 # 5.2 Rank of a Matrix. Set-up Recall block multiplication:

## Presentation on theme: "5.2 Rank of a Matrix. Set-up Recall block multiplication:"— Presentation transcript:

5.2 Rank of a Matrix

Set-up Recall block multiplication:

Theorem 1 The following conditions are equivalent for an n x n matrix A: 1. The rows of A are linearly independent in  n. 2. The rows of A span  n. 3. The columns of A are linearly independent in  n. 4. The columns of A span  n. 5. A is invertible.

Proof of Theorem 1 To prove that 5 statements are equivalent, we can prove that (1)  (2)  (3)  (4)  (5)  (1) or could do: (1)  (2)  (5)  (3)  (4)  (1) which we will do. Proof: (1)  (2) given the rows are linearly independent in  n and we know dim  n = n, we know then that the rows must form a basis, and thus span  n.

Proof of Theorem 1 (cont) (2)  (5) If we can find a matrix B such that BA = I, then we know A is invertible. Let row i of B be K = [k 1 k 2 … k n ] Then I = BA results in: Row i of I = row i of BA = KA = Since the rows of A span  n, we know that such k i exist (we know we can create any vector in  n using linear combinations of the rows in A). Therefore, each row of B can be found making A invertible.

Proof of Theorem 1 (cont) We will now show (5)  (1) and leave (3) and (4) for hw. (5)  (1): Let k 1 R 1 +…+k n R n = 0 where R i is the i th row of A. Let K = [k 1 k 2 … k n ] so KA = k 1 R 1 +…+k n R n = 0 Right multiply by A -1 to get K=0, which means k 1 =…=k n = 0 which means that the rows of A are linearly independent.

A use of Theorem 1 So now to show that an (n x n) matrix is invertible, we just need to show that the rows are linearly independent or that they span  n. Or to show that the rows are linearly independent or that the rows span  n, we could show that the matrix is invertible.

Example 1 Show that {(1,2,-1), (3,1,-4), (1,1,7)} is a basis for  3. We just need to show that the matrix with these vectors as rows is invertible which can be done by showing that it has rank 3, or by showing that the determinant is not zero, etc.

Definition If A is an m x n matrix, the rows of A are vectors in  n, and the subspace of  n spanned by these rows is called the row space of A (row A). The space spanned by the columns of A is called the column space of A (col A).

Theorem 2 A (m x n), U (p x m),V (n x q) 1. row(UA)  row A with equality if U is invertible (& square) 2. col(AV)  col A with equality if V is invertible (& square)

Proof of Theorem 2 Proof: R 1,R 2,…,R m are rows of A row i of mtx U is: row i of UA = U i A= So the rows of UA are in the row space of A, which means that row (UA)  row A. If U is invertible: row A = row[U -1 (UA)]  row UA (same logic as above) so row A = row (UA). (same for part 2)

Theorem 3-Rank Theorem A (m x n) Then, dim(row A) = dim(col A) Also, A can be carried to row echelon form matrix R, and if r is the number of nonzero rows in R, 1. The r nonzero rows are a basis of (row A) 2. If the leading 1’s are in columns j 1,j 2,…,j r, then {j 1,j 2,…,j r } is a basis of (col A)

Theorem 3-Proof R = UA (U is the product of elementary matrices and is invertible). row A = row R (by theorem 2) since U is invertible For a matrix in row echelon form, the rows form a basis for the rows since they span the rows and are linearly independent. Since row A = row R, the nonzero rows also form a basis for A.

Theorem 3-Proof (cont) (2) B={UC j1, UC j2,…,UC jr } is the set of columns of R w/ a leading 1. In homework, you will prove that for a matrix is ref, the columns with leading 1’s form a basis of col R (since they’re li). So B is a basis for R.

Theorem 3-Proof (cont) Since {UC j1, UC j2,…,UC jr } is LI, a So {C j1, C j2,…,C jr } is LI and it spans A so it is a basis of col A. dim(rowA)=r=dim(colA) and is called the rank

Corollary If A can be carried to R in ref by elementary row ops, then the rank of A is equal to the number of nonzero rows of R.

Example Find basis for row and col space of A and find rank Note that the basis of row space comes from ref, and the basis of col space comes from initial state.

Corollaries Cor 2: If A is mxn, then rank A ≤m and rank A≤ n Cor 3: If A is any matrix, rank A = rank A T Cor 4: A (m x n), U (m x m) and V (n x n) invertible, then rank A = rank (UA) = rank (AV) ( by theorem 2) Cor 5: A (n x n) is invertible iff rank A = n.

Example Find basis of U=span {(1,-1,0,3),(2,1,5,1),(4,-2,5,7)} Can just write in matrix form and row reduce.

Theorem 4 A (m x n) has rank r iff there exist invertible matrices U (mxm) and V(nxn) such that Where I r is the r x r identity matrix Proof the row operations that take A to R (rref) also carry I m to Invertible U such that UA = R so R has r nonzero rows, which contain columns of I r.

Theorem 4 Then take RT and put it in rref using U1 so then Let V=U 1 T and we have: And also rank A = r if these U and V exist (by Cor 4)

Use of this Now we can find these U and V which will reduce A to this form. 1) Find U from 2) Find V from

Example Find invertible matrices U and V such that

Definition A (mxn). The set of solutions to AX=0 is a subspace of  n called the null space of A (null A)

Theorem 5 A (m x n) and r = rank A. Then dim (null A) = n-r Proof: Recall that when you solve a homogeneous system, you either get a unique solution if the A is invertible (in which case the set of solutions contains no parameters and thus dim (nullA) = 0. Or, A could reduce to have fewer equations than variables in which case the solution will be of a form like: Which has dim = 2 = # parameters So we just need to find the number of parameters.

Theorem 5-Proof-formal A (m x n) and r = rank A. Then dim (null A) = n-r Proof: null (A) = {X | AX = 0}. Let null(A) = null(UA) since U is invertible dim(null(UA)) = dim(null(UAV)) since V invertible So dim (null(A))=dim(null(UAV)) UAV is in rref so dim (null(UAV)) = n-r (# var-rank)=# parameters.

Example Find the basis of the nullspace of Just find the sol’n to the homogeneous system and then the dimension is the number of parameters, and the vectors will form the basis.`

Theorem 6 The following conditions are equivalent for an (m x n) matrix. 1. Rank A = m 2. The rows of A are LI 3. If YA = 0 with Y in  m, then Y=0 4. AA T is invertible Proof: (1)  (2) rank A = m implies the m rows form a basis for row A which implies that the rows are LI (2)  (3) YA = y 1 R 1 +y 2 R 2 +…+y m R m = 0 implies y 1 =y 2 =..=y m = 0 Since the rows of A are LI, so Y = 0.

Theorem 6 (proof cont) (3)  (4) We will show that Y(AA T ) implies Y=0 (YA)(YA) T = YAA T Y T =0Y T =0 So YA = 0 (since (YA)(YA) T =0 and a AA T =0 implies A=0) so Y = 0 (given in 3) (4)  (1) Just need to show that {R 1,R 2,…,R m } is LI Let x 1 R 1 +…+x m R m = 0. So XA = 0. So XAA T = 0 So X=0 since AA T is invertible So all x’s are 0 and rows are li