Is it possible to find matrix A satisfying certain conditions?

In summary, the conversation discusses the possibility of finding a matrix A, and two vectors b and c, such that Ax = b has no solution and ##A^T## y = c has exactly one solution. It is determined that this is not possible because rank (A) < m and rank (##A^T##) = m, and since rank (A) ##\neq## rank (##A^T##), matrix A cannot exist. This reasoning is considered valid.
  • #1
songoku
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Homework Statement
Is it possible to find A being a m by n matrix, and two vectors b and c, such that Ax = b has no solution and ##A^T## y = c has exactly one solution? Explain why.
Relevant Equations
Maybe Rank
Since Ax = b has no solution, this means rank (A) < m.

Since ##A^T y=c## has exactly one solution, this means rank (##A^T##) = m

Since rank (A) ##\neq## rank (##A^T##) so matrix A can not exist. Is this valid reasoning?

Thanks
 
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  • #2
songoku said:
Homework Statement: Is it possible to find A being a m by n matrix, and two vectors b and c, such that Ax = b has no solution and ##A^T## y = c has exactly one solution? Explain why.
Relevant Equations: Maybe Rank

Since Ax = b has no solution, this means rank (A) < m.

Since ##A^T y=c## has exactly one solution, this means rank (##A^T##) = m

Since rank (A) ##\neq## rank (##A^T##) so matrix A can not exist. Is this valid reasoning?

Thanks
Looks ok to me.
 
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Likes songoku
  • #3
Thank you very much fresh_42
 

FAQ: Is it possible to find matrix A satisfying certain conditions?

Is it possible to find a matrix A such that A^2 = I?

Yes, it is possible to find a matrix A such that A squared equals the identity matrix I. Such a matrix is called an involutory matrix. An example is the identity matrix itself or a diagonal matrix with entries of 1 and -1.

Is it possible to find a matrix A such that A is invertible and A^T = A?

Yes, it is possible to find a matrix A that is both invertible and equal to its transpose. Such matrices are called orthogonal matrices. An example is the identity matrix or any rotation matrix.

Is it possible to find a matrix A such that A is not diagonalizable?

Yes, it is possible to find a matrix A that is not diagonalizable. A common example is a defective matrix, such as a 2x2 matrix with a single eigenvalue and only one linearly independent eigenvector.

Is it possible to find a matrix A such that A is both symmetric and skew-symmetric?

No, it is not possible for a non-zero matrix to be both symmetric and skew-symmetric. A symmetric matrix satisfies A = A^T, while a skew-symmetric matrix satisfies A = -A^T. The only matrix that satisfies both conditions is the zero matrix.

Is it possible to find a matrix A such that det(A) = 0 but A is not the zero matrix?

Yes, it is possible to find a matrix A with a determinant of zero that is not the zero matrix. Such matrices are called singular or degenerate matrices. An example is any matrix with linearly dependent rows or columns.

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