3. And a * is AIJ = - mij, which is composed of the algebraic cofactors of matrix A. mij is the 5th order subformula of a, because the highest order of the non-zero subformula is 3

Let a be a square matrix of order n, | a | = 0, and the algebraic cofactor of an element in a is not zero, then the number of vectors contained in the basic solution system of the solution of the linear equation AX = 0 is the second
Let a be a square matrix of order n, | a | = 0, and the algebraic cofactor of one element in a is not zero, then the number of vectors contained in the basic solution system of the next linear equation system AX = 0 is ()
A 1 B n C n-1 D 2

A
Because | a | = 0, and the algebraic covalent of an element in a is not zero, then the rank of a is n-1, then the solution space of AX = 0 is 1-dimensional

5. Let a be a square matrix of order 5, if rank (a) = 3, then the number of solution vectors contained in the basic solution system of the homogeneous linear equations AX = 0 is__________

The number is 5-3 = 2