To prove the existence of a vector for a matrix with linearly independent rows, you can use the Rank-Nullity Theorem. This theorem states that the rank of a matrix is equal to the number of linearly independent rows, and the nullity is equal to the number of linearly dependent rows. Therefore, if the rank of the matrix is equal to the number of rows, there exists a vector that satisfies the system of linear equations formed by the rows of the matrix.
Yes, consider the matrix A = [1 0 2; 0 1 3; 2 3 4]. The rows of this matrix are linearly independent, as can be seen by solving the system of linear equations formed by the rows (x + 2z = 0, y + 3z = 0, and 2x + 3y + 4z = 0). The only solution is x = y = z = 0, which proves that the rows are linearly independent.
No, if a matrix has linearly independent rows, there must exist a vector that satisfies the system of equations formed by those rows. This is because the Rank-Nullity Theorem guarantees that the rank of the matrix is equal to the number of linearly independent rows, and therefore, there must exist a solution to the system of equations.
A matrix with linearly independent rows is always invertible. This is because having linearly independent rows ensures that the matrix has full rank, which means that it is non-singular and has an inverse. The vector that satisfies the system of equations formed by the rows of the matrix can be used to construct the inverse of the matrix.
Yes, another method is to use the concept of linear independence and span. If the rows of a matrix are linearly independent, it means that they form a basis for the column space of the matrix. Therefore, any vector in the column space can be written as a linear combination of the rows. This shows that there exists a vector that satisfies the system of equations formed by the rows of the matrix.