Section 5 5 Row Space Column Space and
- Slides: 18
Section 5. 5 Row Space, Column Space, and Nullsapce Definition. For an mxn matrix The vectors in Rn formed from the rows of A are called the row vectors of A,
Row Vectors and Column Vectors And the vectors In Rn formed from the columns of A are called the column vectors of A. If A is an mxn matrix, then the subspace of Rn spanned by the row vectors of A is called the row space of A, and the subspace of Rm spanned by the column vectors of A is called the column space of A. The solution space of the homogeneous system of equations Ax=0, which is a subspace of Rn, is called the nullsapce of A.
Nullspace Theorem Elementary row operations do not change the nullspace of a matrix. Example Find a basis for the nullspace of The nullspace of A is the solution space of the homogeneous system Ax=0. 2 x 1 + 2 x 2 – x 3 + x 5 = 0 -x 1 + x 2 + 2 x 3 – 3 x 4 + x 5 = 0 x 1 + x 2 – 2 x 3 – x 5 = 0 x 3+ x 4 + x 5 = 0.
Nullspace Cont. Then by the previous example, we know Form a basis for this space.
Theorems Theorem Elementary row operations do not change the row space of a matrix. Note: Elementary row operations DO change the column space of a matrix. However, we have the following theorem Theorem If A and B are row equivalent matrices, then (a) A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent. (b) A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B.
Theorems Cont. Theorem If a matrix R is in row-echelon form, then the row vectors with the leading 1’s (the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1’s of the row vectors form a basis for the column space of R.
Example Find bases for the row and column spaces of Solution. Since elementary row operations do not change the row space of a matrix, we can find a basis for the row space of A by finding a basis for the row space of any row-echelon form of A.
Example By Theorem, the nonzero row vectors of R form a basis for the row space of R and hence form a basis for the row space of A. These basis vectors are Note that A ad R may have different column spaces, but from Theorem that if we can find a set of column vectors of R that forms a basisi for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A.
Example Note Form a basis for the column space of R; thus the corresponding column vectors of A, Form a basis for the column space of A.
Section 5. 6 Rank and Nullity The four fundamental matrix spaces associated with A are: Row space of A column space of A Nullspace of A nullspace of AT Theorem If A is any matrix, then the row space and column space of A have the same dimension. Definition The common dimension of the row space and column space of a matrix A is called the rank of A and is denoted by rank(A); the dimension of the nullspace of A is called the nullity of A and is denoted by nullity(A).
Example Find the rank and nullity of the matrix Solution. The reduced row-echelon form of A is
Example Since there are two nonzero rows (or, equivalently, two leading 1’s), the row space and column space are both two-dimensional, so rank(A)=2. To find the nullity of A, we must find the dimension of the solution space of the linear system Ax=0. This system can be solved by reducing the augmented matrix to reduced row-echelon form. The corresponding system of equations will be X 1 -4 x 3 -28 x 4 -37 x 5+13 x 6=0 X 2 -2 x 3 -12 x 4 -16 x 5+5 x 6=0 Solving for the leading variables, we have x 1=4 x 3+28 x 4+37 x 5 -13 x 6 X 2=2 x 3+12 x 4 +16 x 5 -5 x 6 It follows that the general solution of the system is
Example Cont. x 1=4 r+28 s+37 t-13 u X 2=2 r+12 s +16 t-5 u X 3=r X 4=s X 5=t X 6=u Equivalently, Because the four vectors on the right side of the equation form a basis for the solution space, nullity(A)=4.
Theorems Theorem If A is any matrix, then rank(A)=rank(AT). Theorem (Dimension Theorem for Matrices) If A is a matrix with n columns, then Rank(A)+nullity(A=n Theorem If A is an mxn matrix, then (a) rank(A)= the number of leading variables in the solution of Ax=0. (b) Nullity(A)= the number of parameters in the general solution of Ax=0.
Theorems Theorem (The Consistency Theorem) If Ax=b is a linear system of m equations in n unknowns, then the following are equivalent. (a) Ax=b is consistent. (b) b is in the column space of A. (c) The coefficient matrix A and the augmented matrix [A | b] have the same rank. Theorem If Ax=b is a linear system of m equations in n unknowns, then the following are equivalent. (a) Ax=b is consistent for every mx 1 matrix b. (b) The column vectors of A span Rm. (c) rank(A)=m. .
Theorems Theorem If Ax=b is a consistent linear system of m equations in n unknowns, and if A has rank r, then the general solution of the system contains n-r parameters. Theorem If A is an mxn matrix, then the following are equivalent. (a) Ax=0 has only the trivial solution. (b) The column vectors of A are linear independent. (c) Ax=b has at most one solution (none or one) for every mx 1 matrix b.
Theorems Theorem (Equivalent Statements) I A is an nxn matrix, and if TA: Rn is multiplication by A, then the following are equivalent. (a) A is invertible. (b) Ax=0 has only the trivial solution. (c) The reduced row-echelon form of A is In. (d) A is expressed as a product of elementary matrices. (e) Ax=b is consistent for every nx 1 matrix b. (f) Ax=b has exactly one solution for every nx 1 matrix b. (g) Det(A) 0. (h) The range of TA is Rn. (i) TA is one-to-one. (j) The column vectors of A are linearly independent.
Theorem Cont. (k) (l) (m) (n) (o) (p) (q) The row vectors of A are linearly independent. The column vectors of A span Rn. The row vectors of A span Rn. The column vectors of A form a basis for Rn. The row vectors of A form a basis for Rn. A has rank n. A has nullity 0.
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